Examples of solving logarithms with an explanation. How to use logarithm formulas: with examples and solutions

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

Types of logarithms

log a b - logarithm of number b to base a (a\u003e 0, a ≠ 1, b\u003e 0);
lg b - decimal logarithm (logarithm base 10, a \u003d 10);
ln b - natural logarithm (logarithm base e, a \u003d e).

How to solve logarithms?

The logarithm of b to base a is an exponent, which requires that base a be raised to b. The result is pronounced like this: “logarithm of b to base a”. The solution to logarithmic problems is that you need to determine a given degree by numbers by the indicated numbers. There are some basic rules for determining or solving the logarithm and transforming the entry itself. Using them, the solution of logarithmic equations is carried out, 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 basic formulas and properties:

For any a; a\u003e 0; a ≠ 1 and for any x; y\u003e 0.

a log a b \u003d b - basic logarithmic identity
log a 1 \u003d 0
log a a \u003d 1
log a (x y) \u003d log a x + log a y
log a x / y \u003d log a x - log a y
log a 1 / x \u003d -log a x
log a x p \u003d p log a x
log a k x \u003d 1 / k log a x, for k ≠ 0
log a x \u003d log a c x c
log a x \u003d log b x / log b a - formula for transition to a new base
log a x \u003d 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 entry is truncated, you get the decimal logarithm. If worth natural number e, then we write down, reducing to the natural logarithm. It means that the result of all logarithms is the degree to which the base number is raised to obtain the number b.

Directly, the solution is to calculate 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.

Adding and subtracting logarithms with two different numbers, but with the same bases, replace with one 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 use expressions to simplify the logarithm, there are some limitations to consider. And that is: the base of the logarithm a is only positive numberbut not equal to one. The number b, like a, must be greater than zero.

There are cases where by simplifying the expression, you cannot calculate the logarithm numerically. 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.

So, we have before us powers of two. If you take the number from the bottom line, then you can easily find the degree to which you have to raise the 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 logarithm base a of the argument x is the power to which the number a must be raised to get the number x.

Notation: log a x \u003d b, where a is the base, x is the argument, b is actually what the logarithm is.

For example, 2 3 \u003d 8 ⇒ log 2 8 \u003d 3 (logarithm base 2 of 8 is three, since 2 3 \u003d 8). With the same success log 2 64 \u003d 6, since 2 6 \u003d 64.

The operation of finding the logarithm of a number in a given base is called the logarithm. 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 \u003d 1	log 2 4 \u003d 2	log 2 8 \u003d 3	log 2 16 \u003d 4	log 2 32 \u003d 5	log 2 64 \u003d 6

Unfortunately, not all logarithms are calculated so easily. For example, try to find log 2 5. Number 5 is not in the table, but logic dictates that the logarithm will lie somewhere on the segment. Because 2 2< 5 < 2 3 , а чем more degree two, the larger the number will be.

Such numbers are called irrational: the numbers after the decimal point can be written indefinitely, 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 the logarithm is an expression with two variables (base and argument). At first, many are confused about where the basis is and where is the argument. To avoid annoying misunderstandings, just take a look at the picture:

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

We figured out the definition - it remains 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:

Argument and radix must always be greater than zero. This follows from the definition of the degree by a rational indicator, to which the definition of the logarithm is reduced.
The base must be different from one, since one is still one to any degree. Because of this, the question “to what degree one should raise one to get a two” is meaningless. There is no such degree!

Such restrictions are called range of valid values (ODZ). It turns out that the ODZ of the logarithm looks like this: log a x \u003d b ⇒ x\u003e 0, a\u003e 0, a ≠ 1.

Note that there is no restriction on the number b (logarithm value). For example, the logarithm may well be negative: log 2 0.5 \u003d −1, because 0.5 \u003d 2 −1.

