Transformation of logarithmic expressions Examples with the solution. Transformation of logarithmic expressions

basic properties.

logax + Logay \u003d Loga (x · y);
logAX - Logay \u003d Loga (X: Y).

same grounds

Log6 4 + Log6 9.

Now a little complicate the task.

Examples of logarithm solutions

What if at the base or argument of logarithm costs a degree? Then the indicator of this extent can be taken out of the logarithm sign according to the following rules:

Of course, all these rules make sense when compliance with the OTZ Logarithm: A\u003e 0, A ≠ 1, X\u003e

A task. Find the value of the expression:

Transition to a new base

Let Logax LogAx. Then for any number C such that C\u003e 0 and C ≠ 1, the equality is true:

A task. Find the value of the expression:

The main properties of logarithm

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

The exhibitor is 2,718281828 .... To remember the exhibitor, you can explore the rule: the exhibitor is 2.7 and twice the year of birth of Leo Nikolayevich Tolstoy.

The main properties of logarithm

Knowing this rule will know the exact value of the exhibitory, and the date of birth of Lion Tolstoy.

Examples on logarithmia

Prologate expressions

Example 1.
but). x \u003d 10As ^ 2 (A\u003e 0, C\u003e 0).

By properties 3.5 calculate

4. Where .

Example 2. Find X if

Example 3. Let the value of logarithms are set

Calculate log (x) if

The main properties of logarithm

Logarithms, like any numbers, can be folded, deduct and convert. But since logarithms are not quite ordinary numbers, there are its own rules that are called basic properties.

These rules must necessarily know - no serious logarithmic task is solved without them. In addition, they are quite a bit - everything can be learned in one day. So, proceed.

Addition and subtraction of logarithms

Consider two logarithm with the same bases: Logax and Logay. Then they can be folded and deducted, and:

logax + Logay \u003d Loga (x · y);
logAX - Logay \u003d Loga (X: Y).

So, the amount of logarithms is equal to the logarithm of the work, and the difference is the logarithm of private. Please note: the key point here is same grounds. If the foundations are different, these rules do not work!

These formulas will help calculate the logarithmic expression even when individual parts are not considered (see the lesson "What is logarithm"). Take a look at the examples - and make sure:

Since the bases in logarithms are the same, we use the sum of the sum:
log6 4 + Log6 9 \u003d log6 (4 · 9) \u003d log6 36 \u003d 2.

A task. Find the value of the expression: Log2 48 - Log2 3.

The foundations are the same, using the difference formula:
log2 48 - Log2 3 \u003d Log2 (48: 3) \u003d log2 16 \u003d 4.

A task. Find the value of the expression: Log3 135 - Log3 5.

Again the foundations are the same, so we have:
log3 135 - Log3 5 \u003d log3 (135: 5) \u003d log3 27 \u003d 3.

As you can see, the initial expressions are made up of "bad" logarithms, which are not separately considered separately. But after transformation, quite normal numbers are obtained. In this fact, many test work are built. But what is the control - such expressions are in full (sometimes - almost unchanged) are offered on the exam.

Executive degree from logarithm

It is easy to see that the last rule follows their first two. But it is better to remember it, in some cases it will significantly reduce the amount of calculations.

Of course, all these rules make sense when compliance with the OTZ Logarithm: A\u003e 0, a ≠ 1, x\u003e 0. And more: learn to apply all formulas not only from left to right, but on the contrary, i.e. You can make numbers facing the logarithm, to the logarithm itself. That is most often required.

A task. Find the value of the expression: log7 496.

Get rid of the extent in the argument on the first formula:
log7 496 \u003d 6 · Log7 49 \u003d 6 · 2 \u003d 12

A task. Find the value of the expression:

Note that in the denominator there is a logarithm, the base and the argument of which are accurate degrees: 16 \u003d 24; 49 \u003d 72. We have:

I think the latest example requires explanation. Where did the logarithms disappeared? Until the last moment, we only work with the denominator.

Formulas logarithms. Logarithms Examples of solutions.

They presented the basis and argument of a logarithm there in the form of degrees and carried out indicators - received a "three-story" fraction.

Now let's look at the basic fraction. In a numerator and denominator, the same number is: log2 7. Since log2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which was done. The result was the answer: 2.

Transition to a new base

Speaking about the rules for the addition and subtraction of logarithms, I specifically emphasized that they work only with the same bases. And what if the foundations are different? What if they are not accurate degrees of the same number?

Formulas for the transition to a new base come to the rescue. We formulate them in the form of theorem:

Let Logax LogAx. Then for any number C such that C\u003e 0 and C ≠ 1, the equality is true:

In particular, if you put C \u003d x, we get:

From the second formula it follows that the base and argument of the logarithm can be changed in places, but at the same time the expression "turns over", i.e. Logarithm turns out to be in the denominator.

These formulas are rare in conventional numerical expressions. Assessing how convenient they are, it is possible only when solving logarithmic equations and inequalities.

However, there are tasks that are generally not solved anywhere as a transition to a new base. Consider a couple of such:

A task. Find the value of the expression: Log5 16 · Log2 25.

Note that the arguments of both logarithms are accurate degrees. Let's take out the indicators: log5 16 \u003d log5 24 \u003d 4Log5 2; Log2 25 \u003d log2 52 \u003d 2log2 5;

And now "invert" the second logarithm:

Since the work does not change from the rearrangement of multipliers, we calmly changed the four and a two, and then sorted out with logarithms.

A task. Find the value of the expression: Log9 100 · LG 3.

The basis and argument of the first logarithm - accurate degrees. We write it and get rid of the indicators:

Now get rid of the decimal logarithm, by turning to the new base:

Basic logarithmic identity

Often, the solution is required to submit a number as a logarithm for a specified base. In this case, formulas will help us:

In the first case, the number N becomes an indicator of the extent in the argument. The number n can be absolutely any, because it is just a logarithm value.

The second formula is actually a paraphrassed definition. It is called :.

In fact, what will happen if the number B is in such a degree that the number B to this extent gives the number a? Right: It turns out this the same number a. Carefully read this paragraph again - many "hang" on it.

Like the transition formulas to a new base, the main logarithmic identity is sometimes the only possible solution.

A task. Find the value of the expression:

Note that log25 64 \u003d log5 8 - just made a square from the base and the logarithm argument. Given the rules for multiplication of degrees with the same base, we get:

If someone is not aware, it was a real task of ege 🙂

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that it is difficult to name the properties - rather, this is the consequence of the definition of logarithm. They are constantly found in tasks and, which is surprising, create problems even for "advanced" students.

logaa \u003d 1 is. Remember times and forever: the logarithm on any base A from the very base is equal to one.
loga 1 \u003d 0 is. The base A may be any sense, but if the argument is a unit - logarithm is zero! Because A0 \u003d 1 is a direct consequence of the definition.

That's all properties. Be sure to practice apply them in practice! Download the crib at the beginning of the lesson, print it - and solve the tasks.

The main properties of logarithm

These properties need to know because, on their basis, almost all tasks are solved and examples are associated with logarithms. The remaining exotic properties can be derived by mathematical manipulations with these formulas

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

In the calculations of the formula of the sum and the difference of logarithms (3.4) are quite common. The rest are somewhat complicated, but in a number of tasks are indispensable to simplify complex expressions and calculate their values.

There are cases of logarithm

One of the common logarithms are such in which the base is smooth ten, exponential or twice.
The logarithm on the basis of ten is customary to call the decimal logarithm and simplifying the LG (X).

From the record it is clear that the foundations in the record are not written. For example

Natural logarithm is a logarithm for which the exhibitor is based on LN (X)).

The exhibitor is 2,718281828 .... To remember the exhibitor, you can explore the rule: the exhibitor is 2.7 and twice the year of birth of Leo Nikolayevich Tolstoy. Knowing this rule will know the exact value of the exhibitory, and the date of birth of Lion Tolstoy.

And one more important logarithm on the base two designate

The derivative of the logarithm function is equal to a unit divided into a variable

Integral or primitive logarithm is determined by addiction

The above material is enough to solve a wide class of tasks associated with logarithms and logarithmation. For mastering the material, I will give only a few common examples from the school program and universities.

Examples on logarithmia

Prologate expressions

Example 1.
but). x \u003d 10As ^ 2 (A\u003e 0, C\u003e 0).

By properties 3.5 calculate

2.
By the properties of the difference logarithms have

3.
Using properties 3.5 find

4. Where .

The form of a complex expression using a number of rules is simplified to mind

Finding the values \u200b\u200bof logarithm

Example 2. Find X if

Decision. For calculation, applicable to the last term of the 3rd and 13 properties

We substitute to write and grieve

Since the grounds are equal, then equating expressions

Logarithmia. First level.

