Solving the simplest logarithmic equations. Logarithmic equations. From simple to complex

Algebra grade 11

Topic: "Methods of solving logarithmic equations »

Lesson objectives:

educational: building knowledge about different ways solutions of logarithmic equations, the ability to apply them in each specific situation and choose any method for solving;

developing: development of skills to observe, compare, apply knowledge in a new situation, identify patterns, generalize; formation of skills of mutual control and self-control;

educational: fostering a responsible attitude to educational work, attentive perception of the material in the lesson, accuracy of keeping records.

Lesson type : a lesson in familiarization with new material.

"The invention of logarithms, by reducing the work of the astronomer, prolonged his life."
The French mathematician and astronomer P.S. Laplace

During the classes

I. Setting the purpose of the lesson

Studied definition of logarithm, properties of logarithms and logarithmic function will allow us to solve logarithmic equations. All logarithmic equations, no matter how complex they are, are solved using unified algorithms. We will consider these algorithms in today's lesson. There are few of them. If you master them, then any equation with logarithms will be within the power of each of you.

Write in a notebook the topic of the lesson: "Methods for solving logarithmic equations." I invite everyone to cooperate.

II. Updating basic knowledge

Let's get ready to study the topic of the lesson. You solve each task and write down the answer; you don't have to write a condition. Work in pairs.

1) For what values ​​of x does the function make sense:

a)

b)

v)

e)

(For each slide, the answers are checked and errors are sorted out)

2) Do the function graphs match?

a) y = x and

b)and

3) Rewrite the equalities as logarithmic equalities:

4) Write the numbers as logarithms with base 2:

4 =

2 =

0,5 =

1 =

5) Calculate :

6) Try to restore or supplement the missing elements in the given equalities.

III. Acquaintance with new material

The statement is displayed on the screen:

"The equation is the golden key that unlocks all mathematical tilings."
Contemporary Polish mathematician S. Koval

Try to formulate the definition of a logarithmic equation. (Equation containing the unknown under the sign of the logarithm ).

Considerthe simplest logarithmic equation: log a x = b (where a> 0, a ≠ 1). Since the logarithmic function increases (or decreases) on the set of positive numbers and takes all real values, then by the root theorem it follows that for any b this equation has, and moreover, only one solution, and it is positive.

Remember the definition of a logarithm. (The logarithm of the number x to the base a is an indicator of the degree to which the base a must be raised to get the number x ). It immediately follows from the definition of the logarithm thata v is such a solution.

Write down the title:Methods for solving logarithmic equations

1. By the definition of the logarithm .

This is how the simplest equations of the form.

ConsiderNo. 514 (a ): Solve the equation

How do you propose to solve it? (By the definition of the logarithm )

Solution . , Hence 2x - 4 = 4; x = 4.

Answer: 4.

In this problem 2x - 4> 0, since> 0, so no extraneous roots can appear, andthere is no need to check ... The condition 2x - 4> 0 in this task does not need to be written out.

2. Potentiation (transition from the logarithm of a given expression to this expression itself).

ConsiderNo. 519 (g): log 5 ( x 2 +8)- log 5 ( x+1)=3 log 5 2

What feature have you noticed?(The bases are the same and the logarithms of the two expressions are equal) ... What can be done?(Potentiate).

It should be borne in mind that any solution is contained among all x for which the logarithmable expression is positive.

Solution: ODZ:

X 2 +8> 0 unnecessary inequality

log 5 ( x 2 +8) = log 5 2 3 + log 5 ( x+1)

log 5 ( x 2 +8)= log 5 (8 x+8)

Potentiating the original equation

x 2 +8= 8 x+8

we get the equationx 2 +8= 8 x+8

We solve it:x 2 -8 x=0

x = 0, x = 8

Answer: 0; eight

In generaltransition to an equivalent system :

The equation

(The system contains a redundant condition - one of the inequalities need not be considered).

