How to deduct logarithms with the same base. Formulas logarithms. Logarithms Examples Solutions

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: Log a. x. and log. a. y.. Then they can be folded and deducted, and:

log. a. x. + Log. a. y. \u003d Log. a. (x. · y.);
log. a. x. - Log. a. y. \u003d Log. a. (x. : y.).

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

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

Log 6 4 + Log 6 9.

Since the bases in logarithms are the same, we use the sum of the sum:
log 6 4 + Log 6 9 \u003d Log 6 (4 · 9) \u003d log 6 36 \u003d 2.

A task. Find the value of the expression: Log 2 48 - Log 2 3.

The foundations are the same, using the difference formula:
log 2 48 - Log 2 3 \u003d Log 2 (48: 3) \u003d log 2 16 \u003d 4.

A task. Find the value of the expression: Log 3 135 - Log 3 5.

Again the foundations are the same, so we have:
log 3 135 - Log 3 5 \u003d log 3 (135: 5) \u003d log 3 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. Many are built on this fact. test papers. But what is the control - such expressions are in full (sometimes - almost unchanged) are offered on the exam.

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.

Of course, all these rules make sense if compliance with the OTZ Logarithm: a. > 0, a. ≠ 1, x. \u003e 0. And also: 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: log 7 49 6.

Get rid of the extent in the argument in the first formula:
log 7 49 6 \u003d 6 · Log 7 49 \u003d 6 · 2 \u003d 12

A task. Find the value of the expression:
[Signature to Figure]

Note that in the denominator there is a logarithm, the base and the argument of which are accurate degrees: 16 \u003d 2 4; 49 \u003d 7 2. We have:

[Signature to Figure]

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.

Now let's look at the basic fraction. The number in the numerator and the denominator is the same number: log 2 7. Since log 2 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 Logarithm Log a. x.. Then for any number c. such that c. \u003e 0 I. c. ≠ 1, true equality:
[Signature to Figure]
In particular, if you put c. = x.We will get:
[Signature to Figure]

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 rarely found in ordinary 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: Log 5 16 · Log 2 25.

Note that the arguments of both logarithms are accurate degrees. I will summarize: log 5 16 \u003d log 5 2 4 \u003d 4Log 5 2; Log 2 25 \u003d log 2 5 2 \u003d 2Log 2 5;

And now "invert" the second logarithm:

[Signature to Figure]

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: Log 9 100 · LG 3.

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

[Signature to Figure]

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

[Signature to Figure]

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 n. It becomes an indicator of the extent in the argument. Number n. It can be absolutely anyone, because it is just a logarithm value.

The second formula is actually a paraphrassed definition. It is called: the main logarithmic identity.

In fact, what will happen if the number b. build in such a degree that the number b. to this extent gives the number a.? Correctly: this is the most 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:
[Signature to Figure]

Note that log 25 64 \u003d log 5 8 - just made a square from the base and the argument of the logarithm. Given the rules for multiplication of degrees with the same base, we get:

[Signature to Figure]

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.

log. a. a. \u003d 1 is a logarithmic unit. Record once and forever: logarithm on any basis a. From the very base is equal to one.
log. a. 1 \u003d 0 is a logarithmic zero. Base a. Maybe somehow, but if the argument is a unit - logarithm is zero! Because a. 0 \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.

Let's start by S. properties logarithm units. Its formulation is as follows: the logarithm unit is zero, that is, log A 1 \u003d 0 For any A\u003e 0, A ≠ 1. The proof does not cause difficulties: since a 0 \u003d 1 for any A, satisfying the conditions specified above A\u003e 0 and A 1, then the provible equality Log A 1 \u003d 0 immediately follows from the definition of logarithm.

We give examples of applying the considered properties: log 3 1 \u003d 0, lg1 \u003d 0 and.

Go to K. next property: the logarithm of the number equal to the base is equal to one, i.e, log A A \u003d 1 With a\u003e 0, a ≠ 1. Indeed, since a 1 \u003d a for any A, then by definition of logarithm Log A a \u003d 1.

