Range of acceptable values ​​(APV) of the logarithm

Now let's talk about restrictions (ODZ - the range of permissible values ​​of variables).

We remember that, for example, the square root cannot be taken from negative numbers; or if we have a fraction, then the denominator cannot be equal to zero. Logarithms have similar limitations:

That is, both the argument and the base must be greater than zero, but the base cannot yet be equal.

Why is that?

Let's start with a simple thing: let's say that. Then, for example, the number does not exist, since no matter what power we raise to, it always turns out. Moreover, it does not exist for anyone. But at the same time it can be equal to anything (for the same reason - equal to any degree). Therefore, the object is of no interest, and it was simply thrown out of mathematics.

We have a similar problem in the case: to any positive power it is, but it cannot be raised to a negative power at all, since this will result in division by zero (let me remind you that).

When we are faced with the problem of raising to a fractional power (which is represented as a root: . For example, (that is), but it does not exist.

Therefore, it is easier to throw away negative reasons than to tinker with them.

Well, since our base a can only be positive, then no matter what power we raise it to, we will always get a strictly positive number. So the argument must be positive. For example, it does not exist, since it will not be a negative number to any degree (or even zero, therefore it also does not exist).

In problems with logarithms, the first thing you need to do is write down the ODZ. Let me give you an example:

Let's solve the equation.

Let's remember the definition: a logarithm is the power to which the base must be raised to obtain an argument. And according to the condition, this degree is equal to: .

We get the usual quadratic equation: . Let's solve it using Vieta's theorem: the sum of the roots is equal, and the product. Easy to pick up, these are numbers and.

But if you immediately take and write both of these numbers in the answer, you can get 0 points for the problem. Why? Let's think about what happens if we substitute these roots into the initial equation?

This is clearly incorrect, since the base cannot be negative, that is, the root is “third party”.

To avoid such unpleasant pitfalls, you need to write down the ODZ even before starting to solve the equation:

Then, having received the roots and, we immediately discard the root and write the correct answer.

Example 1(try to solve it yourself) :

Find the root of the equation. If there are several roots, indicate the smallest of them in your answer.

Solution:

First of all, let’s write the ODZ:

Now let's remember what a logarithm is: to what power do you need to raise the base to get the argument? To the second. That is:

It would seem that the smaller root is equal. But this is not so: according to the ODZ, the root is extraneous, that is, it is not the root of this equation at all. Thus, the equation has only one root: .

Answer: .

Basic logarithmic identity

Let us recall the definition of logarithm in general form:

Let's substitute the logarithm into the second equality:

This equality is called basic logarithmic identity. Although in essence this is equality - just written differently definition of logarithm:

This is the power to which you must raise to get.

For example:

Solve the following examples:

Example 2.

Find the meaning of the expression.

Solution:

Let us remember the rule from the section:, that is, when raising a power to a power, the exponents are multiplied. Let's apply it:

Example 3.

Prove that.

Solution:

Properties of logarithms

Unfortunately, the tasks are not always so simple - often you first need to simplify the expression, bring it to its usual form, and only then will it be possible to calculate the value. This is easiest to do if you know properties of logarithms. So let's learn the basic properties of logarithms. I will prove each of them, because any rule is easier to remember if you know where it comes from.

All these properties must be remembered; without them, most problems with logarithms cannot be solved.

And now about all the properties of logarithms in more detail.

Property 1:

Proof:

Let it be then.

We have: , etc.

Property 2: Sum of logarithms

The sum of logarithms with the same bases is equal to the logarithm of the product: .

Proof:

Let it be then. Let it be then.

Example: Find the meaning of the expression: .

Solution: .

The formula you just learned helps to simplify the sum of logarithms, not the difference, so these logarithms cannot be combined right away. But you can do the opposite - “split” the first logarithm into two: And here is the promised simplification:
.
Why is this necessary? Well, for example: what does it equal?

Now it's obvious that.

Now simplify it yourself:

Tasks:

Answers:

Property 3: Difference of logarithms:

Proof:

Everything is exactly the same as in point 2:

Let it be then.

Let it be then. We have:

The example from the previous paragraph now becomes even simpler:

A more complicated example: . Can you figure out how to solve it yourself?

Here it should be noted that we do not have a single formula about logarithms squared. This is something akin to an expression - it cannot be simplified right away.

Therefore, let’s take a break from formulas about logarithms and think about what kind of formulas we use in mathematics most often? Since 7th grade!

This - . You need to get used to the fact that they are everywhere! They occur in exponential, trigonometric, and irrational problems. Therefore, they must be remembered.