Question to the class : Which of these three solutions did you like best? (Discussion of ways).

You have the right to decide in any way.

3. Introducing a new variable .

ConsiderNo. 520 (g) . .

What have you noticed? (it quadratic equation relative to log3x) Your suggestions? (Introduce new variable)

Solution ... ODZ: x> 0.

Let be, then the equation will take the form:... The discriminant D> 0. Roots by Vieta's theorem:.

Let's go back to the replacement:or.

Having solved the simplest logarithmic equations, we get:

; .

Answer : 27;

4. Logarithm of both sides of the equation.

Solve the equation:.

Solution : ODZ: x> 0, we logarithm both sides of the equation to base 10:

... Let's apply the property of the logarithm of the degree:

(lgx + 3) lgx =

(lgx + 3) lgx = 4

Let lgx = y, then (y + 3) y = 4

, (D> 0) roots by Vieta's theorem: y1 = -4 and y2 = 1.

Let's go back to the replacement, we get: lgx = -4,; lgx = 1,. . It is as follows: if one of the functions y = f (x) increases and the other y = g (x) decreases on the interval X, then the equation f (x) = g (x) has at most one root on the interval X .

If there is a root, then it can be guessed. .

Answer : 2

« Correct use methods can be learned,
only by applying them to various examples. "
Danish historian of mathematics G.G. Zeiten

I V. Homework

P. 39 consider example 3, solve No. 514 (b), No. 529 (b), No. 520 (b), No. 523 (b)

V. Lesson summary

What methods for solving logarithmic equations did we consider in the lesson?

In the next lessons, we'll look at more complex equations. To solve them, the learned methods will be useful.

The last slide is shown:

“What is more than anything else?
Space.
What is the wisest thing?
Time.
What's the nicest thing?
Achieve what you want. "
Thales

I wish everyone to achieve what they want. Thank you for your cooperation and understanding.

Instructions

Write down the given logarithmic expression... If the expression uses the logarithm of 10, then its notation is truncated and looks like this: lg b is the decimal logarithm. If the logarithm has the number e as a base, then write the expression: ln b - natural logarithm... It is understood that the result of any is the power to which the base number must be raised to get the number b.

When finding from the sum of two functions, you just need to differentiate them in turn, and add the results: (u + v) "= u" + v ";

When finding the derivative of the product of two functions, it is necessary to multiply the derivative of the first function by the second and add the derivative of the second function, multiplied by the first function: (u * v) "= u" * v + v "* u;

In order to find the derivative of the quotient of two functions, it is necessary, from the product of the derivative of the dividend multiplied by the divisor function, to subtract the product of the derivative of the divisor multiplied by the function of the dividend, and divide all this by the divisor function squared. (u / v) "= (u" * v-v "* u) / v ^ 2;

If given complex function, then it is necessary to multiply the derivative of the internal function and the derivative of the external one. Let y = u (v (x)), then y "(x) = y" (u) * v "(x).

Using the ones obtained above, you can differentiate almost any function. So, let's look at a few examples:

y = x ^ 4, y "= 4 * x ^ (4-1) = 4 * x ^ 3;

y = 2 * x ^ 3 * (e ^ xx ^ 2 + 6), y "= 2 * (3 * x ^ 2 * (e ^ xx ^ 2 + 6) + x ^ 3 * (e ^ x-2 * x));
There are also problems for calculating the derivative at a point. Let the function y = e ^ (x ^ 2 + 6x + 5) be given, you need to find the value of the function at the point x = 1.
1) Find the derivative of the function: y "= e ^ (x ^ 2-6x + 5) * (2 * x +6).

2) Calculate the value of the function at the given point y "(1) = 8 * e ^ 0 = 8

Related Videos

Helpful advice

Learn the table of elementary derivatives. This will significantly save time.

Sources:

• derivative of a constant

So, what is the difference between an irrational equation and a rational one? If the unknown variable is under the sign square root, then the equation is considered irrational.

