Follows from its definition. And so the logarithm of the number b based on A is defined as the exponent to which a number must be raised a to get the number b(logarithm exists only for positive numbers).

From this formulation it follows that the calculation x=log a b, is equivalent to solving the equation a x =b. For example, log 2 8 = 3 because 8 = 2 3 . The formulation of the logarithm makes it possible to justify that if b=a c, then the logarithm of the number b based on a equals With. It is also clear that the topic of logarithms is closely related to the topic of powers of a number.

With logarithms, as with any numbers, you can do operations of addition, subtraction and transform in every possible way. But due to the fact that logarithms are not entirely ordinary numbers, their own special rules apply here, which are called main properties.

Adding and subtracting logarithms.

Let's take two logarithms with the same bases: log a x And log a y. Then it is possible to perform addition and subtraction operations:

log a x+ log a y= log a (x·y);

log a x - log a y = log a (x:y).

log a(x 1 . x 2 . x 3 ... x k) = log a x 1 + log a x 2 + log a x 3 + ... + log a x k.

From logarithm quotient theorem One more property of the logarithm can be obtained. It is common knowledge that log a 1= 0, therefore

log a 1 /b=log a 1 - log a b= -log a b.

This means there is an equality:

log a 1 / b = - log a b.

Logarithms of two reciprocal numbers for the same reason will differ from each other solely by sign. So:

Log 3 9= - log 3 1 / 9 ; log 5 1 / 125 = -log 5 125.


We continue to study logarithms. In this article we will talk about calculating logarithms, this process is called logarithm. First we will understand the calculation of logarithms by definition. Next, let's look at how the values ​​of logarithms are found using their properties. After this, we will focus on calculating logarithms through the initially specified values ​​of other logarithms. Finally, let's learn how to use logarithm tables. The entire theory is provided with examples with detailed solutions.

Page navigation.

Calculating logarithms by definition

In the simplest cases it is possible to perform quite quickly and easily finding the logarithm by definition. Let's take a closer look at how this process happens.

Its essence is to represent the number b in the form a c, from which, by the definition of a logarithm, the number c is the value of the logarithm. That is, by definition, the following chain of equalities corresponds to finding the logarithm: log a b=log a a c =c.

So, calculating a logarithm by definition comes down to finding a number c such that a c = b, and the number c itself is the desired value of the logarithm.

Taking into account the information in the previous paragraphs, when the number under the logarithm sign is given by a certain power of the logarithm base, you can immediately indicate what the logarithm is equal to - it is equal to the exponent. Let's show solutions to examples.

Example.

Find log 2 2 −3, and also calculate the natural logarithm of the number e 5,3.

Solution.

The definition of the logarithm allows us to immediately say that log 2 2 −3 =−3. Indeed, the number under the logarithm sign is equal to base 2 to the −3 power.

Similarly, we find the second logarithm: lne 5.3 =5.3.

Answer:

log 2 2 −3 =−3 and lne 5,3 =5,3.

If the number b under the logarithm sign is not specified as a power of the base of the logarithm, then you need to carefully look to see if it is possible to come up with a representation of the number b in the form a c . Often this representation is quite obvious, especially when the number under the logarithm sign is equal to the base to the power of 1, or 2, or 3, ...

Example.

Calculate the logarithms log 5 25 , and .

Solution.

It is easy to see that 25=5 2, this allows you to calculate the first logarithm: log 5 25=log 5 5 2 =2.

Let's move on to calculating the second logarithm. The number can be represented as a power of 7: (see if necessary). Hence, .

Let's rewrite the third logarithm in the following form. Now you can see that , from which we conclude that . Therefore, by the definition of logarithm .

Briefly, the solution could be written as follows: .

Answer:

log 5 25=2 , and .

When under the sign of the logarithm there is a sufficiently large natural number, then it wouldn’t hurt to decompose it into prime factors. It often helps to represent such a number as some power of the base of the logarithm, and therefore calculate this logarithm by definition.

Example.

Find the value of the logarithm.

Solution.

Some properties of logarithms allow you to immediately specify the value of logarithms. These properties include the property of the logarithm of one and the property of the logarithm of a number equal to the base: log 1 1=log a a 0 =0 and log a a=log a a 1 =1. That is, when under the sign of the logarithm there is a number 1 or a number a equal to the base of the logarithm, then in these cases the logarithms are equal to 0 and 1, respectively.