Examples of using this property of logarithms are equivals log 5 5 \u003d 1, Log 5.6 5.6 and LNe \u003d 1.

For example, log 2 2 7 \u003d 7, LG10 -4 \u003d -4 and .

Logarithm works of two positive numbers X and Y is equal to the product of the logarithms of these numbers: log A (x · y) \u003d log a x + log a y, A\u003e 0, A ≠ 1. We prove the property of the logarithm of the work. By virtue of the degree a log a x + log a y \u003d a Log A x · a log a y, and since the main logarithmic identity A log a x \u003d x and a log a y \u003d y, then a log a x · a log a y \u003d x · y. Thus, a log a x + log a y \u003d x · y, from where the definition of logarithm implies proven equality.

Let us show examples of using the logarithm properties: log 5 (2 · 3) \u003d log 5 2 + log 5 3 and .

The logarithm property of the work can be generalized on the product of a finite number of N positive numbers x 1, x 2, ..., X n as log A (x 1 · x 2 · ... · x n) \u003d log A x 1 + Log A x 2 + ... + log a x n . This equality is proved without problems.

For example, natural logarithm works can be replaced by the sum of three natural logarithov Numbers 4, E, and.

Logarithm of private two positive numbers X and Y is equal to the difference in the logarithms of these numbers. The properties of the logarithm of the private corresponds to the formula of the form, where a\u003e 0, a ≠ 1, x and y are some positive numbers. The validity of this formula is proved as the logarithm formula: since , By definition of logarithm.

Let us give an example of using this logarithm property: .

Go to K. property of logarithm degree. The logarithm degree is equal to the product of the degree in the logarithm of the module of the base of this degree. We write this property of the logarithm in the formula: log A B P \u003d P · Log A | B |where a\u003e 0, a ≠ 1, b and p such numbers that the degree B p makes sense and b p\u003e 0.

First, we prove this property for positive b. The main logarithmic identity allows us to present the number B as a Log A B, then b p \u003d (a log a b) p, and the resulting expression by virtue of the degree property is a p · log a b. So we come to the equality B p \u003d a p · log a b, from which, by definition of the logarithm, we conclude that Log A B p \u003d p · log a b.

It remains to prove this property for negative b. Here we notice that the expression of the log a b p with a negative B makes sense only at even degree P (since the value of the degree b should be greater than zero, otherwise the logarithm will not make sense), and in this case b p \u003d | b | p. Then b P \u003d | B | P \u003d (A log a | b |) p \u003d a p · log a | b |Where Log A B P \u003d P · Log A | B | .

For example, and ln (-3) 4 \u003d 4 · ln | -3 | \u003d 4 · ln3.

From the previous property flows root logarithm property: the logarithm of the root of n-degree is equal to the product of the fraction 1 / N on the logarithm of the feeding expression, that is, where a\u003e 0, a ≠ 1, n - natural numberMore units, b\u003e 0.

The proof is based on equality (see), which is valid for any positive b, and the logarithm property: .

Here is an example of using this property: .

Now prove the formula for the transition to the new base of logarithm View . To do this, it is enough to prove the validity of the equality Log C B \u003d Log A B · Log C a. The main logarithmic identity allows us the number B to represent as a Log A B, then log c b \u003d log c a b. It remains to take advantage of the property of the logarithm: lOG C A LOG A B \u003d Log A B · Log C A. So proved the equality of log c b \u003d log a b · log c a, and therefore the formula for the transition to the new base of the logarithm is also proved.

Let's show a couple of examples of applying this property of logarithms: and .

The transition formula to a new base allows you to move to work with logarithms that have a "convenient" base. For example, using it, you can go to the natural or decimal logarithms so that you can calculate the logarithm value along the logarithm table. The transition formula to the new base of the logarithm also allows in some cases to find the value of this logarithm, when the values \u200b\u200bof some logarithms with other bases are known.