Instructions

The main method for solving such equations is the method of constructing both parts equations in a square. However. this is natural, the first step is to get rid of the sign. This method is not technically difficult, but sometimes it can get in trouble. For example, the equation v (2x-5) = v (4x-7). By squaring both sides of it, you get 2x-5 = 4x-7. This equation is not difficult to solve; x = 1. But the number 1 will not be the given equations... Why? Substitute 1 in the equation for x, and both the right and the left will contain expressions that don't make sense, that is. This value is not valid for a square root. Therefore, 1 is an extraneous root, and therefore the given equation has no roots.

So, an irrational equation is solved using the method of squaring both sides of it. And having solved the equation, it is imperative to cut off extraneous roots. To do this, substitute the found roots into the original equation.

Consider another one.
2x + vx-3 = 0
Of course, this equation can be solved in the same way as the previous one. Move composite equations that do not have a square root, to the right side and then use the squaring method. solve the resulting rational equation and roots. But also another, more graceful one. Enter a new variable; vx = y. Accordingly, you get an equation of the form 2y2 + y-3 = 0. That is, the usual quadratic equation. Find its roots; y1 = 1 and y2 = -3 / 2. Next, decide two equations vx = 1; vx = -3 / 2. The second equation has no roots, from the first we find that x = 1. Don't forget to check the roots.

Solving identities is easy enough. To do this, you need to do identical transformations until the goal is achieved. Thus, with the help of the simplest arithmetic operations, the task will be solved.

You will need

• - paper;
• - a pen.

Instructions

The simplest of such transformations is algebraic abbreviated multiplication (such as the square of the sum (difference), the difference of squares, the sum (difference), the cube of the sum (difference)). In addition, there are many and trigonometric formulas which are essentially the same identities.

Indeed, the square of the sum of two terms is equal to the square of the first plus twice the product of the first by the second and plus the square of the second, that is, (a + b) ^ 2 = (a + b) (a + b) = a ^ 2 + ab + ba + b ^ 2 = a ^ 2 + 2ab + b ^ 2.

Simplify both

General principles of solution

Review through a textbook on calculus or higher mathematics, which is a definite integral. As you know, the solution definite integral is a function whose derivative will give the integrand. This function is called antiderivative. The basic integrals are constructed according to this principle.
Determine by the type of the integrand which of the tabular integrals is suitable in this case. It is not always possible to determine this immediately. Often, the tabular view becomes noticeable only after several transformations to simplify the integrand.

Variable replacement method

If the integrand is trigonometric function, in the argument of which is some polynomial, then try using the variable replacement method. To do this, replace the polynomial in the argument of the integrand with some new variable. Determine the new limits of integration from the relationship between the new and the old variable. Differentiating this expression, find the new differential in. So you get the new kind the previous integral, close to or even corresponding to some tabular one.

Solution of integrals of the second kind

If the integral is an integral of the second kind, the vector form of the integrand, then you will need to use the rules for passing from these integrals to scalar ones. One of these rules is the Ostrogradsky-Gauss ratio. This law makes it possible to pass from the rotor flux of a certain vector function to a triple integral over the divergence of a given vector field.

Substitution of the limits of integration

After finding the antiderivative, it is necessary to substitute the limits of integration. First, plug the upper limit value into the antiderivative expression. You will get some number. Next, subtract from the resulting number another number obtained from the lower limit to the antiderivative. If one of the limits of integration is infinity, then when substituting it into antiderivative function it is necessary to go to the limit and find what the expression is striving for.
If the integral is two-dimensional or three-dimensional, then you will have to represent geometrically the limits of integration in order to understand how to calculate the integral. Indeed, in the case of, say, a three-dimensional integral, the limits of integration can be entire planes that bound the volume to be integrated.

Solving logarithmic equations. Part 1.

Logarithmic equation is an equation in which the unknown is contained under the sign of the logarithm (in particular, at the base of the logarithm).