Example.

What are logarithms and log10 equal to?

Solution.

Since , then from the definition of logarithm it follows .

In the second example, the number 10 under the logarithm sign coincides with its base, so the decimal logarithm of ten is equal to one, that is, lg10=lg10 1 =1.

Answer:

AND lg10=1 .

Note that the calculation of logarithms by definition (which we discussed in the previous paragraph) implies the use of the equality log a a p =p, which is one of the properties of logarithms.

In practice, when a number under the logarithm sign and the base of the logarithm are easily represented as a power of a certain number, it is very convenient to use the formula , which corresponds to one of the properties of logarithms. Let's look at an example of finding a logarithm that illustrates the use of this formula.

Example.

Calculate the logarithm.

Solution.

Answer:

.

Properties of logarithms not mentioned above are also used in calculations, but we will talk about this in the following paragraphs.

Finding logarithms through other known logarithms

The information in this paragraph continues the topic of using the properties of logarithms when calculating them. But here the main difference is that the properties of logarithms are used to express the original logarithm in terms of another logarithm, the value of which is known. Let's give an example for clarification. Let's say we know that log 2 3≈1.584963, then we can find, for example, log 2 6 by doing a little transformation using the properties of the logarithm: log 2 6=log 2 (2 3)=log 2 2+log 2 3≈ 1+1,584963=2,584963 .

In the above example, it was enough for us to use the property of the logarithm of a product. However, much more often it is necessary to use a wider arsenal of properties of logarithms in order to calculate the original logarithm through the given ones.

Example.

Calculate the logarithm of 27 to base 60 if you know that log 60 2=a and log 60 5=b.

Solution.

So we need to find log 60 27 . It is easy to see that 27 = 3 3, and the original logarithm, due to the property of the logarithm of the power, can be rewritten as 3·log 60 3.

Now let's see how to express log 60 3 in terms of known logarithms. The property of the logarithm of a number equal to the base allows us to write the equality log 60 60=1. On the other hand, log 60 60=log60(2 2 3 5)= log 60 2 2 +log 60 3+log 60 5= 2·log 60 2+log 60 3+log 60 5 . Thus, 2 log 60 2+log 60 3+log 60 5=1. Hence, log 60 3=1−2·log 60 2−log 60 5=1−2·a−b.

Finally, we calculate the original logarithm: log 60 27=3 log 60 3= 3·(1−2·a−b)=3−6·a−3·b.

Answer:

log 60 27=3·(1−2·a−b)=3−6·a−3·b.

Separately, it is worth mentioning the meaning of the formula for transition to a new base of the logarithm of the form. It allows you to move from logarithms with any base to logarithms with a specific base, the values ​​of which are known or it is possible to find them. Usually, from the original logarithm, using the transition formula, they move to logarithms in one of the bases 2, e or 10, since for these bases there are tables of logarithms that allow their values ​​to be calculated with a certain degree of accuracy. In the next paragraph we will show how this is done.

Logarithm tables and their uses

For approximate calculation of logarithm values ​​can be used logarithm tables. The most commonly used base 2 logarithm table is the table natural logarithms and a table of decimal logarithms. When working in the decimal number system, it is convenient to use a table of logarithms based on base ten. With its help we will learn to find the values ​​of logarithms.









The presented table allows you to find the values ​​of the decimal logarithms of numbers from 1,000 to 9,999 (with three decimal places) with an accuracy of one ten-thousandth. We will analyze the principle of finding the value of a logarithm using a table of decimal logarithms using a specific example - it’s clearer this way. Let's find log1.256.

In the left column of the table of decimal logarithms we find the first two digits of the number 1.256, that is, we find 1.2 (this number is circled in blue for clarity). The third digit of the number 1.256 (digit 5) is found in the first or last line to the left of the double line (this number is circled in red). The fourth digit of the original number 1.256 (digit 6) is found in the first or last line to the right of the double line (this number is circled with a green line). Now we find the numbers in the cells of the table of logarithms at the intersection of the marked row and marked columns (these numbers are highlighted orange). The sum of the marked numbers gives the desired value decimal logarithm accurate to the fourth decimal place, that is, log1.236≈0.0969+0.0021=0.0990.

