Sinusitis

Logarithm. Definition of binary logarithm, natural logarithm, decimal logarithm; exponential function exp(x), number e

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Logarithm root of a positive number is equal to the logarithm of the radical expression divided by the exponent of the root:

And in truth, when working with degrees, the dependence is used, therefore, by applying the theorem of the logarithm of degrees, we obtain this formula.

Let's put it into practice, consider example:

At solving problems to find the logarithm Quite often it turns out to be useful from logarithms to one base (for example, A) go to logarithms in a different base (for example, With) . In such situations, the following formula is used:

This means that a, b And With of course positive numbers, and A And With are not equal to one.

To prove this formula, we use basic logarithmic identity:

If positive numbers are equal, then obviously their logarithms to the same base are equal With. That's why:

By applying logarithm of power theorem:

Hence , log a b · log c a = log c b where it comes from formula for changing the base of a logarithm.

The logarithm of a positive number b to base a (a>0, a is not equal to 1) is a number c such that a c = b: log a b = c ⇔ a c = b (a > 0, a ≠ 1, b > 0)

Note that the logarithm of a non-positive number is undefined. In addition, the base of the logarithm must be positive number, not equal to 1. For example, if we square -2, we get the number 4, but this does not mean that the logarithm to the base -2 of 4 is equal to 2.

Basic logarithmic identity

a log a b = b (a > 0, a ≠ 1) (2)

It is important that the scope of definition of the right and left sides of this formula is different. Left side defined only for b>0, a>0 and a ≠ 1. Right side is defined for any b, but does not depend on a at all. Thus, the application of the basic logarithmic “identity” when solving equations and inequalities can lead to a change in the OD.

Two obvious consequences of the definition of logarithm

log a a = 1 (a > 0, a ≠ 1) (3)
log a 1 = 0 (a > 0, a ≠ 1) (4)

Indeed, when raising the number a to the first power, we get the same number, and when raising it to the first power zero degree- one.

Logarithm of the product and logarithm of the quotient

log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) (5)

Log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) (6)

I would like to warn schoolchildren against thoughtlessly using these formulas when solving logarithmic equations and inequalities. When using them “from left to right,” the ODZ narrows, and when moving from the sum or difference of logarithms to the logarithm of the product or quotient, the ODZ expands.

Indeed, the expression log a (f (x) g (x)) is defined in two cases: when both functions are strictly positive or when f (x) and g (x) are both less than zero.

Transforming this expression into the sum log a f (x) + log a g (x), we are forced to limit ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the area acceptable values, and this is categorically unacceptable, because it can lead to loss of solutions. A similar problem exists for formula (6).

The degree can be taken out of the sign of the logarithm

log a b p = p log a b (a > 0, a ≠ 1, b > 0) (7)

And again I would like to call for accuracy. Consider the following example:

Log a (f (x) 2 = 2 log a f (x)

The left side of the equality is obviously defined for all values of f(x) except zero. The right side is only for f(x)>0! By taking the degree out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of acceptable values. All these remarks apply not only to power 2, but also to any even power.

Formula for moving to a new foundation

log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) (8)

That rare case, when the ODZ does not change during the transformation. If you have chosen base c wisely (positive and not equal to 1), the formula for moving to a new base is completely safe.

If we choose the number b as the new base c, we obtain an important special case of formula (8):

Log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) (9)

Some simple examples with logarithms

Example 1. Calculate: log2 + log50.
Solution. log2 + log50 = log100 = 2. We used the formula for the sum of logarithms (5) and the definition decimal logarithm.

Example 2. Calculate: lg125/lg5.
Solution. log125/log5 = log 5 125 = 3. We used the formula for moving to a new base (8).

Table of formulas related to logarithms

a log a b = b (a > 0, a ≠ 1)

log a a = 1 (a > 0, a ≠ 1)

log a 1 = 0 (a > 0, a ≠ 1)

log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0)

log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0)

log a b p = p log a b (a > 0, a ≠ 1, b > 0)

log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1)

log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1)

main properties.

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

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

Task. Find the meaning 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:

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 equal to 2.7 and twice the year of birth of Leo Nikolaevich Tolstoy.

