Examples of solving logarithms with explanation. How to Use Logarithm Formulas: With Examples and Solutions

One of the elements of primitive level algebra is the logarithm. The name came from Greek from the word “number” or “power” and means the power to which it is necessary to raise the number at the base 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);
lnb- natural logarithm(log 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 worth 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.

Adding and subtracting logarithms with two various 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 are using expressions to simplify the logarithm, there are some limitations to be aware of. And that is: the base of the logarithm a - only 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.

So, we have powers of two. If you take the number from the bottom line, then you can easily find the power to which you have to raise a 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 - in fact, the definition of the logarithm:

The logarithm to the 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 equal to.

For example, 2 3 = 8 ⇒ log 2 8 = 3 (the base 2 logarithm of 8 is three because 2 3 = 8). Might as well log 2 64 = 6 because 2 6 = 64 .

The operation of finding the logarithm of a number to a given base is called the logarithm. So let's add a new row 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 considered 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 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 never repeat. If the logarithm turns out to be irrational, it is better to leave it like this: 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 people confuse where the base is and where the argument is. To avoid annoying misunderstandings, just take a look at the picture:

Before us is nothing more than the definition of the logarithm. Remember: the logarithm is the power, to which you need to raise the base to get the argument. It is the base that is raised to a 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 there is no confusion.

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:

The argument and base must always be greater than zero. This follows from the definition of the degree by a rational exponent, to which the definition of the logarithm is reduced.
The base must be different from unity, since a unit to any power is still a unit. 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 valid range(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) is not imposed. For example, the logarithm may well be negative: log 2 0.5 \u003d -1, because 0.5 = 2 −1 .

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

Now consider general scheme logarithm calculations. It consists of three steps:

Express the base a and the argument x as a power with the smallest possible base greater than one. Along the way, it is better to get rid of decimal fractions;
Solve the equation for the variable b: x = 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 that the base be greater than one is very relevant: this reduces the likelihood of error and greatly simplifies calculations. Similar to decimals: if you immediately translate them into ordinary ones, there will be many times less errors.

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

A task. Calculate the logarithm: log 5 25

Let's represent the base and the argument as a power of five: 5 = 5 1 ; 25 = 52;

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

Received an 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 = 2 2 ; 64 = 26;
Let's make and solve the equation:
log 4 64 = b ⇒ (2 2) b = 2 6 ⇒ 2 2b = 2 6 ⇒ 2b = 6 ⇒ b = 3 ;
Received an answer: 3.

A task. Calculate the logarithm: log 16 1

Let's represent the base and the argument as a power of two: 16 = 2 4 ; 1 = 20;
Let's make and solve the equation:
log 16 1 = b ⇒ (2 4) b = 2 0 ⇒ 2 4b = 2 0 ⇒ 4b = 0 ⇒ b = 0 ;
Received a response: 0.

A task. Calculate the logarithm: log 7 14

Let's represent the base and the argument as a power of seven: 7 = 7 1 ; 14 is not represented as a power of seven, because 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 to make sure that a number is not an exact power of another number? Very simple - just decompose it into prime factors. If there are at least two distinct factors in the expansion, the number is not an exact power.

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

8 \u003d 2 2 2 \u003d 2 3 - the exact degree, because there is only one multiplier;
48 = 6 8 = 3 2 2 2 2 = 3 2 4 is not an exact power because there are two factors: 3 and 2;
81 \u003d 9 9 \u003d 3 3 3 3 \u003d 3 4 - exact degree;
35 = 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 the x argument is the base 10 logarithm, i.e. the power to which you need to raise the number 10 to get the number x. Designation: lg x .

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

From now on, when a phrase like “Find lg 0.01” appears in the textbook, know that 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 = log 10 x

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

natural logarithm

There is another logarithm that has its own notation. In a sense, it is even more important than decimal. It's about about the natural logarithm.

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

Many will ask: what else is the number e? This is an irrational number exact value impossible to find and record. Here are just the first numbers:
e = 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 = log e x

Thus ln e = 1 ; log 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 is irrational. Except, of course, unity: ln 1 = 0.

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

basic properties.

logax + logay = log(x y);
logax − logay = log(x: y).

same grounds

log6 4 + log6 9.

Now let's complicate the task a little.

Examples of solving logarithms

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:

Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x >

A task. Find the value of the expression:

Transition to a new foundation

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

A task. Find the value 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 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

Take the logarithm of expressions

Example 1
a). x=10ac^2 (a>0, c>0).

By properties 3,5 we calculate

4. where .

Example 2 Find x if

Example 3. Let the value of logarithms be given

Calculate log(x) if

Basic properties of logarithms

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.

You must 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 base: logax and logay. Then they can be added and subtracted, and:

logax + logay = log(x y);
logax − logay = log(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 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:

Since the bases of logarithms are the same, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 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 = log2 (48: 3) = log2 16 = 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 = log3 (135: 5) = log3 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. Based on this fact, many test papers. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.

Removing the exponent from the logarithm

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.

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

Let's get rid of the degree in the argument according to the first formula:
log7 496 = 6 log7 49 = 6 2 = 12

A task. Find the value of the expression:

Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 24; 49 = 72. We have:

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

Formulas of logarithms. Logarithms are examples of solutions.

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: log2 7. Since log2 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 logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:

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

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:

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

Note that the arguments of both logarithms are exact exponents. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;

Now let's flip the second logarithm:

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

A task. Find the value of the expression: log9 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:

Now let's get rid of decimal logarithm, moving 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's just the value of the logarithm.