The simplest logarithmic equation looks like:

Solution to any logarithmic equation involves the transition from logarithms to expressions under the sign of logarithms. However, this action expands the range of admissible values ​​of the equation and can lead to the appearance of extraneous roots. To avoid the appearance of extraneous roots, you can do one of three ways:

1. Make an equivalent transition from the original equation to the system including

depending on which inequality is or is simpler.

If the equation contains an unknown at the base of the logarithm:

then we go to the system:

2. Find separately the range of admissible values ​​of the equation, then solve the equation and check if the found solutions satisfy the equation.

3. Solve the equation, and then do check: substitute the found solutions into the original equation, and check if we get the correct equality.

The logarithmic equation of any level of complexity ultimately always reduces to the simplest logarithmic equation.

All logarithmic equations can be roughly divided into four types:

1 ... Equations that only contain logarithms to the first degree. With the help of transformations and use they are reduced to the form

Example... Let's solve the equation:

Let's equate the expressions under the logarithm sign:

Let's check if our root satisfies the equation:

Yes, it does.

Answer: x = 5

2 ... Equations that contain logarithms to a degree other than 1 (in particular, in the denominator of a fraction). Such equations are solved using introducing variable change.

Example. Let's solve the equation:

Let's find the ODZ of the equation:

The equation contains logarithms squared, so it is solved by changing the variable.

Important! Before introducing a substitution, it is necessary to "pull apart" the logarithms included in the equation into "bricks" using the properties of the logarithms.

When "pulling" logarithms, it is important to very carefully apply the properties of logarithms:

In addition, there is one more subtle point here, and in order to avoid a common mistake, we will use an intermediate equality: we write the degree of the logarithm in this form:

Similarly,

Substitute the resulting expressions into the original equation. We get:

Now we see that the unknown is contained in the equation in the composition. Let's introduce the replacement:. Since it can take any real value, we do not impose any restrictions on the variable.

basic properties.

1. logax + logay = loga (x y);
2. logax - logay = loga (x: y).

identical grounds

Log6 4 + log6 9.

Now let's complicate the task a little.

Examples of solving logarithms

What if the base or argument of the logarithm is based on a degree? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:

Of course, all these rules make sense if the ODL of the logarithm is observed: a> 0, a ≠ 1, x>

Task. Find the meaning of the expression:

Moving to a new foundation

Let the logarithm be given. Then, for any number c such that c> 0 and c ≠ 1, the following equality holds:

Task. Find the meaning of the expression:

See also:

Basic properties of the logarithm

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

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Lev Nikolaevich Tolstoy.

Basic properties of logarithms

Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.

Examples for logarithms

Logarithm expressions

Example 1.
a). x = 10ac ^ 2 (a> 0, c> 0).

By properties 3.5 we calculate

2.

3.

4. where .

Example 2. Find x if

Example 3. Let the value of the logarithms be given

Evaluate log (x) if

Basic properties of logarithms

Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called basic properties.

It is imperative to know these rules - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.

Addition and subtraction of logarithms

Consider two logarithms with the same bases: logax and logay. Then they can be added and subtracted, and:

1. logax + logay = loga (x y);
2. logax - logay = loga (x: y).

So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Note: key moment here - identical grounds... If the reasons are different, these rules do not work!

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

Since the bases of the logarithms are the same, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.

Task. Find the value of the expression: log2 48 - log2 3.

The bases are the same, we use the difference formula:
log2 48 - log2 3 = log2 (48: 3) = log2 16 = 4.

Task. Find the value of the expression: log3 135 - log3 5.

Again the bases are the same, so we have:
log3 135 - log3 5 = log3 (135: 5) = log3 27 = 3.

As you can see, the original expressions are composed of "bad" logarithms, which are not separately counted. But after transformations, quite normal numbers are obtained. Many are built on this fact. test papers... But what control - such expressions in all seriousness (sometimes - practically unchanged) are offered on the exam.