Basic properties of logarithms

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

Examples for logarithms

Logarithm expressions

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

Using 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 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: logax and logay. Then they can be added and subtracted, and:

logax + logay = loga(x y);
logax − logay = loga (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: key point Here - 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:

Since logarithms have the same bases, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.

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.

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 calculated separately. But after the transformations, completely normal numbers are obtained. Many are built on this fact tests. 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

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

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

Task. Find the value of the expression: log7 496.

Let's get rid of the degree in the argument using the first formula:
log7 496 = 6 log7 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 = 24; 49 = 72. We have:

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

Logarithm formulas. Logarithms examples solutions.

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: 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 is what was done. The result was the answer: 2.

Transition to a new foundation

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

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

Let the logarithm logax be given. 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 when solving logarithmic equations and inequalities.

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

Task. Find the value of the expression: log5 16 log2 25.

Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 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: log9 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 log25 64 = log5 8 - simply took the square from the base and argument of the logarithm. Taking into account the rules for multiplying powers with the same base, we get:

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

Logarithmic unit and logarithmic zero

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

logaa = 1 is. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
loga 1 = 0 is. The base a can be anything, but if the argument contains one, the logarithm is equal to 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

It is necessary to know the above properties, since almost all problems and examples related to logarithms are solved on their basis. Rest exotic properties can be derived by mathematical manipulation of these formulas

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

When calculating the formula for the sum and difference of logarithms (3.4) you come across 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 most common logarithms are those in which the base is equal to ten, exponential or two.
The logarithm to base ten is usually called the decimal logarithm and is simply denoted by lg(x).

It is clear from the recording that the basics are not written in the recording. For example

A natural logarithm is a logarithm whose base is an exponent (denoted by ln(x)).

The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is equal to 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 logarithm to base two is denoted by

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

The integral or antiderivative logarithm is determined by the relationship

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

Examples for logarithms

Logarithm expressions

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

Using properties 3.5 we calculate

2.
By the property of difference of logarithms we have

3.
Using properties 3.5 we find

4. Where .

In appearance complex expression using a number of rules is simplified to form

Finding logarithm values

Example 2. Find x if

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

We put it on record and mourn

Since the bases are equal, we equate the expressions

Logarithms. Entry level.

Let the value of logarithms be given

Calculate log(x) if

Solution: Let's take a logarithm of the variable to write the logarithm through the sum of its terms

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

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.

Adding and subtracting logarithms

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

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

Task. Find the value of the expression: log6 4 + log6 9.

Since logarithms have the same bases, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.

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.

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 calculated separately. But after the transformations, completely normal numbers are obtained. Many tests are based on this fact. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.

Extracting the exponent from the logarithm

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

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

How to solve logarithms

This is what is most often required.

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 = 6 log7 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 = 24; 49 = 72. 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.

Transition to a new foundation

Formulas for transition to a new foundation come to the rescue. Let us 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 set 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 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: log5 16 log2 25.

Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 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: log9 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: .

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 log25 64 = log5 8 - simply took the square from the base and argument of the logarithm. Taking into account the rules for multiplying powers with the same base, we get:

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

Logarithmic unit and logarithmic zero

logaa = 1 is. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
loga 1 = 0 is. The base a can be anything, but if the argument contains one, the logarithm is equal to 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.

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 it is necessary 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 individual species logarithmic expressions:

Natural logarithm ln a, where the base is the Euler number (e = 2.7).
Decimal a, where the base is 10.
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 the even root of 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 large values you will need a table of degrees. 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 answers. numerical values, while when solving the inequality, both the range of permissible values and the breakpoints of this function are determined. 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.

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.
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 prerequisite 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.
The logarithm of the quotient looks like this: log a (s 1/ s 2) = log a s 1 - log a s 2.
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. To enter a university or pass entrance examinations in mathematics, you need to know how to correctly solve such tasks.

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.

The property of the logarithm of a product can be used in tasks where it is necessary to expand great value numbers b into simpler factors. For example, log 2 4 + log 2 128 = log 2 (4*128) = log 2 512. The answer is 9.
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 many logarithmic problems in the Unified State Exam ( state exam for all school leavers). Typically, these tasks are present not only in part A (the easiest test part of the 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.