Let the value of logarithms

Calculate log (x) if

Solution: Progriform the variable to paint logarithm through the sum of the terms

On this acquaintance with logarithms and their properties just begins. Exercise in calculations, enrich practical skills - the knowledge gained will soon be needed to solve logarithmic equations. After studying the basic methods of solving such equations, we will expand your knowledge for another equally important topic - logarithmic inequalities ...

The main properties of logarithm

Logarithms, like any numbers, can be folded, deduct and convert. But since logarithms are not quite ordinary numbers, there are its own rules that are called basic properties.

These rules must necessarily know - no serious logarithmic task is solved without them. In addition, they are quite a bit - everything can be learned in one day. So, proceed.

Addition and subtraction of logarithms

Consider two logarithm with the same bases: Logax and Logay. Then they can be folded and deducted, and:

logax + Logay \u003d Loga (x · y);
logAX - Logay \u003d Loga (X: Y).

These formulas will help calculate the logarithmic expression even when individual parts are not considered (see the lesson "What is logarithm"). Take a look at the examples - and make sure:

A task. Find the value of the expression: Log6 4 + Log6 9.

Since the bases in logarithms are the same, we use the sum of the sum:
log6 4 + Log6 9 \u003d log6 (4 · 9) \u003d log6 36 \u003d 2.

A task. Find the value of the expression: Log2 48 - Log2 3.

The foundations are the same, using the difference formula:
log2 48 - Log2 3 \u003d Log2 (48: 3) \u003d log2 16 \u003d 4.

A task. Find the value of the expression: Log3 135 - Log3 5.

Again the foundations are the same, so we have:
log3 135 - Log3 5 \u003d log3 (135: 5) \u003d log3 27 \u003d 3.

Executive degree from logarithm

Now a little complicate the task. What if at the base or argument of logarithm costs a degree? Then the indicator of this extent can be taken out of the logarithm sign according to the following rules:

It is easy to see that the last rule follows their first two. But it is better to remember it, in some cases it will significantly reduce the amount of calculations.

How to solve logarithm

That is most often required.

A task. Find the value of the expression: log7 496.

Get rid of the extent in the argument on the first formula:
log7 496 \u003d 6 · Log7 49 \u003d 6 · 2 \u003d 12

A task. Find the value of the expression:

Note that in the denominator there is a logarithm, the base and the argument of which are accurate degrees: 16 \u003d 24; 49 \u003d 72. We have:

I think the latest example requires explanation. Where did the logarithms disappeared? Until the last moment, we only work with the denominator. They presented the basis and argument of a logarithm there in the form of degrees and carried out indicators - received a "three-story" fraction.

Transition to a new base

Formulas for the transition to a new base come to the rescue. We formulate them in the form of theorem:

Let Logax LogAx. Then for any number C such that C\u003e 0 and C ≠ 1, the equality is true:

In particular, if you put C \u003d x, we get:

These formulas are rare in conventional numerical expressions. Assessing how convenient they are, it is possible only when solving logarithmic equations and inequalities.

However, there are tasks that are generally not solved anywhere as a transition to a new base. Consider a couple of such:

A task. Find the value of the expression: Log5 16 · Log2 25.

Note that the arguments of both logarithms are accurate degrees. Let's take out the indicators: log5 16 \u003d log5 24 \u003d 4Log5 2; Log2 25 \u003d log2 52 \u003d 2log2 5;

And now "invert" the second logarithm:

Since the work does not change from the rearrangement of multipliers, we calmly changed the four and a two, and then sorted out with logarithms.

A task. Find the value of the expression: Log9 100 · LG 3.

The basis and argument of the first logarithm - accurate degrees. We write it and get rid of the indicators:

Now get rid of the decimal logarithm, by turning to the new base:

Basic logarithmic identity

Often, the solution is required to submit a number as a logarithm for a specified base. In this case, formulas will help us:

In the first case, the number N becomes an indicator of the extent in the argument. The number n can be absolutely any, because it is just a logarithm value.

The second formula is actually a paraphrassed definition. It is called :.

Like the transition formulas to a new base, the main logarithmic identity is sometimes the only possible solution.

A task. Find the value of the expression:

Note that log25 64 \u003d log5 8 - just made a square from the base and the logarithm argument. Given the rules for multiplication of degrees with the same base, we get:

If someone is not aware, it was a real task of ege 🙂

Logarithmic unit and logarithmic zero

logaa \u003d 1 is. Remember times and forever: the logarithm on any base A from the very base is equal to one.
loga 1 \u003d 0 is. The base A may be any sense, but if the argument is a unit - logarithm is zero! Because A0 \u003d 1 is a direct consequence of the definition.

That's all properties. Be sure to practice apply them in practice! Download the crib at the beginning of the lesson, print it - and solve the tasks.

Today we will talk about logarovmov formulas and give indicative examples of solutions.

By themselves imply the decision patterns according to the main properties of logarithms. To first apply the logarithms for solutions to remind you, first all properties:

Now on the basis of these formulas (properties), we will show examples of logarithm solutions.

Examples of logarithms based on the formulas.

Logarithm A positive number B based on A (denoted by Log A B) is an indicator of the degree in which A should be undertaken to obtain B, with the b\u003e 0, a\u003e 0, and 1.

According to the definition of Log A B \u003d x, which is equivalent to a x \u003d b, so Log A A x \u003d x.

LogarithmiaExamples:

log 2 8 \u003d 3, because 2 3 \u003d 8

log 7 49 \u003d 2, because 7 2 \u003d 49

log 5 1/5 \u003d -1, because 5 -1 \u003d 1/5

Decimal logarithm - This is an ordinary logarithm, at the base of which 10. is denoted as LG.

log 10 100 \u003d 2, because 10 2 \u003d 100

Natural logarithm - Also ordinary logarithm logarithm, but already with the basis of E (E \u003d 2,71828 ... - an irrational number). Denotes as ln.

The formulas or properties of logarithms are desirable to remember because they will need in the future when solving logarithms, logarithmic equations and inequalities. Let us work again every formula on the examples.

Basic logarithmic identity
A LOG A B \u003d B
8 2log 8 3 \u003d (8 2log 8 3) 2 \u003d 3 2 \u003d 9
Logarithm works equal to the sum of logarithms
Log A (BC) \u003d Log A B + Log A C
log 3 8.1 + log 3 10 \u003d log 3 (8.1 * 10) \u003d log 3 81 \u003d 4
Logarithm private equal to the difference of logarithms
Log A (B / C) \u003d Log A B - Log A C
9 log 5 50/9 Log 5 2 \u003d 9 log 5 50- log 5 2 \u003d 9 log 5 25 \u003d 9 2 \u003d 81
Properties of the extent of the logarithmatic number and base of logarithm
Indicator of the logarithmatic number Log A B M \u003d Mlog A B

The indicator of the foundation of the logarithm Log A N B \u003d 1 / N * Log A B

log a n b m \u003d m / n * log a b,

if m \u003d n, we get Log a n b n \u003d log a b

log 4 9 \u003d log 2 2 3 2 \u003d log 2 3
Transition to a new base
Log A B \u003d log c b / log c a,
if C \u003d B, we get Log B B \u003d 1

then log a b \u003d 1 / log b a

log 0.8 3 * log 3 1,25 \u003d log 0.8 3 * log 0.8 1,25 / log 0.8 3 \u003d log 0.8 1,25 \u003d log 4/5 5/4 \u003d -1

As you can see, the logarithms are not as complicated as it seems. Now reviewing examples of the solution of logarithms we can move to logarithmic equations. Examples of solving logarithmic equations We will consider in more detail in the article: "". Do not miss!

If you have any questions about the decision, write them in the comments to the article.

Note: We decided to get the formation of another class training abroad as an option for the development of events.

This video I start a long series of lessons about logarithmic equations. Now there are three examples in front of you, on the basis of which we will learn to solve the simplest tasks that are so called - simplest.

log 0.5 (3x - 1) \u003d -3

lG (x + 3) \u003d 3 + 2 LG 5

Let me remind you that the simplest logarithmic equation is the following:

log a f (x) \u003d b

It is important that the variable x is present only inside the argument, i.e. only in the function f (x). A and B numbers are precisely numbers, and in no case are not functions containing a variable x.

Basic methods of solution

There are many ways to solve such structures. For example, most teachers at school offer such a way: to immediately express the function f (x) by the formula f ( x) \u003d. a b. That is, when you meet the simplest design, immediately without additional actions and buildings can go to the solution.