Is it possible, using the table above, to find the values ​​of decimal logarithms of numbers that have more than three digits after the decimal point, as well as those that go beyond the range from 1 to 9.999? Yes, you can. Let's show how this is done with an example.

Let's calculate lg102.76332. First you need to write down number in standard form: 102.76332=1.0276332·10 2. After this, the mantissa should be rounded to the third decimal place, we have 1.0276332 10 2 ≈1.028 10 2, while the original decimal logarithm is approximately equal to the logarithm of the resulting number, that is, we take log102.76332≈lg1.028·10 2. Now we apply the properties of the logarithm: lg1.028·10 2 =lg1.028+lg10 2 =lg1.028+2. Finally, we find the value of the logarithm lg1.028 from the table of decimal logarithms lg1.028≈0.0086+0.0034=0.012. As a result, the entire process of calculating the logarithm looks like this: log102.76332=log1.0276332 10 2 ≈lg1.028 10 2 = log1.028+lg10 2 =log1.028+2≈0.012+2=2.012.

In conclusion, it is worth noting that using a table of decimal logarithms you can calculate the approximate value of any logarithm. To do this, it is enough to use the transition formula to go to decimal logarithms, find their values ​​in the table, and perform the remaining calculations.

For example, let's calculate log 2 3 . According to the formula for transition to a new base of the logarithm, we have . From the table of decimal logarithms we find log3≈0.4771 and log2≈0.3010. Thus, .

Bibliography.

  • Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
  • Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).

Logarithm of the number b (b > 0) to base a (a > 0, a ≠ 1)– exponent to which the number a must be raised to obtain b.

The base 10 logarithm of b can be written as log(b), and the logarithm to base e (natural logarithm) is ln(b).

Often used when solving problems with logarithms:

Properties of logarithms

There are four main properties of logarithms.

Let a > 0, a ≠ 1, x > 0 and y > 0.

Property 1. Logarithm of the product

Logarithm of the product equal to the sum of logarithms:

log a (x ⋅ y) = log a x + log a y

Property 2. Logarithm of the quotient

Logarithm of the quotient equal to the difference of logarithms:

log a (x / y) = log a x – log a y

Property 3. Logarithm of power

Logarithm of degree equal to the product of the power and the logarithm:

If the base of the logarithm is in the degree, then another formula applies:

Property 4. Logarithm of the root

This property can be obtained from the property of the logarithm of a power, since the root of the nth power equal to the power 1/n:

Formula for converting from a logarithm in one base to a logarithm in another base

This formula is also often used when solving various tasks on logarithms:

Special case:

Comparing logarithms (inequalities)

Let us have 2 functions f(x) and g(x) under logarithms with the same bases and between them there is an inequality sign:

To compare them, you need to first look at the base of the logarithms a:

  • If a > 0, then f(x) > g(x) > 0
  • If 0< a < 1, то 0 < f(x) < g(x)

How to solve problems with logarithms: examples

Problems with logarithms included in the Unified State Examination in mathematics for grade 11 in task 5 and task 7, you can find tasks with solutions on our website in the appropriate sections. Also, tasks with logarithms are found in the math task bank. You can find all examples by searching the site.

What is a logarithm

Logarithms have always been considered complex topic V school course mathematics. There are many different definitions logarithm, but for some reason most textbooks use the most complex and unsuccessful of them.

We will define the logarithm simply and clearly. To do this, let's create a table:

So, we have powers of two.

Logarithms - properties, formulas, how to solve

If you take the number from the bottom line, you can easily find the power to which you will have to raise two to get this number. For example, to get 16, you need to raise two to the fourth power. And to get 64, you need to raise two to the sixth power. This can be seen from the table.

And now - actually, the definition of the logarithm:

the base a of the argument x is the power to which the number a must be raised to obtain the number x.

Designation: log a x = b, where a is the base, x is the argument, b is what the logarithm is actually equal to.

For example, 2 3 = 8 ⇒log 2 8 = 3 (the base 2 logarithm of 8 is three because 2 3 = 8). With the same success, log 2 64 = 6, since 2 6 = 64.