Used frequently private case Formulas for the transition to a new base of logarithm at C \u003d B of the species . It can be seen that Log A B and Log B A. For instance, .

Also often used formula which is convenient when finding logarithms. To confirm your words, we show how it is calculated by the value of the logarithm of the view. Have . To prove the formula It suffices to take advantage of the transition to a new base of logarithm A: .

It remains to prove the properties of the comparison of logarithms.

We prove that for any positive numbers B 1 and B 2, B 1 log A B 2, and at a\u003e 1 - inequality log a b 1

Finally, it remains to prove the last of the listed properties of logarithms. We restrict ourselves to the proof of its first part, that is, we prove that if a 1\u003e 1, a 2\u003e 1 and a 1 1 Fair Log A 1 B\u003e Log A 2 b. The remaining statements of this property of logarithms are proved by a similar principle.

We use the method from the opposite. Suppose that at a 1\u003e 1, a 2\u003e 1 and a 1 1 Fair Log A 1 B≤Log A 2 B. According to the properties of logarithms, these inequalities can rewrite as and Accordingly, it follows that Log B A 1 ≤Log B A 2 and Log B A 1 ≥Log B A 2, respectively. Then, according to the properties of degrees with the same bases, equality B log B A 1 ≥B log b a 2 and b log b A 1 ≥B log b a 2, that is, A 1 ≥A 2. So we came to contradiction condition A 1

Bibliography.

Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. et al. Algebra and start analysis: a textbook for 10 - 11 classes of general educational institutions.

Gusev V.A., Mordkovich A.G. Mathematics (allowance for applicants to technical schools).

Logarithm number N. Based on but called an indicator of degree h. in which you need to build but to get a number N.
Provided that
,
,
From the definition of logarithm it follows that
.
- This equality is the main logarithmic identity.
Logarithms based on 10 are called decimal logarithms. Instead
write
.
Logarithmia based on e. called natural and designated
.
The main properties of logarithms.
Logarithm units for any base is zero

The logarithm of the work is equal to the sum of the logarithms of the factors.

3) the logarithm of the private is equal to the difference of logarithms

Factor
called the transition module from logarithms at the base a. to logarithms at the base b. .
Using properties 2-5, it is often possible to reduce the logarithm of a complex expression to the result of simple arithmetic action over logarithms.
For example,
Such transformations of logarithm are called logarithming. Converts inverse logarithming are called potentiation.
Chapter 2. Elements of higher mathematics.
1. Limits
Limit function
is a finite number A, if with the desire xX 0 for each one's defined
there is such a number
that as soon as
T.
.
The function having a limit differs from it to an infinitely low value:
where--- B.M.V., i.e.
.
Example. Consider a function
.
With the desire
function y. She strives for zero:
1.1. The main theorems are about limits.
The constant value limit is equal to this constant value.

.
The limit of the amount (difference) of the final number of functions is equal to the sum of the limits of these functions.

The limit of the finite number of functions is equal to the product of these functions.

The limit of the private two functions is equal to the private limits of these functions, if the limit of the denominator is not zero.

Wonderful limits
,
where
1.2. Examples of calculation limits
However, not all limits are calculated so simple. More often the calculation of the limit is reduced to the disclosure of the uncertainty of the type: or .
.
2. Derivative function
Let we have a function
continuous on segment
.
Argument received some increment
. Then the function will receive increment
.
The meaning of the argument corresponds to the value of the function
.
The meaning of the argument
matches the value of the function.
Hence, .
We will find the limit of this relationship when
. If this limit exists, it is called the derivative of this function.
Definition of the 3-Production This feature
by argument it is called the limit of the relationship of the function of the function to the increment of the argument, when the increment of the argument arbitrarily tends to zero.
Derived function
it can be indicated as follows:
; ; ; .
Determination 4 Operation of finding a derivative of a function called differentiation.
2.1. Mechanical sense derivative.
Consider the straightforward movement of some solid or material point.
Let at some point in time moving point
was at a distance from the initial position
.
After some time
she moved to distance
. Attitude =- average material of the material point
. We find the limit of this relationship, given that
.
Consequently, the definition of the instantaneous velocity of the material point is reduced to finding a derivative from the time.
2.2. The geometric value of the derivative
Let us have a graphically given some function
.
Fig. 1. Geometric meaning derivative
If a
, then point
will move around the curve, approaching the point
.
Hence
. The value of the derivative with this value of the argument it is numerically equal to the tangent of an angle of educated tangent at this point with a positive axis direction.
.
2.3. Table of basic differentiation formulas.
Power function