If you look closely at the first two terms, it becomes clear that this difference of squares:

Answer to check:

Simplify it yourself.

Examples

Answers.

Property 4: Taking the exponent out of the logarithm argument:

Proof: And here we also use the definition of logarithm: let, then. We have: , etc.

This rule can be understood this way:

That is, the degree of the argument is moved ahead of the logarithm as a coefficient.

Example: Find the meaning of the expression.

Solution: .

Decide for yourself:

Examples:

Answers:

Property 5: Taking the exponent from the base of the logarithm:

Proof: Let it be then.

We have: , etc.
Remember: from grounds the degree is expressed as the opposite number, unlike the previous case!

Property 6: Removing the exponent from the base and argument of the logarithm:

Or if the degrees are the same: .

Property 7: Transition to a new base:

Proof: Let it be then.

We have: , etc.

Property 8: Swap the base and argument of the logarithm:

Proof: This is a special case of formula 7: if we substitute, we get: , etc.

Let's look at a few more examples.

Example 4.

Find the meaning of the expression.

We use property of logarithms No. 2 - the sum of logarithms with the same base is equal to the logarithm of the product:

Example 5.

Find the meaning of the expression.

Solution:

We use the property of logarithms No. 3 and No. 4:

Example 6.

Find the meaning of the expression.

Solution:

Let's use property No. 7 - move on to base 2:

Example 7.

Find the meaning of the expression.

Solution:

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On the Unified State Exam and the Unified State Exam and in life in general

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 main properties.

You definitely need to know these rules - without them, not a single serious logarithmic problem can be solved. In addition, there are very few of them - you can learn everything in one day. So let's get started.

Adding and subtracting logarithms

Consider two logarithms with the same bases: log a x and log a y. Then they can be added and subtracted, and:

1. log a x+ log a y=log a (x · y);
2. log a x− log a y=log a (x : y).

So, the sum of logarithms is equal to the logarithm of the product, and the difference is equal to 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 considered (see lesson “What is a logarithm”). Take a look at the examples and see:

Log 6 4 + log 6 9.

Since logarithms have the same bases, we use the sum formula:
log 6 4 + log 6 9 = log 6 (4 9) = log 6 36 = 2.

Task. Find the value of the expression: log 2 48 − log 2 3.

The bases are the same, we use the difference formula:
log 2 48 − log 2 3 = log 2 (48: 3) = log 2 16 = 4.

Task. Find the value of the expression: log 3 135 − log 3 5.

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

As you can see, the original expressions are made up of “bad” logarithms, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many tests are based on this fact. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.

Extracting the exponent from the logarithm

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

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

Of course, all these rules make sense if the ODZ 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 before the logarithm sign into the logarithm itself. This is what is most often required.

Task. Find the value of the expression: log 7 49 6 .

Let's get rid of the degree in the argument using the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12

Task. Find the meaning of the expression:

[Caption for the picture]

Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 2 4 ; 49 = 7 2. We have:

I think the last example requires some clarification. Where have logarithms gone? Until the very last moment we work only with the denominator. We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator contain 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 is what was done. The result was the answer: 2.

Transition to a new foundation

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

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

Let the logarithm log be given a x. Then for any number c such that c> 0 and c≠ 1, the equality is true:

[Caption for the picture]

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

[Caption for the picture]

From the second formula it follows that the base and argument of the logarithm can be swapped, but in this case the entire expression is “turned over”, i.e. the logarithm appears in the denominator.

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

However, there are problems that cannot be solved at all except by moving to a new foundation. Let's look at a couple of these:

Task. Find the value of the expression: log 5 16 log 2 25.

Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2; log 2 25 = log 2 5 2 = 2log 2 5;

Now let’s “reverse” the second logarithm:

Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with logarithms.

Task. Find the value of the expression: log 9 100 lg 3.

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

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

Basic logarithmic identity

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

In the first case, the number n becomes an indicator of the degree standing in the argument. Number n can be absolutely anything, because it’s just a logarithm value.

The second formula is actually a paraphrased definition. That’s what it’s called: the basic logarithmic identity.

In fact, what will happen if the number b raise to such a power that the number b to this power gives the number a? That's right: you get this same number a. Read this paragraph carefully again - many people get stuck on it.

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

Task. Find the meaning of the expression:

[Caption for the picture]

Note that log 25 64 = log 5 8 - simply took the square from the base and argument of the logarithm. Taking into account the rules for multiplying powers with the same base, we get:

If anyone doesn't know, this was a real task from the Unified State Exam :)

Logarithmic unit and logarithmic zero

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

1. log a a= 1 is a logarithmic unit. Remember once and for all: logarithm to any base a from this very base is equal to one.
2. log a 1 = 0 is logarithmic zero. Base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a 0 = 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.