However, now we are considering only numerical expressions, where you do not need to know the ODV of the logarithm. All restrictions have already been taken into account by the task compilers. But when the logarithmic equations and inequalities come in, the DHS requirements will become mandatory. Indeed, at the base and in the argument there can be very strong constructions that do not necessarily correspond to the above restrictions.

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

Represent base a and argument x as a power with the smallest possible base greater than one. Along the way, it's better to get rid of decimals;
Solve the equation for variable b: x \u003d a b;
The resulting number b will be the answer.

That's all! If the logarithm turns out to be irrational, this will be seen already at the first step. The requirement for the base to be greater than one is very relevant: this reduces the likelihood of error and greatly simplifies calculations. Similarly with decimal fractions: if you immediately translate them into ordinary ones, there will be many times less errors.

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

A task. Calculate the logarithm: log 5 25

Let's represent the base and the argument as a power of five: 5 \u003d 5 1; 25 \u003d 5 2;

Let's compose and solve the equation:
log 5 25 \u003d b ⇒ (5 1) b \u003d 5 2 ⇒ 5 b \u003d 5 2 ⇒ b \u003d 2;

Received the answer: 2.

A task. Calculate the logarithm:

A task. Calculate the logarithm: log 4 64

Let's represent the base and the argument as a power of two: 4 \u003d 2 2; 64 \u003d 2 6;
Let's compose and solve the equation:
log 4 64 \u003d b ⇒ (2 2) b \u003d 2 6 ⇒ 2 2b \u003d 2 6 ⇒ 2b \u003d 6 ⇒ b \u003d 3;
Received the answer: 3.

A task. Calculate the logarithm: log 16 1

Let's represent the base and the argument as a power of two: 16 \u003d 2 4; 1 \u003d 2 0;
Let's compose and solve the equation:
log 16 1 \u003d b ⇒ (2 4) b \u003d 2 0 ⇒ 2 4b \u003d 2 0 ⇒ 4b \u003d 0 ⇒ b \u003d 0;
Received the answer: 0.

A task. Calculate the log of: log 7 14

We represent the base and the argument as a power of seven: 7 \u003d 7 1; 14 is not represented as a power of seven, since 7 1< 14 < 7 2 ;
It follows from the previous paragraph that the logarithm is not considered;
The answer is no change: log 7 14.

A small note on the last example. How do you ensure 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.

A task. Find out if the exact powers of the number are: 8; 48; 81; 35; 14 .

8 \u003d 2 2 2 \u003d 2 3 is the exact degree, since there is only one factor;
48 \u003d 6 · 8 \u003d 3 · 2 · 2 · 2 · 2 \u003d 3 · 2 4 - is not an exact degree, since there are two factors: 3 and 2;
81 \u003d 9 9 \u003d 3 3 3 3 3 \u003d 3 4 - the exact degree;
35 \u003d 7 · 5 - again not an exact degree;
14 \u003d 7 2 - again not an exact degree;

Note also that the prime numbers themselves are always exact powers of themselves.

Decimal logarithm

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

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

For example, lg 10 \u003d 1; lg 100 \u003d 2; lg 1000 \u003d 3 - etc.

From now on, when a phrase like "Find lg 0.01" appears in a textbook, you should know: this is not a typo. This is the decimal logarithm. However, if you are not used to such a designation, you can always rewrite it:
log x \u003d log 10 x

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

Natural logarithm

There is another logarithm, which has its own notation. In a way, it is even more important than decimal. It is about the natural logarithm.

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

Many will ask: what else is the number e? This is an irrational number, its exact value cannot be found and recorded. I will give only the first figures:
e \u003d 2.718281828459 ...

We will not delve into what this number is and why it is needed. Just remember that e is the base of the natural logarithm:
ln x \u003d log e x

Thus, ln e \u003d 1; ln e 2 \u003d 2; ln e 16 \u003d 16 - etc. On the other hand, ln 2 is an irrational number. In general, the natural logarithm of any rational number is irrational. Except, of course, units: ln 1 \u003d 0.

