Examples of solutions of logarithms with different. The main property of logarithms and its consequences

Instructions

Write the given logarithmic expression. If the expression uses the logarithm of 10, then its notation is shortened and looks like this: lg b is decimal logarithm. If the logarithm has the number e as its base, then write the expression: ln b – natural logarithm. It is understood that the result of any is the power to which the base number must be raised to obtain the number b.

When finding the sum of two functions, you simply need to differentiate them one by one and add the results: (u+v)" = u"+v";

When finding the derivative of the product of two functions, it is necessary to multiply the derivative of the first function by the second and add the derivative of the second function multiplied by the first function: (u*v)" = u"*v+v"*u;

In order to find the derivative of the quotient of two functions, it is necessary to subtract from the product of the derivative of the dividend multiplied by the divisor function the product of the derivative of the divisor multiplied by the function of the dividend, and divide all this by the divisor function squared. (u/v)" = (u"*v-v"*u)/v^2;

If given complex function, then it is necessary to multiply the derivative of internal function and the derivative of the external one. Let y=u(v(x)), then y"(x)=y"(u)*v"(x).

Using the results obtained above, you can differentiate almost any function. So let's look at a few examples:

y=x^4, y"=4*x^(4-1)=4*x^3;

y=2*x^3*(e^x-x^2+6), y"=2*(3*x^2*(e^x-x^2+6)+x^3*(e^x-2 *x));
There are also problems involving calculating the derivative at a point. Let the function y=e^(x^2+6x+5) be given, you need to find the value of the function at the point x=1.
1) Find the derivative of the function: y"=e^(x^2-6x+5)*(2*x +6).

2) Calculate the value of the function in given point y"(1)=8*e^0=8

Video on the topic

Helpful advice

Learn the table of elementary derivatives. This will significantly save time.

Sources:

• derivative of a constant

So, what's the difference? ir rational equation from the rational? If the unknown variable is under the sign square root, then the equation is considered irrational.

Instructions

The main method for solving such equations is the method of constructing both sides equations into a square. However. this is natural, the first thing you need to do is get rid of the sign. This method is not technically difficult, but sometimes it can lead to trouble. For example, the equation is v(2x-5)=v(4x-7). By squaring both sides you get 2x-5=4x-7. Solving such an equation is not difficult; x=1. But the number 1 will not be given equations. Why? Substitute one into the equation instead of the value of x. And the right and left sides will contain expressions that do not make sense, that is. This value is not valid for a square root. Therefore 1 is an extraneous root, and therefore given equation has no roots.

So, an irrational equation is solved using the method of squaring both its sides. And having solved the equation, it is necessary to cut off extraneous roots. To do this, substitute the found roots into the original equation.

Consider another one.
2х+vх-3=0
Of course, this equation can be solved using the same equation as the previous one. Move Compounds equations, which do not have a square root, in right side and then use the squaring method. solve the resulting rational equation and roots. But also another, more elegant one. Enter a new variable; vх=y. Accordingly, you will receive an equation of the form 2y2+y-3=0. That is, the usual quadratic equation. Find its roots; y1=1 and y2=-3/2. Next, solve two equations vх=1; vх=-3/2. The second equation has no roots; from the first we find that x=1. Don't forget to check the roots.

Solving identities is quite simple. To do this you need to do identity transformations until the goal is achieved. Thus, with the help of simple arithmetic operations, the problem posed will be solved.

You will need

• - paper;
• - pen.

Instructions

The simplest of such transformations are algebraic abbreviated multiplications (such as the square of the sum (difference), difference of squares, sum (difference), cube of the sum (difference)). In addition, there are many and trigonometric formulas, which are essentially the same identities.

Indeed, the square of the sum of two terms is equal to the square of the first plus twice the product of the first by the second and plus the square of the second, that is, (a+b)^2= (a+b)(a+b)=a^2+ab +ba+b ^2=a^2+2ab+b^2.

Simplify both

General principles of the solution

Repeat the textbook on mathematical analysis or higher mathematics, which is a definite integral. As is known, the solution definite integral there is a function whose derivative gives an integrand. This function is called antiderivative. Based on this principle, the main integrals are constructed.
Determine by the form of the integrand which of the table integrals fits in in this case. It is not always possible to determine this immediately. Often, the tabular form becomes noticeable only after several transformations to simplify the integrand.