We will now look at converting expressions containing logarithms from a general perspective. Here we will examine not only the transformation of expressions using the properties of logarithms, but also consider the transformation of expressions with general logarithms, which contain not only logarithms, but also powers, fractions, roots, etc. As usual, we will provide all the material with typical examples with detailed descriptions of solutions.

Page navigation.

Expressions with logarithms and logarithmic expressions

Doing things with fractions

In the previous paragraph, we examined the basic transformations that are carried out with individual fractions containing logarithms. These transformations, of course, can be carried out with each individual fraction that is part of a more complex expression, for example, representing the sum, difference, product and quotient of similar fractions. But in addition to working with individual fractions, transforming expressions of this type often involves performing corresponding operations with fractions. Next we will look at the rules by which these actions are carried out.

Since the 5th-6th grades we have known the rules by which they are carried out. In the article a general look at operations with fractions we have extended these rules with ordinary fractions on a general fraction A/B, where A and B are some numeric, literal expressions or expressions with variables, and B is not identically equal to zero. It is clear that fractions with logarithms are special cases of general fractions. And in this regard, it is clear that operations with fractions that contain logarithms in their notations are carried out according to the same rules. Namely:

• To add or subtract two fractions with the same denominators, you must add or subtract the numerators accordingly, but leave the denominator the same.
• To add or subtract two fractions with different denominators, you need to bring them to a common denominator and perform the appropriate actions according to the previous rule.
• To multiply two fractions, you need to write a fraction whose numerator is the product of the numerators of the original fractions, and the denominator is the product of the denominators.
• To divide a fraction into a fraction, you need to multiply the fraction being divided by the fraction that is the inverse of the divisor, that is, by a fraction with the numerator and denominator swapped.

Here are some examples of how to perform operations with fractions containing logarithms.

Example.

Perform operations with fractions containing logarithms: a) , b) , V) , G) .

Solution.

a) The denominators of the fractions being added are obviously the same. Therefore, according to the rule for adding fractions with the same denominators, we add the numerators, and leave the denominator the same: .

b) Here the denominators are different. Therefore, first you need convert fractions to the same denominator. In our case, the denominators are already presented in the form of products, and all we have to do is take the denominator of the first fraction and add to it the missing factors from the denominator of the second fraction. This way we get a common denominator of the form . In this case, the subtracted fractions are brought to a common denominator using additional factors in the form of a logarithm and the expression x 2 ·(x+1), respectively. After this, all that remains is to subtract fractions with the same denominators, which is not difficult.

So the solution is:

c) It is known that the result of multiplying fractions is a fraction, the numerator of which is the product of the numerators, and the denominator is the product of the denominators, therefore

It’s easy to see that you can reducing a fraction by two and by the decimal logarithm, as a result we have .

d) We move from dividing fractions to multiplication, replacing the divisor fraction with its inverse fraction. So

The numerator of the resulting fraction can be represented as , from which the common factor of the numerator and denominator is clearly visible - factor x, you can reduce the fraction by it:

Answer:

a) , b) , V) , G) .

It should be remembered that operations with fractions are carried out taking into account the order in which the actions are performed: first, multiplication and division, then addition and subtraction, and if there are parentheses, then the actions in parentheses are carried out first.

Example.

Do things with fractions .

Solution.

First, we add the fractions in brackets, after which we will multiply:

Answer:

At this point, it remains to say out loud three rather obvious, but at the same time important points:

Converting Expressions Using Properties of Logarithms

Most often, transforming expressions with logarithms involves the use of identities expressing the definition of logarithm and. For example, turning to the main logarithmic identity a log a b =b , a>0 , a≠1 , b>0 , we can represent the expression x−5 log 5 7 as x−7 , and the formula for moving to a new logarithmic base , where a>0, a≠1, b>0, c>0, c≠1 makes it possible to move from the expression to the difference 1−lnx.

Application of the properties of roots, powers, trigonometric identities, etc.

Expressions with logarithms, in addition to the logarithms themselves, almost always contain powers, roots, trigonometric functions, etc. It is clear that to transform such expressions, along with the properties of logarithms, the properties of powers, roots, etc. may be required. We separately examined the application of each block of properties to the transformation of expressions; links to relevant articles can be found in the section of the website www.site expressions and their transformation. Here we will show the solution to a couple of examples on the use of properties in conjunction with logarithms.

Example.

Simplify an expression .

Solution.