Exponential function

Logarithmic function

Trigonometric function

Reverse trigonometric function

2.4. Differentiation rules.
Derived from

Derived amount (difference) of functions

Derivative work of two functions

Derivative of private two functions

2.5. Derived from complex function.
Let the function be given
such that it can be represented as
and
where the variable is an intermediate argument then
The derivative of the complex function is equal to the product of the derivative of this function by the intermediate argument on the derivative of the intermediate argument by x.
Example1.
Example2.
3. Differential function.
Let it be
differentiable on some segment
let it go w. this function is derived
,
then you can record
(1),
where - Infinitely small value,
since when
Multiplying all members of equality (1) on
we have:
Where
- B.M.V. top order.
Value
called differential function
and denotes
.
3.1. The geometric value of the differential.
Let the function be given
.
Fig.2. Geometric meaning of differential.
.
Obviously, differential function
it is equal to the increment of the ordinate tangent at this point.
3.2. Derivatives and differentials of various orders.
If there
, then
called the first derivative.
The derivative of the first derivative is called a second order derivative and recorded
.
N-th order derivative from function
the derivative (N-1) is called the order and records:
.
Differential from differential function is called the second differential or second-order differential.
.

.
3.3 Solving biological problems with the use of differentiation.
Task1. Studies have shown that the growth of the colony of microorganisms is subject to the law
where N. - the number of microorganisms (in thousands), t. - Large (days).
b) Will there be an increase in or decreased during this period?
Answer. The number of colony will increase.
Task 2. Water in the lake is periodically tested to control the content of pathogenic bacteria. Through t. days after testing The concentration of bacteria is determined by the ratio
.
When will the lake come in the lake a minimum concentration of bacteria and can I swim in it?
Definction reaches MAX or MIN, when its derivative is zero.
,
We define MAX or MIN will be after 6 days. To do this, take the second derivative.

Answer: After 6 days there will be a minimum concentration of bacteria.

Check if there are no negative numbers or a unit under the logarithm sign. This method is applicable to the expressions of the form. Log B \u2061 (x) log b \u2061 (a) (\\ displayStyle (\\ FRAC (\\ log _ (b) (x)) (\\ log _ (b) (a)))). However, it is not suitable for some special occasions:
The logarithm of a negative number is not defined at any base (for example, Log \u2061 (- 3) (\\ DisplayStyle \\ log (-3)) or Log 4 \u2061 (- 5) (\\ DisplayStyle \\ Log _ (4) (- 5))). In this case, write "no solution."

Logarithm zero on any reason is also not defined. If you caught ln \u2061 (0) (\\ displaystyle \\ ln (0)), write down "no solution."

Logarithm units for any reason ( LOG \u2061 (1) (\\ DisplayStyle \\ log (1))) always equal to zero because x 0 \u003d 1 (\\ displaystyle x ^ (0) \u003d 1) For all values x.. Write down instead of such a logarithm 1 and do not use the method below.

If logarithms have different bases, for example L O G 3 (x) L O G 4 (A) (\\ DisplayStyle (\\ FRAC (Log_ (3) (x)) (log_ (4) (a)))), And do not reduce to integer numbers, the expression value cannot be found manually.