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

basic properties.

logax + logay \u003d loga (x y);
logax - logay \u003d loga (x: y).

identical grounds

Log6 4 + log6 9.

Now let's complicate the task a little.

Examples of solving logarithms

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

Of course, all these rules make sense if the ODL of the logarithm is observed: a\u003e 0, a ≠ 1, x\u003e

A task. Find the meaning of the expression:

Moving to a new foundation

Let the logarithm be given. Then for any number c such that c\u003e 0 and c ≠ 1, the following equality holds:

A task. Find the meaning of the expression:

Basic properties of the logarithm

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

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

Basic properties of logarithms

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

Examples for logarithms

Logarithm expressions

Example 1.
and). x \u003d 10ac ^ 2 (a\u003e 0, c\u003e 0).

By properties 3.5 we calculate

4. Where .

Example 2. Find x if

Example 3. Let the value of the logarithms be given

Evaluate log (x) if

Basic properties of logarithms

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

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

Addition and subtraction of logarithms

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

logax + logay \u003d loga (x y);
logax - logay \u003d loga (x: y).

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

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

Since the bases of the logarithms are the same, we use the sum formula:
log6 4 + log6 9 \u003d log6 (4 9) \u003d log6 36 \u003d 2.

A task. Find the value of the expression: log2 48 - log2 3.

The bases are the same, we use the difference formula:
log2 48 - log2 3 \u003d log2 (48: 3) \u003d log2 16 \u003d 4.

A task. Find the value of the expression: log3 135 - log3 5.

Again the bases are the same, so we have:
log3 135 - log3 5 \u003d log3 (135: 5) \u003d log3 27 \u003d 3.

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

Removing the exponent from the logarithm

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

Of course, all these rules make sense if the ODL of the logarithm is observed: a\u003e 0, a ≠ 1, x\u003e 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. it is possible to enter the numbers in front of the sign of the logarithm into the logarithm itself. This is what is most often required.

A task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 \u003d 6 log7 49 \u003d 6 2 \u003d 12

A task. Find the meaning of the expression:

Note that the denominator contains the logarithm, the base and the argument of which are exact powers: 16 \u003d 24; 49 \u003d 72. We have:

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

Formulas for logarithms. Logarithms are examples of solutions.

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

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

Moving to a new foundation

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

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

Let the logarithm be given. Then for any number c such that c\u003e 0 and c ≠ 1, the following equality holds:

In particular, if we put c \u003d x, we get:

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

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

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

A task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms contain exact degrees. Let's take out the indicators: log5 16 \u003d log5 24 \u003d 4log5 2; log2 25 \u003d log2 52 \u003d 2log2 5;

Now let's "flip" the second logarithm:

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

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

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

Now let's get rid of decimal logarithmby going to a new base:

Basic logarithmic identity

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

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

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

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: you get this very number a. Read this paragraph carefully again - many people "hang" on it.

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

A task. Find the meaning of the expression:

Note that log25 64 \u003d log5 8 - just square the base and the logarithm argument. Taking into account the rules for multiplying degrees with the same base, we get:

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

Logarithmic unit and logarithmic zero

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

logaa \u003d 1 is. Remember once and for all: the logarithm to any base a from this base is equal to one.
loga 1 \u003d 0 is. The base a can be anything, but if the argument contains one - the logarithm equals zero! Because a0 \u003d 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.

Basic properties of the logarithm

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

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

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

Common cases of logarithms

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

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

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

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

And another important base two logarithm is

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

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

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

Examples for logarithms

Logarithm expressions

Example 1.
and). x \u003d 10ac ^ 2 (a\u003e 0, c\u003e 0).

By properties 3.5 we calculate

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

3.
Using properties 3,5 we find

4. Where .

A seemingly complex expression using a number of rules is simplified to the form

Finding the values \u200b\u200bof logarithms

Example 2. Find x if

Decision. For the calculation, we apply to the last term 5 and 13 of the properties

Substitute and grieve

Since the bases are equal, we equate the expressions

Logarithms. First level.