Yes, definitely, the solution will be right. However, the problem of this formula is that most students do not understandWhere it comes from and why the letter and we are erected in the letter B.

As a result, I often observe very offensive errors when, for example, these letters are changing in places. This formula must be either understood or tooling, and the second method leads to errors in the most inappropriate and most responsible moments: on exams, control, etc.

That is why all its students, I propose to abandon the standard school formula and use to solve logarithmic equations a second approach, which, as you probably guessed from the name, is called canonical form.

The idea of \u200b\u200bcanonical form is simple. Let's look at our task again: on the left we have Log A, while under the letter A is meant exactly the number, and in no case is it not a function containing the variable x. Consequently, all the limitations are applied to this letter, which are superimposed on the basis of the logarithm. namely:

1 ≠ A\u003e 0

On the other hand, from the same equation, we see that the logarithm should be equal to the number B, and here it does not impose any restrictions on this letter, because it can take any values \u200b\u200b- both positive and negative. It all depends on what values \u200b\u200breceive the function f (x).

And here we remember our remarkable rule that any number B can be represented in the form of a logarithm on the base A of A to the degree B:

b \u003d Log A A B

How to remember this formula? Yes, very simple. Let's write the following design:

b \u003d b · 1 \u003d b · log a a

Of course, while there are all the limitations that we recorded at the beginning. And now let's use the basic property of the logarithm, and make a multiplier B as a degree a. We get:

b \u003d B · 1 \u003d B · Log A A \u003d Log A A B

As a result, the initial equation will rewrite in the following form:

log a f (x) \u003d log a a b → f (x) \u003d a b

That's all. The new function no longer contains logarithm and is solved by standard algebraic techniques.

Of course, someone will now objected: why was it at all invented some canonical formula, why do two additional unnecessary steps, if you could immediately move from the original design to the final formula? Yes, at least then, then the majority of students do not understand where this formula comes from and, as a result, regularly allow errors when applied.

But this sequence of actions consisting of three steps allows you to solve the initial logarithmic equation, even if you do not understand where the very final formula is taken. By the way, the canonical formula is called this record:

lOG A F (X) \u003d Log A A B

The convenience of the canonical form consists also in the fact that it can be used to solve a very wide class of logarithmic equations, and not just the simplest we consider today.

Examples of solutions

Now let's consider real examples. So, we decide:

log 0.5 (3x - 1) \u003d -3

Let's rewrite it as follows:

lOG 0.5 (3X - 1) \u003d log 0.5 0.5 -3

Many students are in a hurry and try to immediately build the number 0.5 to the degree that came to us from the initial task. And indeed, when you are already well trained in solving such tasks, you can immediately perform this step.

However, if now you are just starting to study this topic, it is better not to hurry anywhere in order to prevent offensive mistakes. So, we have a canonical form. We have:

3X - 1 \u003d 0.5 -3

This is no longer a logarithmic equation, but a linear relative to the variable x. To solve it, let's first look at the number of 0.5 V degree -3. Note that 0.5 is 1/2.

(1/2) −3 = (2/1) 3 = 8

All decimal fractions translate into ordinary when you solve the logarithmic equation.

Rewrite and get:

3X - 1 \u003d 8
3X \u003d 9.
x \u003d 3.

All we got the answer. The first task is solved.

Second task

Go to the second task:

As we see, this equation is no longer the simplest. Already at least because on the left there is a difference, and not a single logarithm one for one base.

Consequently, it is necessary to somehow get rid of this difference. In this case, everything is very simple. Let's look attentively on the ground: left is the number under the root:

General Recommendation: In all logarithmic equations, try to get rid of the radicals, i.e., from records with roots and move to power functions, simply because the indicators of these degrees are easily carried out for the logarithm sign and in the final account, such a record simplifies significantly and speeds up the calculations. Here let's write and write down:

Now remember the remarkable property of the logarithm: from the argument, as well as from the base, you can endure degrees. In the case of bases, the following happens:

log A K B \u003d 1 / K Loga B

In other words, the number that stood in the degree of foundation is carried forward and at the same time turns over, i.e. it becomes a different number. In our case, there was a degree of foundation with an indicator 1/2. Consequently, we can take it as 2/1. We get:

5 · 2 log 5 x - log 5 x \u003d 18
10 log 5 x - log 5 x \u003d 18

Please note: in no case can you get rid of logarithms at this step. Remember the Mathematics of 4-5 Class and Procedure: First, multiplication is performed, but only then addition and subtraction. In this case, we subtract one of the same elements:

9 log 5 x \u003d 18
log 5 x \u003d 2

Now our equation looks like. This is the simplest design, and we solve it with the help of a canonical form:

lOG 5 x \u003d log 5 5 2
x \u003d 5 2
x \u003d 25.

That's all. The second task is solved.

Third example

Go to the third task:

lG (x + 3) \u003d 3 + 2 LG 5

Let me remind you to the following formula:

lG B \u003d Log 10 B

If for some reason you are confused by the LG B record, then when performing all the calculations, you can only write Log 10 B. With decimal logarithms, you can work in the same way as with others: to make degrees, fold and represent any numbers in the form of LG 10.

That's it with these properties, we will now use to solve the problem, since it is not the simplest, which we recorded at the very beginning of our lesson.

To begin with, we note that the multiplier 2 facing LG 5 can be made and becomes a degree of foundation 5. In addition, the free term 3 also represents in the form of a logarithm - it is very easy to observe from our record.

Judge for yourself: any number can be represented in the form of LOG based on 10:

3 \u003d log 10 10 3 \u003d lg 10 3

We rewrite the source task, taking into account the changes received:

lG (x - 3) \u003d LG 1000 + LG 25
lG (x - 3) \u003d LG 1000 · 25
lG (x - 3) \u003d LG 25 000

We are again a canonical form, and we got it, bypassing the stage of transformations, i.e. the simplest logarithmic equation did not come across anywhere.

That was what I said at the very beginning of the lesson. The canonical form allows you to solve a wider class of tasks than the standard school formula that most school teachers give.

Well, all, get rid of the sign of the decimal logarithm, and we get a simple linear design:

x + 3 \u003d 25 000
x \u003d 24 997

Everything! The task is solved.

Note on the area of \u200b\u200bdefinition

Here I would like to bring an important remark about the area of \u200b\u200bdefinition. Surely now there will be students and teachers who will say: "When we solve expressions with logarithms, it is necessary to remember that the argument f (x) should be greater than zero!" In this regard, there is a logical question: why didn't we require in one of the considered tasks that this inequality is carried out?

Do not worry. No extra roots in these cases will arise. And this is another wonderful trick that allows you to speed up the decision. Just know that if the task variable x is found only in one place (or rather - in a single argument of a single logarithm), and nowhere else in our case there is no variable, then write the definition area not necessarybecause it will be performed automatically.

Judge for yourself: in the first equation, we received that 3x - 1, that is, the argument should be equal to 8. This automatically means that 3 - 1 will be greater than zero.

With the same success, we can write down that in the second case x should be 5 2, that is, he is obviously more zero. And in the third case, where x + 3 \u003d 25,000, i.e., again, more than more zero. In other words, the definition area is performed automatically, but only under the condition that X is found only in the argument of only one logarithm.

That's all you need to know to solve the simplest tasks. Already one of this rule, together with the conversion rules, will allow you to solve a very wide class of tasks.

But let's be honest: in order to finally deal with this technique to learn how to apply the canonical form of the logarithmic equation, it is not enough to simply see one video tutorial. Therefore, right now download options for an independent solution that are attached to this video language and start solving at least one of these two independent work.

You will leave you literally a few minutes. But the effect of such training will be much higher compared to the time you simply viewed this video tutorial.

I hope this lesson will help you deal with logarithmic equations. Use the canonical form, simplify expressions using the rules of working with logarithms - and no tasks will not be terrible. And I have everything today.

Accounting of the definition area

Now let's talk about the field of determining the logarithmic function, as well as how it affects the solution of logarithmic equations. Consider the design of the view

log a f (x) \u003d b

This expression is called the simplest - in it only one function, and the number A and B is precisely numbers, and in no case is not a function depending on the variable x. It is solved very simple. Just use the formula only:

b \u003d Log A A B

This formula is one of the key properties of the logarithm, and when substituting in our initial expression, we will receive the following:

lOG A F (X) \u003d Log A A B

f (x) \u003d a b

This is already familiar formula from school textbooks. Many students will certainly have a question: since in the initial expression, the function f (x) is under the Log sign, the following restrictions are superimposed on it:

f (x)\u003e 0

This restriction acts because the logarithm from negative numbers does not exist. So, maybe, as a result, the restrictions should be imposed on the answers? Maybe they need to be substituted into the source?