Convert an expression to one logarithm. If the expression does not apply to the above particular cases, it can be represented as a single logarithm. Use for this the following formula: log b \u2061 (x) log b \u2061 (a) \u003d log a \u2061 (x) (\\ displayStyle (\\ FRAC (\\ log _ (b) (x)) (\\ log _ (b) (a))) \u003d \\ Example 1: Consider the expression.
Log \u2061 16 log \u2061 2 (\\ DisplayStyle (\\ FRAC (\\ log (16)) (\\ log (2)))) To begin with, we will submit an expression in the form of one logarithm with the help of the above formula:.
log \u2061 16 log \u2061 2 \u003d log 2 \u2061 (16) (\\ DisplayStyle (\\ FRAC (\\ LOG (16)) (\\ log (2))) \u003d \\ log _ (2) (16)) This formula "base replacement" logarithm is derived from the main properties of logarithms..

If possible, calculate the value of the expression manually.

To find log a \u2061 (x) (\\ displaystyle \\ log _ (a) (x)) , imagine an expression "A? \u003d x (\\ displaystyle a ^ (?) \u003d x) ", that is, ask the following question:" What degree you need to build, To obtain a.? ". To answer this question, you may need a calculator, but if you are lucky, you can find it manually. x.Example 1 (continued): Rewrite as
2? \u003d 16 (\\ displaystyle 2 ^ (?) \u003d 16) . It is necessary to find which number should be instead of the sign "?". This can be done by samples and errors:2 2 \u003d 2 * 2 \u003d 4 (\\ displaystyle 2 ^ (2) \u003d 2 * 2 \u003d 4)
2 3 \u003d 4 * 2 \u003d 8 (\\ displaystyle 2 ^ (3) \u003d 4 * 2 \u003d 8)
2 4 \u003d 8 * 2 \u003d 16 (\\ displaystyle 2 ^ (4) \u003d 8 * 2 \u003d 16)
So, the desired number is 4:
Log 2 \u2061 (16) (\\ DisplayStyle \\ log _ (2) (16)) Leave the answer in the logarithmic form if you cannot simplify it. = 4 .

Many logarithms are very difficult to calculate manually. In this case, to get an accurate answer, you will need a calculator. However, if you decide the task in the lesson, the teacher is likely to satisfy the answer in the logarithmic form. Below, the method under consideration is used to solve a more complex example: Example 2: What is equal
log 3 \u2061 (58) log 3 \u2061 (7) (\\ DisplayStyle (\\ FRAC (\\ Log _ (3) (58)) (\\ log _ (3) (7)))) We transform this expression into one logarithm:?

log 3 \u2061 (58) log 3 \u2061 (7) \u003d log 7 \u2061 (58) (\\ DisplayStyle (\\ FRAC (\\ Log _ (3) (58)) (\\ log _ (3) (7)) \u003d \\ . Please note that the base for both logarithms 3 disappears; This is true for any reason. Rewrite an expression in the form7? \u003d 58 (\\ displayStyle 7 ^ (?) \u003d 58)

And try to find a value?: {!LANG-a1da3834ae5faba39a0df81ba57fb000!}{!LANG-e22540a34ac2edc8c7f8f1a01d9bbc66!}
7 2 \u003d 7 * 7 \u003d 49 (\\ displaystyle 7 ^ (2) \u003d 7 * 7 \u003d 49)
7 3 \u003d 49 * 7 \u003d 343 (\\ displaystyle 7 ^ (3) \u003d 49 * 7 \u003d 343)
Since 58 is between these two numbers, is not expressed in an integer.

Leave an answer in logarithmic form: log 7 \u2061 (58) (\\ DisplayStyle \\ Log _ (7) (58)).