The second formula is actually a paraphrased definition. It's called like this:

Indeed, what will happen if the number b is raised to such a degree that the number b in this degree gives the 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.

A task. Find the value of the expression:

Note that log25 64 = log5 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:

If someone is not in the know, this was a real task from the Unified State Examination 🙂

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.

logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
loga 1 = 0 is. The base a can be anything, but if the argument is one - the logarithm zero! Because a0 = 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 above properties must be known, since, on their basis, almost all problems and examples are solved based on logarithms. The remaining exotic properties can be derived 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 deuce.
The base ten logarithm is usually called the base ten logarithm and is simply denoted lg(x).

It can be seen from the record that the basics are not written in the record. For example

The natural logarithm is the logarithm whose basis is the exponent (denoted 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 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 logarithm is determined by the dependence

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

Examples for logarithms

Take the logarithm of expressions

Example 1
a). x=10ac^2 (a>0, c>0).

By properties 3,5 we calculate

2.
By the difference property of logarithms, we have

3.
Using properties 3.5 we find

4. where .

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

Finding Logarithm Values

Example 2 Find x if

Solution. For the calculation, we apply properties 5 and 13 up to the last term

Substitute in the record and mourn

Since the bases are equal, we equate the expressions

Logarithms. First level.

Let the value of the logarithms be given

Calculate log(x) if

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

This is just the beginning of acquaintance with logarithms and their properties. Practice calculations, enrich your practical skills - you will soon need the acquired 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

You must 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 base: logax and logay. Then they can be added and subtracted, and:

logax + logay = log(x y);
logax − logay = log(x: y).

These formulas will help 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:

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

Since the bases of logarithms are the same, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 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 = log2 (48: 3) = log2 16 = 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 = log3 (135: 5) = log3 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.

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 according to the first formula:
log7 496 = 6 log7 49 = 6 2 = 12

A task. Find the value of the expression:

Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 24; 49 = 72. We have:

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.

Transition to a new foundation

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

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

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

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:

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

Note that the arguments of both logarithms are exact exponents. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;

Now let's flip the second logarithm:

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

A task. Find the value of the expression: log9 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:

Now let's get rid of the decimal logarithm by moving 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's just the value of the logarithm.

The second formula is actually a paraphrased definition. It's called like this:

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

A task. Find the value of the expression:

Note that log25 64 = log5 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:

If someone is not in the know, this was a real task from the Unified State Examination 🙂

Logarithmic unit and logarithmic zero

logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
loga 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 = 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, solution of examples. In this article, we will consider problems related to solving logarithms. The tasks raise the question of finding the value of the expression. It should be noted that the concept of the logarithm is used in many tasks and it is extremely important to understand its meaning. As for the USE, the logarithm is used in solving equations, in applied problems, and also in tasks related to the study of functions.

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

Basic logarithmic identity:

Properties of logarithms that you must always remember:

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

* * *

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

* * *

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

* * *

*Transition to new base

* * *

More properties:

* * *

Computing logarithms is closely related to using the properties of exponents.

We list some of them:

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

Consequence of this property:

* * *

When raising a power to a power, the base remains the same, but the exponents are multiplied.

* * *

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

Practice, solve the simplest examples from the math course first, then move on to more complex ones. In the future, I will definitely show how the “ugly” logarithms are solved, there will be no such ones at the exam, but they are of interest, do not miss it!

That's all! Good luck to you!

Sincerely, Alexander Krutitskikh

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

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

In other words logarithm numbers b by reason a formulated as an exponent to which a number must be raised a to get the number b(the logarithm exists only for positive numbers).

From this formulation it follows that the calculation x= log α b, is equivalent to solving the equation a x =b.

For example:

log 2 8 = 3 because 8=2 3 .

We note that the indicated formulation of the logarithm makes it possible to immediately determine logarithm value when the number under the sign of the logarithm is a certain power of the base. Indeed, the formulation of the logarithm makes it possible to justify that if b=a c, then the logarithm of the number b by reason a equals With. It is also clear that the topic of logarithm is closely related to the topic degree of number.

The calculation of the logarithm is referred to logarithm. Logarithm is the mathematical operation of taking a logarithm. When taking a logarithm, the products of factors are transformed into sums of terms.

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

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

At this stage, it is worth considering samples of logarithms log 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 - a negative number in the base, and in the third - and a negative number under the sign of the logarithm and a unit in the base.

Conditions for determining the logarithm.

It is worth considering separately the conditions a > 0, a ≠ 1, b > 0. definition of a logarithm. Let's consider why these restrictions are taken. This will help us with an equality of the form x = log α b, called the basic logarithmic identity, which directly follows from the definition of the logarithm given above.

Take the condition a≠1. Since one is equal to one to any power, then the equality x=log α b can only exist when b=1, but 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>0. At a=0 according to the formulation of the logarithm, can only exist when b=0. And then accordingly log 0 0 can be any non-zero real number, since zero to any non-zero power 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 of the logarithm, since the exponent with rational and irrational exponent is defined only for non-negative bases. It is for this reason that the condition a>0.

And the last condition b>0 follows from the inequality a>0, because x=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 greatly facilitate painstaking calculations. In the transition "to the world of logarithms", multiplication is transformed into much easier addition, division into subtraction, and exponentiation and root extraction are transformed into multiplication and division by the exponent, respectively.

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