The operation of finding the logarithm of a number to a given base is called. So, let's add a new line to our table:

2 1 2 2 2 3 2 4 2 5 2 6
2 4 8 16 32 64
log 2 2 = 1 log 2 4 = 2 log 2 8 = 3 log 2 16 = 4 log 2 32 = 5 log 2 64 = 6

Unfortunately, not all logarithms are calculated so easily. For example, try to find log 2 5. The number 5 is not in the table, but logic dictates that the logarithm will lie somewhere on the interval. Because 2 2< 5 < 2 3 , а чем больше степень двойки, тем больше получится число.

Such numbers are called irrational: the numbers after the decimal point can be written ad infinitum, and they are never repeated. If the logarithm turns out to be irrational, it is better to leave it that way: log 2 5, log 3 8, log 5 100.

It is important to understand that a logarithm is an expression with two variables (the base and the argument). At first, many people confuse where the basis is and where the argument is. To avoid annoying misunderstandings, just look at the picture:

Before us is nothing more than the definition of a logarithm. Remember: logarithm is a power, into which the base must be built in order to obtain an argument. It is the base that is raised to a power - it is highlighted in red in the picture. It turns out that the base is always at the bottom! I tell my students this wonderful rule at the very first lesson - and no confusion arises.

How to count logarithms

We've figured out the definition - all that's left is to learn how to count logarithms, i.e. get rid of the "log" sign. To begin with, we note that two important facts follow from the definition:

  1. The argument and the base must always be greater than zero. This follows from the definition of the degree rational indicator, to which the definition of a logarithm comes down.
  2. The base must be different from one, since one to any degree still remains one. Because of this, the question “to what power must one be raised to get two” is meaningless. There is no such degree!

Such restrictions are called range of acceptable values(ODZ). It turns out that the ODZ of the logarithm looks like this: log a x = b ⇒x > 0, a > 0, a ≠ 1.

Note that there are no restrictions on the number b (the value of the logarithm). For example, the logarithm may well be negative: log 2 0.5 = −1, because 0.5 = 2 −1.

However, now we are considering only numerical expressions, where it is not required to know the VA of the logarithm. All restrictions have already been taken into account by the authors of the tasks. But when logarithmic equations and inequalities come into play, DL requirements will become mandatory. After all, the basis and argument may contain very strong constructions that do not necessarily correspond to the above restrictions.

Now let's consider general scheme calculating logarithms. It consists of three steps:

  1. Express the base a and the argument x as a power with the minimum possible base greater than one. Along the way, it’s better to get rid of decimals;
  2. Solve the equation for variable b: x = a b ;
  3. The resulting number b will be the answer.

That's all! If the logarithm turns out to be irrational, this will be visible already in the first step. The requirement that the base be greater than one is very important: this reduces the likelihood of error and greatly simplifies the calculations. Same with decimals: if you immediately convert them to regular ones, there will be many fewer errors.

Let's see how this scheme works using specific examples:

Task. Calculate the logarithm: log 5 25

  1. Let's imagine the base and argument as a power of five: 5 = 5 1 ; 25 = 5 2 ;

    2. Let's create and solve the equation:
    log 5 25 = b ⇒(5 1) b = 5 2 ⇒5 b = 5 2 ⇒ b = 2;

  2. We received the answer: 2.

Task. Calculate the logarithm:

Task. Calculate the logarithm: log 4 64

  1. Let's imagine the base and argument as a power of two: 4 = 2 2 ; 64 = 2 6 ;
  2. Let's create and solve the equation:
    log 4 64 = b ⇒(2 2) b = 2 6 ⇒2 2b = 2 6 ⇒2b = 6 ⇒ b = 3;
  3. We received the answer: 3.

Task. Calculate the logarithm: log 16 1

  1. Let's imagine the base and argument as a power of two: 16 = 2 4 ; 1 = 2 0 ;
  2. Let's create and solve the equation:
    log 16 1 = b ⇒(2 4) b = 2 0 ⇒2 4b = 2 0 ⇒4b = 0 ⇒ b = 0;
  3. We received the answer: 0.

Task. Calculate the logarithm: log 7 14

  1. Let's imagine the base and argument as a power of seven: 7 = 7 1 ; 14 cannot be represented as a power of seven, since 7 1< 14 < 7 2 ;
  2. From the previous paragraph it follows that the logarithm does not count;
  3. The answer is no change: log 7 14.