Variable Replacement Method

If the integrand function is trigonometric function, whose argument contains some polynomial, then try using the variable replacement method. In order to do this, replace the polynomial in the argument of the integrand with some new variable. Based on the relationship between the new and old variables, determine the new limits of integration. By differentiating this expression, find the new differential in . So you will get the new kind of the previous integral, close to or even corresponding to any tabular one.

Solving integrals of the second kind

If the integral is an integral of the second kind, a vector form of the integrand, then you will need to use the rules for the transition from these integrals to scalar ones. One such rule is the Ostrogradsky-Gauss relation. This law allows you to move from the rotor flow to some vector function to the triple integral over the divergence of a given vector field.

Substitution of integration limits

After finding the antiderivative, it is necessary to substitute the limits of integration. First, substitute the value of the upper limit into the expression for the antiderivative. You will get some number. Next, subtract from the resulting number another number obtained from the lower limit into the antiderivative. If one of the limits of integration is infinity, then when substituting it into antiderivative function it is necessary to go to the limit and find what the expression strives for.
If the integral is two-dimensional or three-dimensional, then you will have to represent the limits of integration geometrically to understand how to evaluate the integral. Indeed, in the case of, say, a three-dimensional integral, the limits of integration can be entire planes that limit the volume being integrated.

Tasks whose solution is converting logarithmic expressions, are quite common on the Unified State Examination.

In order to successfully cope with them with minimal time, in addition to the basic logarithmic identities, you need to know and correctly use some more formulas.

This is: a log a b = b, where a, b > 0, a ≠ 1 (It follows directly from the definition of the logarithm).

log a b = log c b / log c a or log a b = 1/log b a
where a, b, c > 0; a, c ≠ 1.

log a m b n = (m/n) log |a| |b|
where a, b > 0, a ≠ 1, m, n Є R, n ≠ 0.

a log c b = b log c a
where a, b, c > 0 and a, b, c ≠ 1

To show the validity of the fourth equality, let’s take the logarithm of the left and right sides to base a. We get log a (a log with b) = log a (b log with a) or log with b = log with a · log a b; log c b = log c a · (log c b / log c a); log with b = log with b.

We have proven the equality of logarithms, which means that the expressions under the logarithms are also equal. Formula 4 has been proven.

Example 1.

Calculate 81 log 27 5 log 5 4 .

Solution.

81 = 3 4 , 27 = 3 3 .

log 27 5 = 1/3 log 3 5, log 5 4 = log 3 4 / log 3 5. Therefore,

log 27 5 log 5 4 = 1/3 log 3 5 (log 3 4 / log 3 5) = 1/3 log 3 4.

Then 81 log 27 5 log 5 4 = (3 4) 1/3 log 3 4 = (3 log 3 4) 4/3 = (4) 4/3 = 4 3 √4.

You can complete the following task yourself.

Calculate (8 log 2 3 + 3 1/ log 2 3) - log 0.2 5.

As a hint, 0.2 = 1/5 = 5 -1 ; log 0.2 5 = -1.

Answer: 5.

Example 2.

Calculate (√11) log √3 9- log 121 81 .

Solution.

Let's change the expressions: 9 = 3 2, √3 = 3 1/2, log √3 9 = 4,

121 = 11 2, 81 = 3 4, log 121 81 = 2 log 11 3 (formula 3 was used).

Then (√11) log √3 9- log 121 81 = (11 1/2) 4-2 log 11 3 = (11) 2- log 11 3 = 11 2 / (11) log 11 3 = 11 2 / ( 11 log 11 3) = 121/3.

Example 3.

Calculate log 2 24 / log 96 2 - log 2 192 / log 12 2.

Solution.

We replace the logarithms contained in the example with logarithms with base 2.

log 96 2 = 1/log 2 96 = 1/log 2 (2 5 3) = 1/(log 2 2 5 + log 2 3) = 1/(5 + log 2 3);

log 2 192 = log 2 (2 6 3) = (log 2 2 6 + log 2 3) = (6 + log 2 3);

log 2 24 = log 2 (2 3 3) = (log 2 2 3 + log 2 3) = (3 + log 2 3);

log 12 2 = 1/log 2 12 = 1/log 2 (2 2 3) = 1/(log 2 2 2 + log 2 3) = 1/(2 + log 2 3).

Then log 2 24 / log 96 2 – log 2 192 / log 12 2 = (3 + log 2 3) / (1/(5 + log 2 3)) – ((6 + log 2 3) / (1/( 2 + log 2 3)) =

= (3 + log 2 3) · (5 + log 2 3) – (6 + log 2 3)(2 + log 2 3).