Let the value of the logarithms be given

Evaluate log (x) if

Solution: We logarithm the variable to write the logarithm through the sum of terms

This is where the acquaintance with logarithms and their properties just begins. Practice calculations, enrich your practical skills - you will soon need this knowledge to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge for another no less important topic - logarithmic inequalities ...

Basic properties of logarithms

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

Addition and subtraction of logarithms

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

logax + logay \u003d loga (x y);
logax - logay \u003d loga (x: y).

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

A task. Find the value of the expression: log6 4 + log6 9.

Since the bases of the logarithms are the same, we use the sum formula:
log6 4 + log6 9 \u003d log6 (4 9) \u003d log6 36 \u003d 2.

A task. Find the value of the expression: log2 48 - log2 3.

The bases are the same, we use the difference formula:
log2 48 - log2 3 \u003d log2 (48: 3) \u003d log2 16 \u003d 4.

A task. Find the value of the expression: log3 135 - log3 5.

Again the bases are the same, so we have:
log3 135 - log3 5 \u003d log3 (135: 5) \u003d log3 27 \u003d 3.

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

Removing the exponent from the logarithm

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

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

How to solve logarithms

This is what is most often required.

A task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 \u003d 6 log7 49 \u003d 6 2 \u003d 12

A task. Find the meaning of the expression:

Note that the denominator contains the logarithm, the base and the argument of which are exact powers: 16 \u003d 24; 49 \u003d 72. We have:

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

Moving to a new foundation

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

Let the logarithm be given. Then for any number c such that c\u003e 0 and c ≠ 1, the following equality holds:

In particular, if we put c \u003d x, we get:

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

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

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

A task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms contain exact degrees. Let's take out the indicators: log5 16 \u003d log5 24 \u003d 4log5 2; log2 25 \u003d log2 52 \u003d 2log2 5;

Now let's "flip" the second logarithm:

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

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

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

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

Basic logarithmic identity

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

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

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

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

A task. Find the meaning of the expression:

Note that log25 64 \u003d log5 8 - just square the base and the logarithm argument. Taking into account the rules for multiplying degrees with the same base, we get:

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

Logarithmic unit and logarithmic zero

logaa \u003d 1 is. Remember once and for all: the logarithm to any base a from this base is equal to one.
loga 1 \u003d 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 \u003d 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.

Logarithmic expressions, solving examples. In this article we will look at the problems associated with solving logarithms. In the tasks, the question of finding the meaning of the expression is raised. It should be noted that the concept of a logarithm is used in many tasks and it is extremely important to understand its meaning. As for the exam, the logarithm is used when solving equations, in applied problems, and also in tasks related to the study of functions.

Here are some examples to understand the very meaning of the logarithm:

Basic logarithmic identity:

Properties of logarithms that must always be remembered:

* The logarithm of the product is the sum of the logarithms of the factors.

* * *

* The logarithm of the quotient (fraction) is equal to the difference between the logarithms of the factors.

* * *

* Logarithm of degree is equal to the product exponent per logarithm of its base.

* * *

* Transition to a new base

* * *

More properties:

* * *

The calculation of logarithms is closely related to the use of the properties of exponents.

Here are some of them:

The essence of this property is that when the numerator is transferred to the denominator and vice versa, the sign of the exponent changes to the opposite. For instance:

Consequence of this property:

* * *

When raising a power to a power, the base remains the same, and the indicators are multiplied.

* * *

As you can see, the very concept of a logarithm is simple. The main thing is what is needed good practice, which gives a certain skill. Of course, knowledge of the formulas is required. If the skill in converting elementary logarithms is not formed, then when solving simple tasks, you can easily make a mistake.