Removing the exponent from the logarithm

It's easy to see that the last rule follows the first two. But it is better to remember it all the same - in some cases it will significantly reduce the amount of computation.

Of course, all these rules make sense if the ODL of the logarithm is observed: a> 0, a ≠ 1, x> 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers in front of the sign of the logarithm into the logarithm itself. This is what is most often required.

Task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 = 6 log7 49 = 6 2 = 12

Task. Find the meaning of the expression:

Note that the denominator contains the logarithm, the base and argument of which are exact powers: 16 = 24; 49 = 72. We have:

I think the last example needs some clarification. Where did the logarithms disappear? Until the very last moment, we work only with the denominator.

Formulas for logarithms. Logarithms are examples of solutions.

We presented the base and the argument of the logarithm standing there in the form of degrees and brought out the indicators - we got a "three-story" fraction.

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

Moving to a new foundation

Speaking about the rules for addition and subtraction of logarithms, I specifically emphasized that they only work for the same bases. What if the reasons are different? What if they are not exact powers of the same number?

Formulas for the transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:

Let the logarithm be given. Then, for any number c such that c> 0 and c ≠ 1, the following equality holds:

In particular, if we put c = x, we get:

From the second formula it follows that it is possible to swap the base and the argument of the logarithm, but in this case the whole expression is "reversed", i.e. the logarithm appears in the denominator.

These formulas are rarely found in conventional numerical expressions... It is possible to estimate how convenient they are only when solving logarithmic equations and inequalities.

However, there are tasks that are generally not solved except by the transition to a new foundation. Consider a couple of these:

Task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms contain exact degrees. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;

Now let's "flip" the second logarithm:

Since the product does not change from the permutation of the factors, we calmly multiplied the four and two, and then dealt with the logarithms.

Task. Find the value of the expression: log9 100 · lg 3.

The base and argument of the first logarithm are exact degrees. Let's write this down and get rid of the metrics:

Now let's get rid of the decimal logarithm by moving to the new base:

Basic logarithmic identity

Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:

In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it is just the value of the logarithm.

The second formula is actually a paraphrased definition. It is called that:.

Indeed, what happens if the number b is raised to such a power that the number b to this power gives the number a? That's right: you get this very number a. Read this paragraph carefully again - many people "hang" on it.

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

Task. Find the meaning of the expression:

Note that log25 64 = log5 8 - just moved the square out of the base and the logarithm argument. Considering the rules for multiplying powers with on the same basis, we get:

If someone is not in the know, it was a real problem from the exam 🙂

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that can hardly be called properties - rather, these are consequences of the definition of the logarithm. They are constantly encountered in problems and, surprisingly, create problems even for "advanced" students.

1. logaa = 1 is. Remember once and for all: the logarithm to any base a from this base is equal to one.
2. loga 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 = 1 is a direct consequence of the definition.

That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.

See also:

The logarithm of b to base a denotes an expression. To calculate the logarithm means to find such a power of x () at which the equality

Basic properties of the logarithm

The given properties need to be known, since, on their basis, almost all problems and examples are associated with logarithms are solved. The rest of the exotic properties can be deduced by mathematical manipulations with these formulas

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

When calculating the formulas for the sum and difference of logarithms (3.4) are encountered quite often. The rest are somewhat complex, but in a number of tasks they are indispensable for simplifying complex expressions and calculating their values.

Common cases of logarithms

Some of the common logarithms are those in which the base is even ten, exponential or two.
The base ten logarithm is usually called the decimal logarithm and is simply denoted lg (x).

From the recording it is clear that the basics are not written in the recording. For example

The natural logarithm is the logarithm based on the exponent (denoted by ln (x)).

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Lev Nikolaevich Tolstoy. Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.