The logarithm of the positive number B for the base A (A\u003e 0, A is not equal to 1) they call such a number with that A C \u003d B: Log A B \u003d C ⇔ A C \u003d B (A\u003e 0, A ≠ 1, B\u003e \u200b\u200b0) & NBSP & NBSP & NBSP & NBSP & NBSP & NBSP

Please note: the logarithm from an inadequate number is not defined. In addition, at the base of the logarithm should be a positive number, not equal to 1. For example, if we are erected in a square, we obtain the number 4, but this does not mean that the logarithm on the base is -2 from 4 is 2.
Basic logarithmic identity
a log a b \u003d b (a\u003e 0, a ≠ 1) (2)
It is important that the areas of determining the right and left parts of this formula are different. The left part is defined only at b\u003e 0, a\u003e 0 and a ≠ 1. The right side is defined at any b, and it does not depend on A at all. Thus, the use of the main logarithmic "identity" in solving equations and inequalities can lead to a change in the OTZ.
Two obvious consequences of the definition of logarithm
Log A A \u003d 1 (A\u003e 0, A ≠ 1) (3)
Log A 1 \u003d 0 (A\u003e 0, A ≠ 1) (4)
Indeed, when the number A is erected in the first degree, we will get the same number, and when it is erected into a zero degree.
Logarithm works and logarithm private
Log A (B C) \u003d Log A B + Log A C (A\u003e 0, A ≠ 1, B\u003e \u200b\u200b0, C\u003e 0) (5)
Log a b c \u003d log a b - log a c (a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0) (6)
I would like to warn schoolchildren from thoughtless application of these formulas in solving logarithmic equations and inequalities. When using them, "from left to right" there is a narrowing of OTZ, and in the transition from the amount or difference of logarithms to the logarithm of the work or private - the expansion of OTZ.

Indeed, the expression Log A (F (X) G (X)) is defined in two cases: when both functions are strictly positive or when f (x) and g (x) are less than zero.

Converting this expression in the amount of log a f (x) + Log A G (x), we are forced to limitate only by the case when f (x)\u003e 0 and g (x)\u003e 0. There is a narrowing area of \u200b\u200bpermissible values, and this is categorically unacceptable, since it can lead to loss of decisions. A similar problem exists for formula (6).
The degree can be made for the logarithm sign
Log A B P \u003d P Log A B (A\u003e 0, A ≠ 1, b\u003e 0) (7)
And again I would like to call for accuracy. Consider the following example:
Log A (F (X) 2 \u003d 2 Log A F (X)
The left part of equality is determined, obviously, with all values \u200b\u200bof F (x), except for zero. Right part - only at F (X)\u003e 0! After making a degree from the logarithm, we suvain the OTZ. The reverse procedure leads to expanding the area of \u200b\u200bpermissible values. All these comments refer not only to degree 2, but also to any even degree.
Formula of the transition to a new base
Log a B \u003d log c b log c a (a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0, c ≠ 1) (8)
The rare case when OTZ does not change when converting. If you wisely chose the base with (positive and not equal to 1), the transition formula to a new base is absolutely safe.

If as a new base with choose the number B, we get an important special case of formula (8):
Log A B \u003d 1 Log B A (A\u003e 0, A ≠ 1, B\u003e \u200b\u200b0, B ≠ 1) (9)
Some simple examples with logarithms

Example 1. Calculate: LG2 + LG50.
Decision. LG2 + LG50 \u003d LG100 \u003d 2. We used the formula sum of logarithms (5) and the determination of the decimal logarithm.

Example 2. Calculate: LG125 / LG5.
Decision. LG125 / LG5 \u003d log 5 125 \u003d 3. We used the transition to a new base (8).
Table formulas related to logarithms
A log a b \u003d b (a\u003e 0, a ≠ 1)
Log A A \u003d 1 (A\u003e 0, A ≠ 1)
Log A 1 \u003d 0 (A\u003e 0, A ≠ 1)
log a (b c) \u003d log a b + log a c (a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0)
Log A B C \u003d Log A B - Log A C (A\u003e 0, A ≠ 1, B\u003e \u200b\u200b0, C\u003e 0)
Log A B P \u003d P Log A B (A\u003e 0, A ≠ 1, b\u003e 0)
Log a B \u003d log c b log c a (a\u003e 0, a ≠ 1, b\u003e 0, c\u003e 0, c ≠ 1)
Log A B \u003d 1 Log B A (A\u003e 0, A ≠ 1, b\u003e 0, b ≠ 1)