Practice, solve the simplest examples from the math course first, then move on to more difficult ones. In the future I will definitely show you how the "ugly" logarithms are solved, there will not be such ones on the exam, but they are of interest, do not miss it!

That's all! Success to you!

Best regards, Alexander Krutitskikh

P.S: I would be grateful if you could tell us about the site on social networks.

(from the Greek λόγος - "word", "relation" and ἀριθμός - "number") numbers b by reason a (log α b) is called such a number cand b= a c, that is, log α b=c and b \u003d a c are equivalent. The logarithm makes sense if a\u003e 0, and ≠ 1, b\u003e 0.

In other words logarithm numbers b by reason andis formulated as an indicator of the degree to which the number must be raised ato get the number b(only positive numbers have a logarithm).

This formulation implies that the computation x \u003d log α b, is equivalent to solving the equation a x \u003d b.

For instance:

log 2 8 \u003d 3 because 8 \u003d 2 3.

We emphasize that the above formulation of the logarithm makes it possible to immediately determine logarithm value, when the number under the sign of the logarithm is some degree of the base. And in truth, the formulation of the logarithm makes it possible to prove that if b \u003d a c, then the logarithm of the number b by reason a is equal from... It is also clear that the topic of logarithm is closely related to the topic degree of number.

Calculation of the logarithm is called by taking the logarithm... Taking the logarithm is a mathematical operation of taking the logarithm. When taking the logarithm, the product of the factors is transformed into the sum of the terms.

Potentiation is a mathematical operation inverse to logarithm. In potentiation, the given base is raised to the power of the expression over which the potentiation is performed. In this case, the sums of the members are transformed into the product of the factors.

Real logarithms with bases 2 (binary), e Euler's number e ≈ 2.718 (natural logarithm) and 10 (decimal) are often used.

At this stage, it is advisable to consider samples of logarithmslog 7 2 , ln √ 5, lg0.0001.

And the entries lg (-3), log -3 3.2, log -1 -4.3 do not make sense, since in the first of them a negative number is placed under the sign of the logarithm, in the second - negative number at the base, and in the third - both a negative number under the sign of the logarithm and one at the base.

Conditions for determining the logarithm.

It is worth considering separately the conditions a\u003e 0, a ≠ 1, b\u003e 0 under which definition of the logarithm. Let's consider why these restrictions are taken. An equality of the form x \u003d log α b , called the basic logarithmic identity, which directly follows from the definition of a logarithm given above.

Let's take the condition a ≠ 1... Since one is equal to one to any degree, the equality x \u003d log α b can exist only when b \u003d 1but log 1 1 will be any real number. To eliminate this ambiguity, we take a ≠ 1.

Let us prove the necessity of the condition a\u003e 0... When a \u003d 0 according to the formulation of the logarithm, it can only exist for b \u003d 0... And accordingly then log 0 0can be any nonzero real number, since zero in any nonzero degree is zero. To eliminate this ambiguity, the condition a ≠ 0... And when a<0 we would have to reject the analysis of rational and irrational values \u200b\u200bof the logarithm, since a degree with a rational and irrational exponent is defined only for non-negative grounds. It is for this reason that the condition is stipulated a\u003e 0.

And the last condition b\u003e 0 follows from the inequality a\u003e 0since x \u003d log α b, and the value of the degree with a positive base a always positive.

Features of logarithms.

Logarithms characterized by distinctive features, which led to their widespread use to significantly facilitate painstaking calculations. In the transition "to the world of logarithms," multiplication is transformed into a much easier addition, division into subtraction, and exponentiation and root extraction are transformed, respectively, into multiplication and division by an exponent.

Formulation of logarithms and a table of their values \u200b\u200b(for trigonometric functions) was first published in 1614 by the Scottish mathematician John Napier. Logarithmic tables, magnified and detailed by other scientists, were widely used in scientific and engineering calculations, and remained relevant until electronic calculators and computers were used.