And another important base two logarithm is

The derivative of the logarithm of the function is equal to one divided by the variable

The integral or antiderivative of the logarithm is determined by the dependence

The given material is enough for you to solve a wide class of problems related to logarithms and logarithms. To assimilate the material, I will give only a few common examples from school curriculum and universities.

Examples for logarithms

Logarithm expressions

Example 1.
a). x = 10ac ^ 2 (a> 0, c> 0).

By properties 3.5 we calculate

2.
By the property of the difference of logarithms, we have

3.
Using properties 3,5 we find

4. where .

By the look complex expression using a number of rules is simplified to the form

Finding the values ​​of logarithms

Example 2. Find x if

Solution. To calculate, we apply to the last term 5 and 13 properties

Substitute and grieve

Since the bases are equal, we equate the expressions

Logarithms. First level.

Let the value of the logarithms be given

Evaluate log (x) if

Solution: Let us logarithm the variable to write the logarithm through the sum of terms

This is where the acquaintance with logarithms and their properties just begins. Practice calculations, enrich your practical skills - you will soon need this knowledge to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge for another equally important topic - logarithmic inequalities ...

Basic properties of logarithms

Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called basic properties.

It is imperative to know these rules - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.

Addition and subtraction of logarithms

Consider two logarithms with the same bases: logax and logay. Then they can be added and subtracted, and:

1. logax + logay = loga (x y);
2. logax - logay = loga (x: y).

So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Please note, the key point here is - identical grounds... If the reasons are different, these rules do not work!

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

Task. Find the value of the expression: log6 4 + log6 9.

Since the bases of the logarithms are the same, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.

Task. Find the value of the expression: log2 48 - log2 3.

The bases are the same, we use the difference formula:
log2 48 - log2 3 = log2 (48: 3) = log2 16 = 4.

Task. Find the value of the expression: log3 135 - log3 5.

Again the bases are the same, so we have:
log3 135 - log3 5 = log3 (135: 5) = log3 27 = 3.

As you can see, the original expressions are composed of "bad" logarithms, which are not separately counted. But after transformations, quite normal numbers are obtained. Many tests are based on this fact. But what control - such expressions in all seriousness (sometimes - practically unchanged) are offered on the exam.

Removing the exponent from the logarithm

Now let's complicate the task a little. What if the base or argument of the logarithm is based on a degree? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:

It's easy to see that the last rule follows the first two. But it is better to remember it all the same - in some cases it will significantly reduce the amount of computation.

Of course, all these rules make sense if the ODL of the logarithm is observed: a> 0, a ≠ 1, x> 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers in front of the sign of the logarithm into the logarithm itself.

How to solve logarithms

This is what is most often required.

Task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 = 6 log7 49 = 6 2 = 12

Task. Find the meaning of the expression:

Note that the denominator contains the logarithm, the base and argument of which are exact powers: 16 = 24; 49 = 72. We have:

I think the last example needs some clarification. Where did the logarithms disappear? Until the very last moment, we work only with the denominator. We presented the base and the argument of the logarithm standing there in the form of degrees and brought out the indicators - we got a "three-story" fraction.

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

Moving to a new foundation

Speaking about the rules for addition and subtraction of logarithms, I specifically emphasized that they only work for the same bases. What if the reasons are different? What if they are not exact powers of the same number?

Formulas for the transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:

Let the logarithm be given. Then, for any number c such that c> 0 and c ≠ 1, the following equality holds:

In particular, if we put c = x, we get:

From the second formula it follows that it is possible to swap the base and the argument of the logarithm, but in this case the whole expression is "reversed", i.e. the logarithm appears in the denominator.

These formulas are rarely found in conventional numeric expressions. It is possible to estimate how convenient they are only when solving logarithmic equations and inequalities.

However, there are tasks that are generally not solved except by the transition to a new foundation. Consider a couple of these:

Task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms contain exact degrees. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;

Now let's "flip" the second logarithm:

Since the product does not change from the permutation of the factors, we calmly multiplied the four and two, and then dealt with the logarithms.