A small note to last example. How can you be sure that a number is not an exact power of another number? It’s very simple - just factor it into prime factors. If the expansion has at least two different factors, the number is not an exact power.

Task. Find out whether the numbers are exact powers: 8; 48; 81; 35; 14.

8 = 2 · 2 · 2 = 2 3 - exact degree, because there is only one multiplier;
48 = 6 · 8 = 3 · 2 · 2 · 2 · 2 = 3 · 2 4 - is not an exact power, since there are two factors: 3 and 2;
81 = 9 · 9 = 3 · 3 · 3 · 3 = 3 4 - exact degree;
35 = 7 · 5 - again not an exact power;
14 = 7 · 2 - again not an exact degree;

Let us also note that we ourselves prime numbers are always exact degrees of themselves.

Decimal logarithm

Some logarithms are so common that they have a special name and symbol.

of the argument x is the logarithm to base 10, i.e. The power to which the number 10 must be raised to obtain the number x. Designation: lg x.

For example, log 10 = 1; lg 100 = 2; lg 1000 = 3 - etc.

From now on, when a phrase like “Find lg 0.01” appears in a textbook, know: this is not a typo. This is a decimal logarithm. However, if you are unfamiliar with this notation, you can always rewrite it:
log x = log 10 x

Everything that is true for ordinary logarithms is also true for decimal logarithms.

Natural logarithm

There is another logarithm that has its own designation. In some ways, it's even more important than decimal. We are talking about the natural logarithm.

of the argument x is the logarithm to base e, i.e. the power to which the number e must be raised to obtain the number x. Designation: ln x.

Many people will ask: what is the number e? This is an irrational number; its exact value cannot be found and written down. I will give only the first figures:
e = 2.718281828459…

We will not go into detail about what this number is and why it is needed. Just remember that e is the base of the natural logarithm:
ln x = log e x

Thus ln e = 1; ln e 2 = 2; ln e 16 = 16 - etc. On the other hand, ln 2 is an irrational number. In general, the natural logarithm of any rational number irrational. Except, of course, for one: ln 1 = 0.

For natural logarithms, all the rules that are true for ordinary logarithms are valid.

See also:

Logarithm. Properties of the logarithm (power of the logarithm).

How to represent a number as a logarithm?

We use the definition of logarithm.

A logarithm is an exponent to which the base must be raised to obtain the number under the logarithm sign.

Thus, in order to represent a certain number c as a logarithm to base a, you need to put a power with the same base as the base of the logarithm under the sign of the logarithm, and write this number c as the exponent:

Absolutely any number can be represented as a logarithm - positive, negative, integer, fractional, rational, irrational:

In order not to confuse a and c under stressful conditions of a test or exam, you can use the following memorization rule:

what is below goes down, what is above goes up.

For example, you need to represent the number 2 as a logarithm to base 3.

We have two numbers - 2 and 3. These numbers are the base and the exponent, which we will write under the sign of the logarithm. It remains to determine which of these numbers should be written down, to the base of the power, and which – up, to the exponent.

The base 3 in the notation of a logarithm is at the bottom, which means that when we represent two as a logarithm to the base 3, we will also write 3 down to the base.

2 is higher than three. And in notation of the degree two we write above the three, that is, as an exponent:

Logarithms. First level.

Logarithms

Logarithm positive number b based on a, Where a > 0, a ≠ 1, is called the exponent to which the number must be raised a, To obtain b.

Definition of logarithm can be briefly written like this:

This equality is valid for b > 0, a > 0, a ≠ 1. It is usually called logarithmic identity.
The action of finding the logarithm of a number is called by logarithm.

Properties of logarithms:

Logarithm of the product:

Logarithm of the quotient:

Replacing the logarithm base:

Logarithm of degree:

Logarithm of the root:

Logarithm with power base:





Decimal and natural logarithms.

Decimal logarithm numbers call the logarithm of this number to base 10 and write   lg b
Natural logarithm numbers are called the logarithm of that number to the base e, Where e- an irrational number approximately equal to 2.7. At the same time they write ln b.

Other notes on algebra and geometry

Basic properties of logarithms

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 main properties.

You definitely need to know these rules - not a single serious logarithmic problem can be solved without them. 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 the 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 are built on this fact test papers. 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.