No, in the simplest logarithmic equations, an additional check of excess. And that's why. Take a look at our final formula:

f (x) \u003d a b

The fact is that the number A in any case is more than 0 - this requirement is also superimposed by logarithm. The number A is the basis. At the same time, the number b no restrictions is superimposed. But it does not matter, because in which a degree we would have erected a positive number, we still receive a positive number at the exit. Thus, the requirement F (x)\u003e 0 is performed automatically.

What is really worth checking is the area of \u200b\u200bdefining a function under the Log sign. There may be pretty difficult designs, and in the process of solving it is necessary to follow them. Let's see.

First task:

The first step: we transform the fraction on the right. We get:

We get rid of the sign of the logarithm and get the usual irrational equation:

From the resulting roots we are satisfied with only the first, since the second root is less than zero. The only answer will be the number 9. All, the task is solved. No additional checks that the expression under the logarithm sign is greater than 0 is not required, because it is not just greater than 0, but by the condition of the equation it is equal to 2. Consequently, the requirement "greater than zero" is performed automatically.

Go to the second task:

Here is all the same. Rewrite the design by replacing the top three:

Get rid of logarithm signs and get an irrational equation:

We build both parts into a square with regard to restrictions and get:

4 - 6x - x 2 \u003d (x - 4) 2

4 - 6x - x 2 \u003d x 2 + 8x + 16

x 2 + 8x + 16 -4 + 6x + x 2 \u003d 0

2x 2 + 14x + 12 \u003d 0 |: 2

x 2 + 7x + 6 \u003d 0

We solve the obtained equation through the discriminant:

D \u003d 49 - 24 \u003d 25

x 1 \u003d -1

x 2 \u003d -6

But x \u003d -6 does not suit us, because if we substitute this number in our inequality, we will get:

−6 + 4 = −2 < 0

In our case, it is required that there is more than 0 or in a pinch is equal. But x \u003d -1 suits us:

−1 + 4 = 3 > 0

The only answer in our case will be x \u003d -1. That's all the decision. Let's return to the very beginning of our calculations.

The main conclusion from this lesson: Checking the limitations for a function in the simplest logarithmic equations is not required. Because in the process of solving, all restrictions are performed automatically.

However, this does not mean that you can forget about checking. In the process of working on the logarithmic equation, it may well go to the irrational, in which there will be its limitations and requirements for the right part, in which we have been sent to two different examples.

Boldly solve such tasks and be particularly attentive if the root stands in the argument.

Logarithmic equations with different bases

We continue to study the logarithmic equations and we will analyze two more fairly interesting techniques, with the help of which it is fashionable to solve more complex structures. But first, let's remember how simple tasks are solved:

log a f (x) \u003d b

In this record, A and B are precisely numbers, and in the function f (x) the variable x should be present, and only there, i.e. x should be only in the argument. We will convert such logarithmic equations using a canonical form. To do this, we note that

b \u003d Log A A B

And a B is the argument. Let's rewrite this expression as follows:

lOG A F (X) \u003d Log A A B

We just do this and seek on the left, and on the right there was a logarithm on the basis of a. In this case, we can, figuratively speaking, cross the signs of Log, and from the point of view of mathematics we can say that we simply equate the arguments:

f (x) \u003d a b

As a result, we will get a new expression that will be solved much easier. Let's apply this rule to our today's tasks.

So, the first design:

First of all, I note that the right is a fraction, in the denominator which is Log. When you see such an expression, it will not be superfluous to recall the remarkable property of logarithms:

Transferring to Russian, this means that any logarithm can be represented as a private two logarithms with any base with. Of course, 0< с ≠ 1.

So: this formula has one wonderful case when the variable C is equal to the variable b. In this case, we will get the design of the form:

It is such a design that we observe from the sign to the right in our equation. Let's replace this design on Log A B, we get:

In other words, in comparison with the initial task, we changed the argument and the base of the logarithm. In exchange, we had to turn the fraction.

We remember that any degree can be made from the ground according to the following rule:

In other words, the K coefficient, which is the degree of foundation, is carried out as an inverted fraction. Let's bring it out as an inverted fraction:

The fractional multiplier cannot be left in front, because in this case we will not be able to submit this entry as a canonical form (because in canonical form before the second logarithm is no additional multiplier). Consequently, let's make a fraction of 1/4 into an argument in the form of a degree:

Now we equate the arguments, the foundations of which are the same (and the foundations we are really the same), and write:

x + 5 \u003d 1

x \u003d -4.

That's all. We received an answer to the first logarithmic equation. Please note: in the source task, the variable x is found only in one LOG, and it is in its argument. Consequently, checking the definition area is not required, and our number x \u003d -4 is indeed the answer.

Now go to the second expression:

lG 56 \u003d LG 2 LOG 2 7 - 3LG (X + 4)

Here, in addition to ordinary logarithms, we will have to work with LG F (X). How to solve such an equation? An unprepared student may seem that it is some kind of tin, but in fact everything is solved elementary.

Carefully look at the term LG 2 log 2 7. What can we say about it? The bases and arguments of the LOG and LG coincide, and it should make some thoughts. Let's remember once again how degrees from the logarithm sign:

log A B n \u003d nlog a b

In other words, what was the degree from number B in the argument becomes a multiplier to the Log itself. Let's apply this formula to expressions LG 2 log 2 7. Let you not scare LG 2 - this is the most common expression. You can rewrite it as follows:

All rules that act for any other logarithm are fair for him. In particular, the multiplier standing in front can be made to the degree of argument. Let's write:

Very often, students do not see this action, because it is not good to enter one log under the sign of another. In fact, nothing criminal in it. Moreover, we get a formula that is easily considered if you remember the important rule:

This formula can be considered as a definition, and as one of its properties. In any case, if you convert the logarithmic equation, this formula you should know exactly the same as the representation of any number in the form of log.

Return to our task. Rewrite it, taking into account the fact that the first term to the right of the equality sign will be just LG 7. We have:

lG 56 \u003d LG 7 - 3LG (X + 4)

Let's transfer LG 7 to the left, we get:

lG 56 - LG 7 \u003d -3LG (X + 4)

We subtract the expression on the left, because they have the same base:

lG (56/7) \u003d -3lg (x + 4)

Now let's look at the equation that we got. It is practically a canonical form, but the multiplier is present to the right. Let's make it in the right LG argument:

lG 8 \u003d LG (X + 4) -3

We have the canonical form of the logarithmic equation, so we strike the LG signs and equate the arguments:

(x + 4) -3 \u003d 8

x + 4 \u003d 0.5

That's all! We solved the second logarithmic equation. At the same time, no additional checks are required, because in the initial task it was visited only in one argument.

I will list the key points of this lesson again.

The main formula that is studied in all lessons on this page dedicated to the solution of logarithmic equations is a canonical form. And let you not frighten that in most school textbooks you are taught to solve such tasks in a different way. This tool works very effectively and allows you to solve a much wider class of tasks, rather than the simplest, which we studied at the very beginning of our lesson.

In addition, to solve logarithmic equations it will be useful to know the basic properties. Namely:

The transition formula to one base and a special case when we turn the Log (it was very useful to us in the first task);
The formula for making and making degrees from under the sign of the logarithm. Here, many students hang and emphasize that the submission and contribution itself may contain Log F (x). Nothing wrong with that. We can make one log on another sign and at the same time to significantly simplify the solution of the problem, which we observe in the second case.

In conclusion, I would like to add that it is not necessary to check the definition area in each of these cases, because everywhere the variable x is present only in one LOG sign, and at the same time it is in its argument. As a result, all the requirements of the definition area are performed automatically.

Variable base tasks

Today we will look at the logarithmic equations that for many students seem non-standard, and even unresolved at all. We are talking about expressions, at the base of which are not numbers, but variables and even functions. We will solve such structures with the help of our standard reception, namely through the canonical form.

To begin with, let's remember how the simplest tasks are solved, at the base of which there are ordinary numbers. So, the simplest is called the design of the type

log a f (x) \u003d b

To solve such tasks, we can use the following formula:

b \u003d Log A A B

We rewrite our initial expression and get:

lOG A F (X) \u003d Log A A B

Then we equate the arguments, i.e. write:

f (x) \u003d a b

Thus, we get rid of the sign of the Log and we decide the usual task. At the same time, the roots obtained in solving the root and will be the roots of the original logarithmic equation. In addition, the entry, when to the left, and the right is one and the same logarithm with the same base, just called the canonical form. It is to such a record that we will try to reduce today's designs. So, let's go.