First, let's transform expressions with roots. On the ODZ of the variable x for the original expression (which in our case is the set of positive real numbers) from roots you can go to powers with fractional exponents, and then use the property of multiplying powers with the same bases: . Thus,

Now we represent the numerator as (what the property of a degree to a degree allows us to do, if necessary, see the transformation of expressions using the properties of degrees, as well as the representation of a number, which allows us to replace the sum of the squares of the sine and cosine of the same argument with one. This way we get one under the sign of the logarithm. A, As you know, the logarithm of unity is zero.

Let's write down the transformations made:

Zero cubed is zero, so let's move on to the expression .

A fraction whose numerator is zero and whose denominator is different from zero (in our case this is indeed the case, because it is easy to justify that the value of the expression under the sign of the natural logarithm is different from one) is equal to zero. Thus,

Further transformations are carried out based on the definition of an odd root of a negative number: .

Since 2 15 is a positive number, we can apply the properties of roots that lead to the final result: .

Answer:

Problem B7 gives some expression that needs to be simplified. The result should be a regular number that can be written down on your answer sheet. All expressions are conventionally divided into three types:

1. Logarithmic,
2. Indicative,
3. Combined.

Exponential and logarithmic expressions in pure form practically never occur. However, knowing how they are calculated is absolutely necessary.

In general, problem B7 is solved quite simply and is quite within the capabilities of the average graduate. The lack of clear algorithms is compensated for by its standardization and monotony. You can learn to solve such problems simply through a lot of training.

Logarithmic Expressions

The vast majority of B7 problems involve logarithms in one form or another. This topic is traditionally considered difficult, since its study usually occurs in the 11th grade - the era of mass preparation for final exams. As a result, many graduates have a very vague understanding of logarithms.

But in this task no one requires deep theoretical knowledge. We will encounter only the simplest expressions that require simple reasoning and can easily be mastered independently. Below are the basic formulas you need to know to cope with logarithms:

In addition, you must be able to replace roots and fractions with powers with a rational exponent, otherwise in some expressions there will simply be nothing to take out from under the logarithm sign. Replacement formulas:

Task. Find the meaning of expressions:
log 6 270 − log 6 7.5
log 5 775 − log 5 6.2

The first two expressions are converted as the difference of logarithms:
log 6 270 − log 6 7.5 = log 6 (270: 7.5) = log 6 36 = 2;
log 5 775 − log 5 6.2 = log 5 (775: 6.2) = log 5 125 = 3.

To calculate the third expression, you will have to isolate powers - both in the base and in the argument. First, let's find the internal logarithm:

Then - external:

Constructions of the form log a log b x seem complex and misunderstood to many. Meanwhile, this is just a logarithm of the logarithm, i.e. log a (log b x ). First, the internal logarithm is calculated (put log b x = c), and then the external one: log a c.

Demonstrative Expressions

We will call an exponential expression any construction of the form a k, where the numbers a and k are arbitrary constants, and a > 0. Methods for working with such expressions are quite simple and are discussed in 8th grade algebra lessons.

Below are the basic formulas that you definitely need to know. The application of these formulas in practice, as a rule, does not cause problems.

1. a n · a m = a n + m ;
2. a n / a m = a n − m ;
3. (a n ) m = a n · m ;
4. (a · b ) n = a n · b n ;
5. (a : b ) n = a n : b n .

If you come across a complex expression with powers, and it is not clear how to approach it, use a universal technique - decomposition into simple factors. As a result, large numbers in the bases of powers are replaced by simple and understandable elements. Then all that remains is to apply the above formulas - and the problem will be solved.

Task. Find the values ​​of the expressions: 7 9 · 3 11: 21 8, 24 7: 3 6: 16 5, 30 6: 6 5: 25 2.

Solution. Let's decompose all the bases of powers into simple factors:
7 9 3 11: 21 8 = 7 9 3 11: (7 3) 8 = 7 9 3 11: (7 8 3 8) = 7 9 3 11: 7 8: 3 8 = 7 3 3 = 189.
24 7: 3 6: 16 5 = (3 2 3) 7: 3 6: (2 4) 5 = 3 7 2 21: 3 6: 2 20 = 3 2 = 6.
30 6: 6 5: 25 2 = (5 3 2) 6: (3 2) 5: (5 2) 2 = 5 6 3 6 2 6: 3 5: 2 5: 5 4 = 5 2 3 2 = 150.

Combined tasks

If you know the formulas, then all exponential and logarithmic expressions can be solved literally in one line. However, in Problem B7 powers and logarithms can be combined to form quite strong combinations.