How to solve logarithms

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:

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

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

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 conventional numerical expressions. It is possible to evaluate how convenient they are only by deciding 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 the exponent in the argument. The number n can be absolutely anything, because it is just a logarithm value.

The second formula is actually a paraphrased definition. That's what it's called: .

In fact, 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: the result is the 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:

Note that log 25 64 = log 5 8 - we 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. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
  2. log a 1 = 0 is. The 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.

As you know, when multiplying expressions with powers, their exponents always add up (a b *a c = a b+c). This mathematical law was derived by Archimedes, and later, in the 8th century, the mathematician Virasen created a table of integer exponents. It was they who served for the further discovery of logarithms. Examples of using this function can be found almost everywhere where you need to simplify cumbersome multiplication by simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. In simple and accessible language.

Definition in mathematics

A logarithm is an expression of the following form: log a b=c, that is, the logarithm of any non-negative number (that is, any positive) “b” to its base “a” is considered to be the power “c” to which it is necessary to raise the base “a” in order to ultimately get the value "b". Let's analyze the logarithm using examples, let's say there is an expression log 2 8. How to find the answer? It’s very simple, you need to find a power such that from 2 to the required power you get 8. After doing some calculations in your head, we get the number 3! And that’s true, because 2 to the power of 3 gives the answer as 8.

Types of logarithms

For many pupils and students, this topic seems complicated and incomprehensible, but in fact logarithms are not so scary, the main thing is to understand their general meaning and remember their properties and some rules. There are three separate types of logarithmic expressions:

  1. Natural logarithm ln a, where the base is the Euler number (e = 2.7).
  2. Decimal a, where the base is 10.
  3. Logarithm of any number b to base a>1.

Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to a single logarithm using logarithmic theorems. To obtain the correct values ​​of logarithms, you should remember their properties and the sequence of actions when solving them.

Rules and some restrictions

In mathematics, there are several rules-constraints that are accepted as an axiom, that is, they are not subject to discussion and are the truth. For example, it is impossible to divide numbers by zero, and it is also impossible to extract an even root from negative numbers. Logarithms also have their own rules, following which you can easily learn to work even with long and capacious logarithmic expressions:

  • The base “a” must always be greater than zero, and not equal to 1, otherwise the expression will lose its meaning, because “1” and “0” to any degree are always equal to their values;
  • if a > 0, then a b >0, it turns out that “c” must also be greater than zero.

How to solve logarithms?

For example, the task is given to find the answer to the equation 10 x = 100. This is very easy, you need to choose a power by raising the number ten to which we get 100. This, of course, is 10 2 = 100.

Now let's represent this expression in logarithmic form. We get log 10 100 = 2. When solving logarithms, all actions practically converge to find the power to which it is necessary to enter the base of the logarithm in order to obtain a given number.

To accurately determine the value of an unknown degree, you need to learn how to work with a table of degrees. It looks like this:

As you can see, some exponents can be guessed intuitively if you have a technical mind and knowledge of the multiplication table. However, for larger values ​​you will need a power table. It can be used even by those who know nothing at all about complex mathematical topics. The left column contains numbers (base a), the top row of numbers is the value of the power c to which the number a is raised. At the intersection, the cells contain the number values ​​that are the answer (a c =b). Let's take, for example, the very first cell with the number 10 and square it, we get the value 100, which is indicated at the intersection of our two cells. Everything is so simple and easy that even the most true humanist will understand!

Equations and inequalities

It turns out that under certain conditions the exponent is the logarithm. Therefore, any mathematical numerical expressions can be written as a logarithmic equality. For example, 3 4 =81 can be written as the base 3 logarithm of 81 equal to four (log 3 81 = 4). For negative powers the rules are the same: 2 -5 = 1/32 we write it as a logarithm, we get log 2 (1/32) = -5. One of the most fascinating sections of mathematics is the topic of “logarithms”. We will look at examples and solutions of equations below, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.

Given an expression of the following form: log 2 (x-1) > 3 - it is logarithmic inequality, since the unknown value "x" is under the sign of the logarithm. And also in the expression two quantities are compared: the logarithm of the desired number to base two is greater than the number three.