First task:

log X - 2 (2x 2 - 13x + 18) \u003d 1

We replace 1 on log x - 2 (x - 2) 1. The extent that we observe the argument is, in fact, the number B that stood to the right of the sign of equality. Thus, we rewrite our expression. We get:

lOG X - 2 (2X 2 - 13X + 18) \u003d LOG X - 2 (X - 2)

What do we see? We have the canonical form of the logarithmic equation, so we can safely equate the arguments. We get:

2x 2 - 13x + 18 \u003d x - 2

But this decision does not end, because this equation is not equivalent to the initial one. After all, the resulting design consists of functions that are defined on the entire numerical line, and our initial logarithms are not defined everywhere and not always.

Therefore, we must separately write the definition area. Let's not wise and write all the requirements for the beginning:

First, the argument of each of the logarithms should be greater than 0:

2x 2 - 13x + 18\u003e 0

x - 2\u003e 0

Secondly, the base should be not only more than 0, but also different from 1:

x - 2 ≠ 1

As a result, we get the system:

But you do not be afraid: when processing logarithmic equations, such a system can be significantly simplified.

Judge for yourself: on the one hand, we require the quadratic function to be greater than zero, and on the other hand, this quadratic function is equal to a linear expression, which also requires it to be greater than zero.

In this case, if we require x - 2\u003e 0, then the requirement 2x 2 - 13x + 18\u003e 0 will automatically be performed. Therefore, we can safely cross the inequality containing a quadratic function. Thus, the number of expressions that is contained in our system will decrease to three.

Of course, with the same success, we could cross the linear inequality, i.e., delete x - 2\u003e 0 and demand that 2x 2 - 13x + 18\u003e 0. But agree that it is much faster to solve the simplest linear inequality much faster and easier, The quadratic, even if, if, as a result of the solution of the whole system, we will get the same roots.

In general, if possible, try to optimize calculations. And in the case of logarithmic equations, strike out the most complex inequalities.

Let's rewrite our system:

Here is a system of three expressions, with two of which we, in fact, have already figured out. Let's separately discard the square equation and solve it:

2x 2 - 14x + 20 \u003d 0

x 2 - 7x + 10 \u003d 0

We have a given square three-half-one and, therefore, we can take advantage of the formulas of the Vieta. We get:

(x - 5) (x - 2) \u003d 0

x 1 \u003d 5

x 2 \u003d 2

And now we return to our system and we find that x \u003d 2 does not suit us, because we need from it to be strictly greater than 2.

But x \u003d 5 we are quite satisfied with: the number 5 is greater than 2, and at the same time 5 is not 3. Therefore, the only solution of this system will be x \u003d 5.

Everything, the task is solved, including, taking into account the OTZ. Go to the second equation. Here we are waiting for more interesting and meaningful calculations:

The first step: as last time, we give all this case to canonical form. For this, the number 9 we can write as follows:

The base with the root can not be touched, but the argument is better to convert. Let's go from the root into a degree with a rational indicator. We write:

Let me not rewrite all our large logarithmic equation, but simply equalize the arguments immediately:

x 3 + 10x 2 + 31x + 30 \u003d x 3 + 9x 2 + 27x + 27

x 2 + 4x + 3 \u003d 0

We have a newly reduced three-half-one in front of us, we use the formulas of the Vieta and write:

(x + 3) (x + 1) \u003d 0

x 1 \u003d -3

x 2 \u003d -1

So, we got roots, but no one guarantees us that they would suit the initial logarithmic equation. After all, Log signs impose additional restrictions (here we would have to write down the system, but because of the bulkness of the whole design, I decided to calculate the definition area separately).

First of all, we remember that the arguments should be greater than 0, namely:

These are the requirements imposed by the definition area.

Immediately note that since we equate the first two expressions of the system to each other, then any of them we can delete. Let us cross the first because it looks more threatening than the second.

In addition, we note that the decision of the second and third inequality will be the same (cube of some number more than zero, if the very number is greater than zero; similarly and with the root of the third degree - these inequalities are completely similar, so we can cross it out).

But with the third inequality it will not pass. Get rid of the sign of the radical standing on the left, for which they erect both parts in the cube. We get:

So, we obtain the following requirements:

- 2 ≠ x\u003e -3

Which of our roots: x 1 \u003d -3 or x 2 \u003d -1 meet these requirements? Obviously, only x \u003d -1, because x \u003d -3 does not satisfy the first inequality (for inequality we have strict). Total returning to our task, we get one root: x \u003d -1. That's all, the task is solved.

Once again the key points of this task:

Feel free to apply and solve logarithmic equations using a canonical form. Students who make such a record, and do not pass directly from the initial problem to the design of the Log A f (x) \u003d b design, allow much less errors than those that are in a rush somewhere, passing intermediate steps of calculations;
As soon as an alternating base appears in logarithm, the task stops being the simplest. Consequently, when it is decided, it is necessary to take into account the field of definition: the arguments must be greater than zero, and the base is not only greater than 0, but they should not be equal to 1.

You can enter the latest requirements for final answers in different ways. For example, you can solve a whole system containing all the requirements for the definition area. On the other hand, you can first solve the task itself, and then remember about the definition area, separately work it in the form of the system and impose on the roots received.

What is the way to choose when solving a specific logarithmic equation, solve only you. In any case, the answer will turn out the same.

Listed equality in converting expressions with logarithms are used both to the right left and left to right.

It is worth noting that to memorize the effects from the properties is optional: when carrying out transformations, it is possible to do with the main properties of logarithms and other facts (for example, in that with b≥0), of which the corresponding consequences flow. The "side effect" of this approach is only manifested that the decision will be slightly longer. For example, to do without the investigation, which is expressed by the formula And repelly only from the main properties of logarithms, you will have to carry out a chain of transformations of the following type: .

The same can be said about the last property from the above list, which corresponds to the formula Since it also follows from the main properties of logarithms. The main thing to understand that there is always the possibility of a positive number with a logarithm in the indicator to change the foundation of the degree and the number under the logarithm sign. For the sake of justice, we note that examples that imply the implementation of transformations of such a kind are rare in practice. We give a few examples below the text.

Transformation of numerical expressions with logarithms

The properties of logarithms remembered, now it's time to learn to apply them in practice to convert expressions. Naturally start with the transformation of numerical expressions, and not expressions with variables, as they are more convenient and easier to know the basics. So we will do, and start with very simple examples to learn how to choose the desired property of the logarithm, but we will gradually complicate examples, up to the moment when you need to use several properties in a row to get the final result.

Selection of the desired properties of logarithms

The properties of logarithms are not so little, and it is clear that you need to be able to choose from them the appropriate, which in this particular case will lead to the desired result. It is usually difficult to do this, comparing the type of transformed logarithm or expression with the views of the left and right parts of the formulas expressing the properties of logarithms. If the left or right side of one of the formulas coincides with a given logarithm or expression, then most likely it is this property that needs to be applied when converting. The following examples are clearly demonstrated.

Let's start with examples of converting expressions using the definition of a logarithm that corresponds to the formula A Log A B \u003d B, A\u003e 0, A ≠ 1, B\u003e \u200b\u200b0.

Example.

Calculate if possible: a) 5 log 5 4, b) 10 lg (1 + 2 · π), B) , d) 2 log 2 (-7), e).

Decision.

In the example, under the letter a), the structure A Log A b is clearly visible, where a \u003d 5, b \u003d 4. These numbers satisfy the conditions a\u003e 0, a ≠ 1, b\u003e 0, so you can use the equality a Log A b \u003d b. We have 5 log 5 4 \u003d 4.

b) here a \u003d 10, b \u003d 1 + 2 · π, conditions a\u003e 0, a ≠ 1, b\u003e 0 are made. In this case, there is an equality of 10 lg (1 + 2 · π) \u003d 1 + 2 · π.

c) and in this example we are dealing with a degree of type A log a b, where and b \u003d ln15. So .

Despite the belonging to the same type of A Log A B (here a \u003d 2, b \u003d -7), the expression under the letter d) cannot be converted by the formula A Log A B \u003d B. The reason is that it does not make sense, as it contains a negative number under the sign of the logarithm. Moreover, the number B \u003d -7 does not satisfy the condition B\u003e 0, which does not allow to resort to the formula A log a B \u003d B, since it requires the fulfillment of the conditions a\u003e 0, a ≠ 1, b\u003e 0. So, it is impossible to talk about the calculation of the value of 2 log 2 (-7). In this case, recording 2 log 2 (-7) \u003d -7 will be an error.

Similarly, in the example under the letter D), the solution cannot be brought Since the initial expression does not make sense.

Answer:

a) 5 log 5 4 \u003d 4, b) 10 lg (1 + 2 · π) \u003d 1 + 2 · π, c) , d), e) expressions do not make sense.