After opening the parentheses and bringing similar terms, we get the number 3. (When simplifying the expression, we can denote log 2 3 by n and simplify the expression

(3 + n) · (5 + n) – (6 + n)(2 + n)).

Answer: 3.

You can complete the following task yourself:

Calculate (log 3 4 + log 4 3 + 2) log 3 16 log 2 144 3.

Here it is necessary to make the transition to base 3 logarithms and decompose into prime factors large numbers.

Answer:1/2

Example 4.

Given three numbers A = 1/(log 3 0.5), B = 1/(log 0.5 3), C = log 0.5 12 – log 0.5 3. Arrange them in ascending order.

Solution.

Let's transform the numbers A = 1/(log 3 0.5) = log 0.5 3; C = log 0.5 12 – log 0.5 3 = log 0.5 12/3 = log 0.5 4 = -2.

Let's compare them

log 0.5 3 > log 0.5 4 = -2 and log 0.5 3< -1 = log 0,5 2, так как функция у = log 0,5 х – убывающая.

Or 2< log 0,5 3 < -1. Тогда -1 < 1/(log 0,5 3) < -1/2.

Answer. Therefore, the order of placing the numbers is: C; A; IN.

Example 5.

How many integers are in the interval (log 3 1 / 16 ; log 2 6 48).

Solution.

Let us determine between which powers of the number 3 the number 1/16 is located. We get 1/27< 1 / 16 < 1 / 9 .

Since the function y = log 3 x is increasing, then log 3 (1 / 27)< log 3 (1 / 16) < log 3 (1 / 9); -3 < log 3 (1 / 16) < -2.

log 6 48 = log 6 (36 4 / 3) = log 6 36 + log 6 (4 / 3) = 2 + log 6 (4 / 3). Let's compare log 6 (4 / 3) and 1 / 5. And for this we compare the numbers 4/3 and 6 1/5. Let's raise both numbers to the 5th power. We get (4 / 3) 5 = 1024 / 243 = 4 52 / 243< 6. Следовательно,

log 6 (4 / 3)< 1 / 5 . 2 < log 6 48 < 2 1 / 5 . Числа, входящие в двойное неравенство, положительные. Их можно возводить в квадрат. Знаки неравенства при этом не изменятся. Тогда 4 < log 6 2 48 < 4 21 / 25.

Therefore, the interval (log 3 1 / 16 ; log 6 48) includes the interval [-2; 4] and the integers -2 are placed on it; -1; 0; 1; 2; 3; 4.

Answer: 7 integers.

Example 6.

Calculate 3 lglg 2/ lg 3 - lg20.

Solution.

3 lg lg 2/ lg 3 = (3 1/ lg3) lg lg 2 = (3 lо g 3 10) lg lg 2 = 10 lg lg 2 = lg2.

Then 3 lglg2/lg3 - lg 20 = lg 2 – lg 20 = lg 0.1 = -1.

Answer: -1.

Example 7.

It is known that log 2 (√3 + 1) + log 2 (√6 – 2) = A. Find log 2 (√3 –1) + log 2 (√6 + 2).

Solution.

Numbers (√3 + 1) and (√3 – 1); (√6 – 2) and (√6 + 2) are conjugate.

Let us carry out the following transformation of expressions

√3 – 1 = (√3 – 1) · (√3 + 1)) / (√3 + 1) = 2/(√3 + 1);

√6 + 2 = (√6 + 2) · (√6 – 2)) / (√6 – 2) = 2/(√6 – 2).

Then log 2 (√3 – 1) + log 2 (√6 + 2) = log 2 (2/(√3 + 1)) + log 2 (2/(√6 – 2)) =

Log 2 2 – log 2 (√3 + 1) + log 2 2 – log 2 (√6 – 2) = 1 – log 2 (√3 + 1) + 1 – log 2 (√6 – 2) =

2 – log 2 (√3 + 1) – log 2 (√6 – 2) = 2 – A.

Answer: 2 – A.

Example 8.

Simplify and find the approximate value of the expression (log 3 2 log 4 3 log 5 4 log 6 5 ... log 10 9.

Solution.

We reduce all logarithms to common ground 10.

(log 3 2 log 4 3 log 5 4 log 6 5 ... log 10 9 = (lg 2 / lg 3) (lg 3 / lg 4) (lg 4 / lg 5) (lg 5 / lg 6) · … · (lg 8 / lg 9) · lg 9 = lg 2 ≈ 0.3010 (The approximate value of lg 2 can be found using a table, slide rule or calculator).

