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

These rules must be known - 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 base: 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 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 - same grounds. If the bases are different, these rules do not work!

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

log 6 4 + log 6 9.

Since the bases of logarithms are the same, 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 considered separately. But after transformations quite normal numbers turn out. Many tests are based on this fact. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.

Removing the exponent from the logarithm

Now let's complicate the task a little. What if there is a degree in the base or argument of the logarithm? 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 their 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 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 sign of the logarithm 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 according to the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12

Task. Find the value of the expression:

[Figure caption]

Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 2 4 ; 49 = 72. We have:

[Figure caption]

I think the last example needs clarification. Where have logarithms gone? Until the very last moment, we work only with the denominator. They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator have 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 is 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 bases are different? What if they are not exact powers of the same number?

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

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

[Figure caption]

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

[Figure caption]

It follows from the second formula that it is possible to interchange the base and the argument of the logarithm, but in this case the whole expression is “turned over”, i.e. the logarithm is 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 tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:

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

Note that the arguments of both logarithms are exact exponents. 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 flip the second logarithm:

[Figure caption]

Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the 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 it down and get rid of the indicators:

[Figure caption]

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

[Figure caption]

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 of the argument. Number n can be absolutely anything, because it's just the value of the logarithm.

The second formula is actually a paraphrased definition. It's called the basic logarithmic identity.

Indeed, what will happen if the number b raise to the power so that b to this extent gives a number a? That's right: this is the same number a. Read this paragraph carefully again - many people “hang” on it.

Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.

Task. Find the value of the expression:

[Figure caption]

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

[Figure caption]

If someone is not in the know, this was a real task from the exam :)

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that are difficult to call properties - rather, these are consequences from the definition of the logarithm. They are constantly found in problems and, surprisingly, create problems even for "advanced" students.

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

The focus of this article is logarithm. Here we will give the definition of the logarithm, show the accepted notation, give examples of logarithms, and talk about natural and decimal logarithms. After that, consider the basic logarithmic identity.

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Definition of logarithm

The concept of a logarithm arises when solving a problem in a certain sense inverse, when you need to find the exponent from a known value of the degree and a known base.

But enough preamble, it's time to answer the question "what is a logarithm"? Let us give an appropriate definition.

Definition.

Logarithm of b to base a, where a>0 , a≠1 and b>0 is the exponent to which you need to raise the number a to get b as a result.

At this stage, we note that the spoken word "logarithm" should immediately raise two ensuing questions: "what number" and "on what basis." In other words, there is simply no logarithm, but there is only the logarithm of a number in some base.

We will immediately introduce logarithm notation: the logarithm of the number b to the base a is usually denoted as log a b . The logarithm of the number b to the base e and the logarithm to the base 10 have their own special designations lnb and lgb respectively, that is, they write not log e b , but lnb , and not log 10 b , but lgb .

Now you can bring: .
And the records do not make sense, since in the first of them there is a negative number under the sign of the logarithm, in the second - a negative number in the base, and in the third - both a negative number under the sign of the logarithm and a unit in the base.

Now let's talk about rules for reading logarithms. The entry log a b is read as "logarithm of b to base a". For example, log 2 3 is the logarithm of three to base 2, and is the logarithm of two integer two base thirds of the square root of five. The logarithm to base e is called natural logarithm, and the notation lnb is read as "the natural logarithm of b". For example, ln7 is the natural logarithm of seven, and we will read it as the natural logarithm of pi. The logarithm to base 10 also has a special name - decimal logarithm, and the notation lgb is read as "decimal logarithm b". For example, lg1 is the decimal logarithm of one, and lg2.75 is the decimal logarithm of two point seventy-five hundredths.

It is worth dwelling separately on the conditions a>0, a≠1 and b>0, under which the definition of the logarithm is given. Let us explain where these restrictions come from. To do this, we will be helped by an equality of the form, called , which directly follows from the definition of the logarithm given above.

Let's start with a≠1 . Since one is equal to one to any power, the equality can only be true for b=1, but log 1 1 can be any real number. To avoid this ambiguity, a≠1 is accepted.

Let us substantiate the expediency of the condition a>0 . With a=0, by the definition of the logarithm, we would have equality , which is possible only with b=0 . But then log 0 0 can be any non-zero real number, since zero to any non-zero power is zero. This ambiguity can be avoided by the condition a≠0 . And for a<0 нам бы пришлось отказаться от рассмотрения рациональных и иррациональных значений логарифма, так как степень с рациональным и иррациональным показателем определена лишь для неотрицательных оснований. Поэтому и принимается условие a>0 .

Finally, the condition b>0 follows from the inequality a>0 , since , and the value of the degree with a positive base a is always positive.

In conclusion of this paragraph, we say that the voiced definition of the logarithm allows you to immediately indicate the value of the logarithm when the number under the sign of the logarithm is a certain degree of base. Indeed, the definition of the logarithm allows us to assert that if b=a p , then the logarithm of the number b to the base a is equal to p . That is, the equality log a a p =p is true. For example, we know that 2 3 =8 , then log 2 8=3 . We will talk more about this in the article.

One of the elements of primitive level algebra is the logarithm. The name comes from the Greek language from the word “number” or “degree” and means the degree to which it is necessary to raise the number at the base in order to find the final number.

Types of logarithms

• log a b is the logarithm of the number b to the base a (a > 0, a ≠ 1, b > 0);
• lg b - decimal logarithm (logarithm base 10, a = 10);
• ln b - natural logarithm (logarithm base e, a = e).

How to solve logarithms?

The logarithm of the number b to the base a is an exponent, which requires that the base a be raised to the number b. The result is pronounced like this: “logarithm of b to the base of a”. The solution to logarithmic problems is that you need to determine the given degree by the numbers by the specified numbers. There are some basic rules for determining or solving the logarithm, as well as transforming the notation itself. Using them, logarithmic equations are solved, derivatives are found, integrals are solved, and many other operations are carried out. Basically, the solution to the logarithm itself is its simplified notation. Below are the main formulas and properties:

For any a ; a > 0; a ≠ 1 and for any x ; y > 0.

• a log a b = b is the basic logarithmic identity
• log a 1 = 0
• log a a = 1
• log a (x y ) = log a x + log a y
• log a x/ y = log a x – log a y
• log a 1/x = -log a x
• log a x p = p log a x
• log a k x = 1/k log a x , for k ≠ 0
• log a x = log a c x c
• log a x \u003d log b x / log b a - formula for the transition to a new base
• log a x = 1/log x a

How to solve logarithms - step by step instructions for solving

• First, write down the required equation.

Please note: if the base logarithm is 10, then the record is shortened, a decimal logarithm is obtained. If there is a natural number e, then we write down, reducing to a natural logarithm. It means that the result of all logarithms is the power to which the base number is raised to obtain the number b.

Directly, the solution lies in the calculation of this degree. Before solving an expression with a logarithm, it must be simplified according to the rule, that is, using formulas. You can find the main identities by going back a little in the article.

When adding and subtracting logarithms with two different numbers but with the same base, replace with a single logarithm with the product or division of the numbers b and c, respectively. In this case, you can apply the transition formula to another base (see above).

If you are using expressions to simplify the logarithm, there are some limitations to be aware of. And that is: the base of the logarithm a is only a positive number, but not equal to one. The number b, like a, must be greater than zero.

There are cases when, having simplified the expression, you will not be able to calculate the logarithm in numerical form. It happens that such an expression does not make sense, because many degrees are irrational numbers. Under this condition, leave the power of the number as a logarithm.