Task. Find the value of the expression: log9 100 · lg 3.

The base and argument of the first logarithm are exact degrees. Let's write this down and get rid of the metrics:

Now let's get rid of the decimal logarithm by moving to the new base:

Basic logarithmic identity

Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:

In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it is just the value of the logarithm.

The second formula is actually a paraphrased definition. It is called that:.

Indeed, what happens if the number b is raised to such a power that the number b to this power gives the number a? That's right: you get this very number a. Read this paragraph carefully again - many people "hang" on it.

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

Task. Find the meaning of the expression:

Note that log25 64 = log5 8 - just moved the square out of the base and the logarithm argument. Taking into account the rules for multiplying degrees with the same base, we get:

If someone is not in the know, it was a real problem from the exam 🙂

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that can hardly be called properties - rather, these are consequences of the definition of the logarithm. They are constantly encountered in problems and, surprisingly, create problems even for "advanced" students.

1. logaa = 1 is. Remember once and for all: the logarithm to any base a from this base is equal to one.
2. loga 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 = 1 is a direct consequence of the definition.

That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.

In this lesson, we will review the basic theoretical facts about logarithms and consider solving the simplest logarithmic equations.

Let us recall the central definition - the definition of the logarithm. It is related to the decision exponential equation... This equation has a single root, it is called the logarithm of b to base a:

Definition:

The logarithm of the number b to the base a is the exponent to which the base a must be raised to get the number b.

Recall basic logarithmic identity.

Expression (expression 1) is the root of the equation (expression 2). Substitute the value of x from expression 1 instead of x into expression 2 and get the basic logarithmic identity:

So we see that each value is assigned a value. We denote b by x (), c by y, and thus we obtain a logarithmic function:

For example:

Let's recall the main properties of the logarithmic function.

Let's pay attention once again, here, because under the logarithm there can be a strictly positive expression, as the base of the logarithm.

Rice. 1. Graph of the logarithmic function at various bases

The function graph for is shown in black. Rice. 1. If the argument increases from zero to infinity, the function increases from minus to plus infinity.

The function graph for is shown in red. Rice. 1.

Properties of this function:

Domain: ;

Range of values:;

The function is monotonic throughout its entire domain of definition. When monotonically (strictly) increases, more meaning the argument matches the larger value of the function. When monotonically (strictly) decreases, the larger value of the argument corresponds to the smaller value of the function.

The properties of the logarithmic function are the key to solving a variety of logarithmic equations.

Consider the simplest logarithmic equation, all other logarithmic equations, as a rule, are reduced to this form.

Since the bases of the logarithms and the logarithms themselves are equal, the functions under the logarithm are also equal, but we must not miss the domain of definition. Only a positive number can stand under the logarithm, we have:

We found out that the functions f and g are equal, so it is enough to choose any one inequality in order to comply with the DHS.

So we got mixed system, in which there is an equation and inequality:

As a rule, it is not necessary to solve an inequality, it is enough to solve the equation and substitute the found roots into the inequality, thus performing a check.

Let us formulate a method for solving the simplest logarithmic equations:

Equalize the bases of logarithms;

Equate sub-logarithmic functions;

Check.

Let's look at specific examples.

Example 1 - Solve the equation:

The bases of the logarithms are initially equal, we have the right to equate sub-logarithmic expressions, do not forget about the ODZ, we will choose the first logarithm to compose the inequality:

Example 2 - Solve the equation:

This equation differs from the previous one in that the bases of the logarithms are less than one, but this does not affect the solution in any way:

Find the root and substitute it into the inequality:

We got the wrong inequality, which means that the found root does not satisfy the ODV.

Example 3 - Solve the equation:

The bases of the logarithms are initially equal, we have the right to equate sub-logarithmic expressions, do not forget about the ODZ, we will choose the second logarithm to compose the inequality:

Find the root and substitute it into the inequality:

Obviously, only the first root satisfies the ODV.