It is often useful for the conversion at which a positive number is presented in the form of a degree of any positive and different number with a logarithm in the indicator. It is based on the same definition of logarithm a log a b \u003d b, a\u003e 0, a ≠ 1, b\u003e 0, but the formula is applied to the right left, that is, in the form B \u003d a log a b. For example, 3 \u003d E ln3 or 5 \u003d 5 log 5 5.

Go to the application of the properties of logarithms to convert expressions.

Example.

Find the value of the expression: a) log -2 1, b) log 1 1, c) log 0 1, d) log 7 1, e) ln1, e) lg1, g) log 3,75, s) log 5 · π 7 1.

Decision.

In the examples under the letters a), b) and c) the expressions of Log -2 1, Log 1 1, log 0 1, which does not make sense, because at the base of the logarithm should not be a negative number, zero or unit, because we determined Logarithm only for positive and different from the base unit. Therefore, in examples a) - c) there can be no question of finding the expression value.

In all other tasks, it is obvious that there are positive and different numbers from the unit 7, E, 10, 3.75 and 5 · π 7, respectively, and under the signs of logarithms everywhere there are units. And we know the property of the logarithm unit: Log A 1 \u003d 0 for any A\u003e 0, A ≠ 1. Thus, the values \u200b\u200bof expressions b) - e) are equal to zero.

Answer:

a), b), c) expressions do not make sense, d) log 7 1 \u003d 0, d) ln1 \u003d 0, e) lg1 \u003d 0, g) log 3,75 1 \u003d 0, h) log 5 · E 7 1 \u003d 0.

Example.

Calculate: a), b) lne, c) LG10, D) log 5 · π 3 -2 (5 · π 3 -2), e) log -3 (-3), e) log 1 1.

Decision.

It is clear that we have to take advantage of the property of the logarithm of the base, which corresponds to the formula Log A A \u003d 1 at a\u003e 0, a ≠ 1. Indeed, in the tasks under all the letters, the number under the sign of the logarithm coincides with its basis. Thus, I want to immediately say that the meaning of each of the specified expressions is 1. However, it is not necessary to hurry with the conclusions: in the tasks under the letters a) - d) the values \u200b\u200bof expressions are truly equal to one, and in the tasks D) and E) the initial expressions do not make sense, therefore it cannot be said that the values \u200b\u200bof these expressions are 1.

Answer:

a), b) lne \u003d 1, c) lg10 \u003d 1, d) log 5 · π 3 -2 (5 · π 3 -2) \u003d 1, D), e) expressions do not make sense.

Example.

Find a value: a) log 3 3 11, b) , c), d) log -10 (-10) 6.

Decision.

Obviously, under the signs of logarithms there are some degrees of foundation. Based on this, we understand that it is useful to us here the degree of the foundation: Log A A P \u003d P, where a\u003e 0, a ≠ 1 and p is any valid number. Given this, we have the following results: a) Log 3 3 11 \u003d 11, b) , in) . Is it possible to record similar equality for the example under the letter d) of the type of log -10 (-10) 6 \u003d 6? No, it is impossible, since the expression Log -10 (-10) 6 does not make sense.

Answer:

a) log 3 3 11 \u003d 11, b) , in) , d) the expression does not make sense.

Example.

Imagine an expression in the form of a sum or the difference of logarithms on the same basis: a) , b), c) lg ((- 5) · (-12)).

Decision.

a) under the sign of the logarithm is a work, and we know the logarithm property of the work of Log A (x · y) \u003d log a x + log a y, a\u003e 0, a ≠ 1, x\u003e 0, y\u003e 0. In our case, the number at the base of the logarithm and the number in the work is positive, that is, satisfy the conditions of the selected property, so we can calmly apply it: .

b) Here we use the property of the logarithm of the private, where a\u003e 0, a ≠ 1, x\u003e 0, y\u003e 0. In our case, the base of the logarithm is a positive number E, the numerator and denominator π are positive, which means that the conditions of the property are satisfying, so we have the right to use the chosen formula: .

c) First, we note that the expression Lg ((- 5) · (-12)) makes sense. But at the same time, for him, we do not have the right to apply the logarithm formula of the work of Log A (x · y) \u003d log a x + log ay, a\u003e 0, a ≠ 1, x\u003e 0, y\u003e 0, since the numbers -5 and -12 - negative and do not satisfy the conditions x\u003e 0, y\u003e 0. That is, it is impossible to conduct such a conversion: lG ((- 5) · (-12)) \u003d lg (-5) + lg (-12). And what to do? In such cases, the initial expression needs a preliminary transformation that allows you to get away from negative numbers. We will talk about such cases of transformation of expressions with negative numbers under the sign of logarithm in detail in one of the following examples, which is understandable, and without explanation: lG ((- 5) · (-12)) \u003d lg (5 · 12) \u003d LG5 + LG12.

Answer:

but) b) , c) lg ((- 5) · (-12)) \u003d LG5 + LG12.

Example.

Simplify the expression: a) log 3 0.25 + log 3 16 + log 3 0.5, b).

Decision.

Here we will help all the same properties of the logarithm of the work and logarithm of the private, which we used in previous examples, only now we will apply them to right to left. That is, the amount of logarithms is transforming into the logarithm of the work, and the difference between logarithms - in the logarithm of private. Have
but) log 3 0.25 + log 3 16 + log 3 0.5 \u003d log 3 (0.25 · 16 · 0.5) \u003d log 3 2.
b) .

Answer:

but) log 3 0.25 + log 3 16 + Log 3 0.5 \u003d log 3 2b) .

Example.

Get rid of the extent under the sign of the logarithm: a) log 0.7 5 11, b) , c) log 3 (-5) 6.

Decision.

It is easy to see that we are dealing with expressions of the Log A B p. The corresponding property of the logarithm has the kind of log a B p \u003d p · log a b, where a\u003e 0, a ≠ 1, b\u003e 0, p is any valid number. That is, when performing conditions a\u003e 0, a ≠ 1, b\u003e 0 from the logarithm of the degree of log a b p, we can move to the product P · Log A B. We will conduct this conversion with specified expressions.

a) In this case, a \u003d 0.7, b \u003d 5 and p \u003d 11. So log 0.7 5 11 \u003d 11 · Log 0.7 5.

b) Here, conditions a\u003e 0, a ≠ 1, b\u003e 0 are performed. therefore

c) The expression Log 3 (-5) 6 has the same structure log a b p, a \u003d 3, b \u003d -5, p \u003d 6. But for B, the condition b\u003e 0 is not satisfied, which makes it impossible to use the Log A B P \u003d P · Log A B. So, it is impossible to cope with the task? It is possible, but a pre-converting expression is required, we will talk about in detail below in the heading point. The decision will be: log 3 (-5) 6 \u003d log 3 5 6 \u003d 6 · LOG 3 5.

Answer:

a) log 0.7 5 11 \u003d 11 · log 0.7 5,
b)
c) Log 3 (-5) 6 \u003d 6 · Log 3 5.

Quite often, the logarithm formula of the degree during transformation is necessary to apply right to left as p · log a b \u003d log a b p (this requires the performance of the same conditions for A, B and P). For example, 3 · ln5 \u003d ln5 3 and lg2 · log 2 3 \u003d log 2 3 LG2.

Example.

a) Calculate the value of Log 2 5, if it is known that LG2≈0,3010 and LG5≈0,6990. b) Present a fraction in the form of a logarithm based on 3.

Decision.

a) The formula for the transition to a new base of the logarithm allows this logarithm to represent in the form of a ratio of decimal logarithms, whose values \u200b\u200bare known to us :. It remains only to carry out calculations, we have .

b) here it is enough to take advantage of the transition to a new base, and apply it to the right left, that is, in the form of . Receive .

Answer:

a) log 2 5≈2,3223, b) .

At this stage, we sufficiently scrupulously considered the conversion of the most simple expressions using the main properties of logarithms and the definition of logarithm. In these examples, we had to apply some kind of property and nothing more. Now with a calm conscience, you can move to examples, the transformation of which requires the use of several properties of logarithms and other additional transformations. We will go in the next paragraph. But before that, in short, we will briefly focus on the examples of the consequences of the main properties of logarithms.

Example.

a) Get rid of the root under the sign of the logarithm. b) Convert fraction in the logarithm on the base 5. c) Frequently from the degrees under the logarithm sign and in its foundation. d) Calculate the value of the expression . e) replace the expression of the degree with the base 3.

Decision.

a) if you remember about the consequence of the property of the logarithm You can immediately answer: .

b) we use the formula right to left we have .

c) In this case, the result leads the formula . Receive .

d) and here it is enough to apply a consequence that the formula is responsible . So .

e) property logarithm Allows us to achieve the desired result: .