The most important difference between logarithmic equations and inequalities is that equations with logarithms (for example, the logarithm 2 x = √9) imply one or more specific numerical values ​​in the answer, while when solving an inequality, both the range of acceptable values ​​​​and the points are determined breaking this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer to an equation, but a continuous series or set of numbers.

Basic theorems about logarithms

When solving primitive tasks of finding the values ​​of the logarithm, its properties may not be known. However, when it comes to logarithmic equations or inequalities, first of all, it is necessary to clearly understand and apply in practice all the basic properties of logarithms. We will look at examples of equations later; let's first look at each property in more detail.

  1. The main identity looks like this: a logaB =B. It applies only when a is greater than 0, not equal to one, and B is greater than zero.
  2. The logarithm of the product can be represented in the following formula: log d (s 1 * s 2) = log d s 1 + log d s 2. In this case, the mandatory condition is: d, s 1 and s 2 > 0; a≠1. You can give a proof for this logarithmic formula, with examples and solution. Let log a s 1 = f 1 and log a s 2 = f 2, then a f1 = s 1, a f2 = s 2. We obtain that s 1 * s 2 = a f1 *a f2 = a f1+f2 (properties of degrees ), and then by definition: log a (s 1 * s 2) = f 1 + f 2 = log a s1 + log a s 2, which is what needed to be proven.
  3. The logarithm of the quotient looks like this: log a (s 1/ s 2) = log a s 1 - log a s 2.
  4. The theorem in the form of a formula takes the following form: log a q b n = n/q log a b.

This formula is called the “property of the degree of logarithm.” It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics is based on natural postulates. Let's look at the proof.

Let log a b = t, it turns out a t =b. If we raise both parts to the power m: a tn = b n ;

but since a tn = (a q) nt/q = b n, therefore log a q b n = (n*t)/t, then log a q b n = n/q log a b. The theorem has been proven.

Examples of problems and inequalities

The most common types of problems on logarithms are examples of equations and inequalities. They are found in almost all problem books, and are also a required part of mathematics exams. For admission to university or passing entrance examinations in mathematics you need to know how to solve such problems correctly.

Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, but certain rules can be applied to each mathematical inequality or logarithmic equation. First of all, you should find out whether the expression can be simplified or lead to general appearance. Simplify long ones logarithmic expressions possible if you use their properties correctly. Let's get to know them quickly.

When solving logarithmic equations, we must determine what type of logarithm we have: an example expression may contain a natural logarithm or a decimal one.

Here are examples ln100, ln1026. Their solution boils down to the fact that they need to determine the power to which the base 10 will be equal to 100 and 1026, respectively. To solve natural logarithms, you need to apply logarithmic identities or their properties. Let's look at examples of solving logarithmic problems of various types.

How to Use Logarithm Formulas: With Examples and Solutions

So, let's look at examples of using the basic theorems about logarithms.

  1. The property of the logarithm of a product can be used in tasks where it is necessary to expand great importance numbers b into simpler factors. For example, log 2 4 + log 2 128 = log 2 (4*128) = log 2 512. The answer is 9.
  2. log 4 8 = log 2 2 2 3 = 3/2 log 2 2 = 1.5 - as you can see, using the fourth property of the logarithm power, we managed to solve a seemingly complex and unsolvable expression. You just need to factor the base and then take the exponent values ​​out of the sign of the logarithm.

Assignments from the Unified State Exam

Logarithms are often found in entrance exams, especially a lot of logarithmic problems in the Unified State Exam (state exam for all school graduates). Usually these tasks are present not only in part A (the easiest test part exam), but also in part C (the most complex and voluminous tasks). The exam requires accurate and perfect knowledge of the topic “Natural logarithms”.

Examples and solutions to problems are taken from official Unified State Exam options. Let's see how such tasks are solved.

Given log 2 (2x-1) = 4. Solution:
let's rewrite the expression, simplifying it a little log 2 (2x-1) = 2 2, by the definition of the logarithm we get that 2x-1 = 2 4, therefore 2x = 17; x = 8.5.

  • It is best to reduce all logarithms to the same base so that the solution is not cumbersome and confusing.
  • All expressions under the logarithm sign are indicated as positive, therefore, when the exponent of an expression that is under the logarithm sign and as its base is taken out as a multiplier, the expression remaining under the logarithm must be positive.