Answer:

but) . b) . in) . d) . e) .

Sequential use of several properties

Real tasks for the transformation of expressions using the properties of logarithms are usually more complicated by those we engaged in the previous paragraph. In them, as a rule, the result is not one step, and the solution is already in the consistent application of one property after another, along with additional identity transformations, such as disclosure of brackets, bringing similar terms, reduction of fractions, etc. So let's get closer to such examples. There is nothing difficult in this, the main thing is to act neatly and consistently, observing the procedure for performing actions.

Example.

Calculate the value of the expression (Log 3 15-Log 3 5) · 7 log 7 5.

Decision.

The difference of logarithms in brackets for the property of the logarithm of a private can be replaced with logarithm Log 3 (15: 5), and further calculate its value log 3 (15: 5) \u003d log 3 3 \u003d 1. And the value of expression 7 log 7 5 by definition of logarithm is equal to 5. Substitute these results in the original expression, we get (log 3 15-Log 3 5) · 7 log 7 5 \u003d 1 · 5 \u003d 5.

Let us give a solution without explanation:
(Log 3 15-log 3 5) · 7 log 7 5 \u003d log 3 (15: 5) · 5 \u003d
\u003d log 3 3 · 5 \u003d 1 · 5 \u003d 5.

Answer:

(log 3 15-Log 3 5) · 7 log 7 5 \u003d 5.

Example.

What is the value of the numerical expression log 3 log 2 2 3 -1?

Decision.

We first transform the logarithm, which is located under the sign of the logarithm, according to the logarithm formula: Log 2 2 3 \u003d 3. Thus, log 3 log 2 2 3 \u003d Log 3 3 and further log 3 3 \u003d 1. So log 3 log 2 2 3 -1 \u003d 1-1 \u003d 0.

Answer:

log 3 log 2 2 3 -1 \u003d 0.

Example.

Simplify the expression.

Decision.

The transition formula to the new base of the logarithm allows the relationship of logarithms to one base to be represented as Log 3 5. In this case, the initial expression will take the form. By definition of logarithm 3 Log 3 5 \u003d 5, that is , And the value of the expression obtained, due to the same definition of the logarithm, is two.

Here is a brief version of the solution that is usually given: .

Answer:

For a smooth transition to the following item information, let's take a look at the expressions 5 2 + LOG 5 3, and LG0.01. Their structure is not suitable for any of the properties of logarithms. So what happens, they cannot be converted using the properties of logarithms? It is possible if you can conduct preliminary transformations that prepare these expressions to the application of the properties of logarithms. So 5 2 + LOG 5 3 \u003d 5 2 · 5 log 5 3 \u003d 25 · 3 \u003d 75, and LG0.01 \u003d LG10 -2 \u003d -2. Then we will understand in detail how such training of expressions is carried out.

Preparation of expressions to the application of the properties of logarithms

Logarithms in the composition of the transformed expression very often differ from the left and right parts of the formulas corresponding to the properties of logarithms. But no less often the transformation of these expressions implies the use of the properties of logarithms: to use them only requires preliminary preparation. And this preparation is in carrying out certain identical transformations leading logarithms to the form, convenient to apply properties.

For fairness, we note that almost any transformations of expressions can act as preliminary transformations, from the banal actuator of such terms to the use of trigonometric formulas. This is understandable, since the transformed expressions may contain any mathematical objects: brackets, modules, fractions, roots, degrees, etc. Thus, you need to be ready to perform any needed conversion to further be able to use the properties of logarithms.

Immediately, let's say that in this point we do not set ourselves the task to classify and disassemble all imaginable preliminary transformations, which further apply the properties of logarithms or the definition of logarithm. Here we will dwell only on four of them, which are most characteristic and most often found in practice.

And now in detail about each of them, after which, as part of our topic, it will only remain to deal with the transformation of expressions with variables under the signs of logarithms.

Selection of degrees under the sign of logarithm and in its foundation

Let's start immediately from the example. Let us be logarithm. Obviously, in this form, its structure does not have to use the properties of logarithms. Is it possible to somehow convert this expression to simplify it, and even better calculate its value? To answer this question, let's look carefully in numbers 81 and 1/9 in the context of our example. It is easy to notic here that these numbers allow the representation of the degree of number 3, indeed, 81 \u003d 3 4 and 1/9 \u003d 3 -2. In this case, the initial logarithm is presented in the form and the possibility of applying the formula . So, .

An analysis of the disassembled example creates the following thought: If possible, you can try to highlight the degree under the sign of the logarithm and in its foundation to apply the property of the logarithm or its consequence. It remains only to find out how to allocate these degrees. Let's give some recommendations on this issue.

Sometimes it is rather obvious that the number under the sign of the logarithm and / or in its foundation is some of the whole degree as in the example above. Practically constantly have to deal with detects of twos, which were well thought away: 4 \u003d 2 2, 8 \u003d 2 3, 16 \u003d 2 4, 32 \u003d 2 5, 64 \u003d 2 6, 128 \u003d 2 7, 256 \u003d 2 8, 512 \u003d 2 9, 1024 \u003d 2 10. This can be said about the degree of triple: 9 \u003d 3 2, 27 \u003d 3 3, 81 \u003d 3 4, 243 \u003d 3 5, ... in general, it does not hurt if it will be before our eyes table of degrees of natural numbers within a dozen. It is also not difficult to work with integer degrees of ten, hundred, thousands, etc.

Example.

Calculate the value or simplify the expression: a) log 6 216, b), c) log 0.000001 0.001.

Decision.

a) It is obvious that 216 \u003d 6 3, therefore Log 6 216 \u003d log 6 6 3 \u003d 3.

b) Table of degrees of natural numbers allows you to present numbers 343 and 1/243 in the form of degrees 7 3 and 3 -4, respectively. Therefore, it is possible to follow the following transformation of a given logarithm:

c) as 0.000001 \u003d 10 -6 and 0.001 \u003d 10 -3, then lOG 0.000001 0.001 \u003d log 10 -6 10 -3 \u003d (- 3) / (- 6) \u003d 1/2.

Answer:

a) log 6 216 \u003d 3, b) , c) log 0.000001 0.001 \u003d 1/2.

In more complex cases, to highlight the degrees of numbers have to resort to.

Example.

Convert the expression to a simpler type of Log 3 648 · Log 2 3.

Decision.

Let's see what is a decomposition of a number of 648 per simple factors:

That is, 648 \u003d 2 3 · 3 4. In this way, log 3 648 · Log 2 3 \u003d log 3 (2 3 · 3 4) · log 2 3.

Now the logarithm of works is transforming in the amount of logarithms, after which the properties of the logarithm of the degree are applicable:
log 3 (2 3 · 3 4) · log 2 3 \u003d (log 3 2 3 + log 3 3 4) · log 2 3 \u003d
\u003d (3 · Log 3 2 + 4) · Log 2 3.

Due to the investigation from the property of the logarithm to which the formula is responsible The product log32 · log23 is a work, and it is known to be one. Considering it, we get 3 · log 3 2 · log 2 3 + 4 · log 2 3 \u003d 3 · 1 + 4 · log 2 3 \u003d 3 + 4 · LOG 2 3.

Answer:

log 3 648 · Log 2 3 \u003d 3 + 4 · Log 2 3.

Quite often, expressions under the logarithm sign and in its foundation are works or ratios of roots and / or degrees of some numbers, for example,. Such expressions can be represented as a degree. For this, the transition from roots to degrees, and applied. These conversions allow you to highlight degrees under the logarithm sign and in its base, after which you apply the properties of logarithms.

Example.

Calculate: a) , b).

Decision.

a) the expression at the base of the logarithm is the product of degrees with the same bases, according to the appropriate property of degrees, we have 5 2 · 5 -0,5 · 5 -1 \u003d 5 2-0.5-1 \u003d 5 0.5.

Now we transform the fraction under the sign of the logarithm: we turn from the root to the degree, after which we will use the property of degrees with the same grounds: .

It remains to substitute the results obtained into the initial expression, use the formula and finish transformations:

b) Since 729 \u003d 3 6, and 1/9 \u003d 3 -2, then the initial expression can be rewritten in the form.

Next, apply the property of the root from the degree, we carry out the transition from the root to the degree and use the degree ratio property to convert the logarithm to the degree: .

Considering the last result, we have .

Answer:

but) , b).

It is clear that in general, to obtain degrees under the sign of the logarithm and, in its foundation, various transformations of various expressions may be required. We give a couple of examples.

Example.

What is the value of the expression: a) b) .

Decision.

Therefore, we note that the specified expression has the form of Log A B P, where a \u003d 2, b \u003d x + 1 and p \u003d 4. Numeric expressions of such kind We were converted by the property of the logarithm of the extent Log A B P \u003d p · Log A B, therefore, with a given expression, I want to do the same as, and from Log 2 (X + 1) 4 go to 4 · log 2 (x + 1). And now let's calculate the value of the initial expression and the expression obtained after the transformation, for example, with x \u003d -2. Have log 2 (-2 + 1) 4 \u003d log 2 1 \u003d 0, and 4 · Log 2 (-2 + 1) \u003d 4 · log 2 (-1) - Not meaning expression. This causes a natural question: "What did we do wrong"?

And the reason is as follows: we performed the transformation log 2 (x + 1) 4 \u003d 4 · log 2 (x + 1), based on the formula Log ABP \u003d P · Log AB, but we have the right to apply this formula only when condition A \u003e 0, a ≠ 1, b\u003e 0, p - any valid number. That is, the conversion done by us takes place if X + 1\u003e 0, which is the same x\u003e -1 (for A and P - the conditions are made). However, in our case, the OTZ variable x for the initial expression consists not only from the interval x\u003e -1, but also from the period x<−1 . Но для x<−1 мы не имели права осуществлять преобразование по выбранной формуле.

The need to take into account ...

We will continue to disassemble the transformation of the Log 2 (X + 1) 4 selected expressions by us, and now let's see what happens with OTZ when moving to expression 4 · log 2 (x + 1). In the previous paragraph, we found even the source expression - this is a set (-∞, -1) ∪ (-1, + ∞). Now we find the area of \u200b\u200bpermissible values \u200b\u200bof the variable x for expression 4 · log 2 (x + 1). It is determined by the condition X + 1\u003e 0, which corresponds to the set (-1, + ∞). Obviously, when moving from Log 2 (X + 1) 4 to 4 · Log 2 (X + 1), the area of \u200b\u200bvalid values \u200b\u200boccurs. And we agreed to avoid transformations leading to the narrowing of OTZ, as this can lead to various negative consequences.

It is worth noting here for yourself that it is useful to control the OTZ at every step of the transformation and prevent its narrowing. And if suddenly, at some stage of the transformation, there was a narrowing of OST, then it is worth looking very carefully, and whether this transformation is permissible and whether we have the right to carry out it.

For example, let's say that in practice, it is usually necessary to work with expressions, whose OTZ variables are such that, when carrying out transformations, use the properties of logarithms without restrictions in the form already known to us, and both from left to right and right to left. You quickly get used to it, and you begin to carry out transformations mechanically, without thinking, and whether it was possible to conduct them. And at such moments, as discharged, slipper more complex examples in which the inaccient use of the properties of logarithms leads to errors. So you need to always be on a check, and follow that there is no narrowing of OTZ.

It does not hurt separately select the main transformations based on the properties of logarithms that need to be carried out very carefully, which can lead to a narrowing of OTZ, and as a result - to errors:

Some transformations of expressions according to the properties of logarithms can lead to the inverse - expansion of OTZ. For example, the transition from 4 · log 2 (x + 1) to Log 2 (X + 1) 4 expands odd from the set (-1, + ∞) to (-∞, -1) ∪ (-1, + ∞). Such transformations occur if remain within the ODZD for the initial expression. So the only mentioned conversion 4 · log 2 (x + 1) \u003d log 2 (x + 1) 4 takes place on the OTZ variable x for the original expression 4 · log 2 (x + 1), that is, with x + 1\u003e 0, which is the same (-1, + ∞).

Now that we discussed the nuances for which you need to pay attention to when converting expressions with variables using the properties of logarithms, it remains to figure out how correctly these transformations need to be carried out.

X + 2\u003e 0. Does it work in our case? To answer this question, take a look at the OTZ variable x. It is determined by the inequality system which is equivalent to the condition x + 2\u003e 0 (if necessary, see the article solving systems of inequality). Thus, we can calmly apply the logarithm property.

Have
3 · lg (x + 2) 7 -lg (x + 2) -5 · lg (x + 2) 4 \u003d
\u003d 3 · 7 · lg (x + 2) -lg (x + 2) -5 · 4 · lg (x + 2) \u003d
\u003d 21 · lg (x + 2) -lg (x + 2) -20 · lg (x + 2) \u003d
\u003d (21-1-20) · Lg (x + 2) \u003d 0.

You can act and otherwise, the benefit of OTZ allows it to do, for example:

Answer:

3 · lg (x + 2) 7 -lg (x + 2) -5 · lg (x + 2) 4 \u003d 0.

And what to do when the conditions for the accompanying properties of logarithms are not satisfied? We will deal with this on the examples.

Suppose from us to simplify the expression Lg (x + 2) 4 -lg (x + 2) 2. The transformation of this expression, in contrast to the expression from the previous example, does not allow the log of the logarithm degree. Why? OTZ variable x in this case is a combination of two gaps x\u003e -2 and x<−2 . При x>-2 We can calmly apply the logarithm property and act as disassembled above: lg (x + 2) 4 -lg (x + 2) 2 \u003d 4 · lg (x + 2) -2 · lg (x + 2) \u003d 2 · lg (x + 2). But OTZ contains another period of x + 2<0 , для которого последнее преобразование будет некорректно. Что же делать при x+2<0 ? В подобных случаях на помощь приходит . Определение модуля позволяет выражение x+2 при x+2<0 представить как −|x+2| . Тогда при x+2<0 от lg(x+2) 4 −lg(x+2) 2 переходим к lG (- | x + 2 |) 4 -LG (- | x + 2 |) 2 And further by force of degree properties to LG | X + 2 | 4 -LG | X + 2 | 2. The resulting expression can be converted by the logarithm property, since | x + 2 |\u003e 0 for any values \u200b\u200bof the variable. Have lG | X + 2 | 4 -LG | X + 2 | 2 \u003d 4 · lg | x + 2 | -2 · lg | x + 2 | \u003d 2 · lg | x + 2 |. Now you can free yourself from the module, as he did his job. Since we are conducting conversion at x + 2<0 , то 2·lg|x+2|=2·lg(−(x+2)) . Итак, можно считать, что мы справились с поставленной задачей. Ответ: . Полученный результат можно записать компактно с использованием модуля как .

Consider another example so that work with modules has become familiar. Let we have conceived from the expression Go to the sum and difference between the logarithms of linear bounces X-1, X-2 and X-3. First we find ...

At the interval (3, + ∞) the values \u200b\u200bof expressions X-1, X-2 and X-3 are positive, so we calmly apply the properties of the logarithm of sums and differences:

And on the interval (1, 2), the values \u200b\u200bof the expression X-1 are positive, and the values \u200b\u200bof expressions X-2 and X-3 are negative. Therefore, in the interval under consideration, we present X-2 and X-3 using the module as - | x-2 | and - | x-3 | respectively. Wherein

Now you can apply the properties of the logarithm of the work and private, as it is on the interval (1, 2) the values \u200b\u200bof expressions X-1, | x-2 | and | x-3 | - Positive.

Have

The results can be combined:

In general, similar arguments allow the logarithm formulas on the basis of the logarithm, relationships and degrees to obtain three practically useful results, which are quite convenient to use:

The logarithm works of two arbitrary expressions X and Y of the type of log a (x · y) can be replaced with the summable logarithms log a | x | + log a | y | , A\u003e 0, A ≠ 1.
Logarithm Private Log A (X: Y) Can be replaced with the difference between logarithms Log A | X | -Log A | Y | , a\u003e 0, a ≠ 1, x and y - arbitrary expressions.
From the logarithm of some expression B in an even degree p of the Log A B P form you can go to the expression P · log a | b | , where a\u003e 0, a ≠ 1, p is an even number and b - an arbitrary expression.

Similar results are given, for example, in instructions to solve indicative and logarithmic equations in the collection of problems in mathematics for applicants to universities under the editors of M. I. Scanavi.

Example.

Simplify expression .

Decision.

It would be good to apply the properties of the logarithm, amounts and differences. But can we do it here? To answer this question, we need to know OTZ.

We define it:

It is rather obvious that the expressions X + 4, X-2 and (X + 4) 13 on the values \u200b\u200bof the permissible values \u200b\u200bof the variable x can take both positive and negative values. Therefore, we will have to act through modules.

The properties of the module allow you to rewrite as, therefore

Also nothing prevents from the property of the logarithm degree, then bring similar terms to:

The other sequence of transformations leads to the same result:

and since the expression X-2 can take both positive and negative values, then when submitting an even degree rate of 14