Answer: 0.3010.

Example 9.

Calculate log a 2 b 3 √(a 11 b -3) if log √ a b 3 = 1. (In this example, a 2 b 3 is the base of the logarithm).

Solution.

If log √ a b 3 = 1, then 3/(0.5 log a b = 1. And log a b = 1/6.

Then log a 2 b 3√(a 11 b -3) = 1/2 log a 2 b 3 (a 11 b -3) = log a (a 11 b -3) / (2log a (a 2 b 3) ) = (log a a 11 + log a b -3) / (2(log a a 2 + log a b 3)) = (11 – 3log a b) / (2(2 + 3log a b)) Considering that that log a b = 1/6 we get (11 – 3 1 / 6) / (2(2 + 3 1 / 6)) = 10.5/5 = 2.1.

Answer: 2.1.

You can complete the following task yourself:

Calculate log √3 6 √2.1 if log 0.7 27 = a.

Answer: (3 + a) / (3a).

Example 10.

Calculate 6.5 4/ log 3 169 · 3 1/ log 4 13 + log125.

Solution.

6.5 4/ log 3 169 · 3 1/ log 4 13 + log 125 = (13/2) 4/2 log 3 13 · 3 2/ log 2 13 + 2log 5 5 3 = (13/2) 2 log 13 3 3 2 log 13 2 + 6 = (13 log 13 3 / 2 log 13 3) 2 (3 log 13 2) 2 + 6 = (3/2 log 13 3) 2 (3 log 13 2) 2 + 6 = (3 2 /(2 log 13 3) 2) · (2 ​​log 13 3) 2 + 6.

(2 log 13 3 = 3 log 13 2 (formula 4))

We get 9 + 6 = 15.

Answer: 15.

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Logarithmic expressions, solving examples. In this article we will look at problems related to solving logarithms. The tasks ask the question of finding the meaning of an expression. It should be noted that the concept of logarithm is used in many tasks and understanding its meaning is extremely important. As for the Unified State Exam, the logarithm is used when solving equations, in applied problems, and also in tasks related to the study of functions.

Let us give 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 equal to the sum of the logarithms of the factors.

* * *

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

* * *

*The logarithm of an exponent is equal to the product of the exponent and the logarithm of its base.

* * *

*Transition to a new foundation

* * *

More properties:

* * *

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

Let's list 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 example:

A corollary from this property:

* * *

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

* * *

As you have seen, the concept of a logarithm itself is simple. The main thing is what is needed good practice, which gives a certain skill. Of course, knowledge of formulas is required. If the skill in converting elementary logarithms has not been developed, then when solving simple tasks It's easy to make a mistake.

Practice, solve the simplest examples from the mathematics course first, then move on to more complex ones. In the future, I will definitely show how “scary” logarithms are solved; they won’t appear on the Unified State Examination, but they are of interest, don’t miss them!

That's all! Good luck to you!

Sincerely, Alexander Krutitskikh

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

What is a logarithm?

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

What is a logarithm? How to solve logarithms? These questions confuse many graduates. Traditionally, the topic of logarithms is considered complex, incomprehensible and scary. Especially equations with logarithms.

This is absolutely not true. Absolutely! Don't believe me? Fine. Now, in just 10 - 20 minutes you:

1. You will understand what is a logarithm.

2. Learn to solve a whole class exponential equations. Even if you haven't heard anything about them.

3. Learn to calculate simple logarithms.

Moreover, for this you will only need to know the multiplication table and how to raise a number to a power...

I feel like you have doubts... Well, okay, mark the time! Go!

First, solve this equation in your head:

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

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

Definition in mathematics

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

Types of logarithms

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

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

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

Rules and some restrictions

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

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

How to solve logarithms?

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

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

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

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

Equations and inequalities

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

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

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

Basic theorems about logarithms

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

1. The main identity looks like this: a logaB =B. It applies only when a is greater than 0, not equal to one, and B is greater than zero.
2. The logarithm of the product can be represented in the following formula: log d (s 1 *s 2) = log d s 1 + log d s 2. In this case 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.
3. The logarithm of the quotient looks like this: log a (s 1/ s 2) = log a s 1 - log a s 2.
4. The theorem in the form of a formula takes the following form: log a q b n = n/q log a b.

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

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

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

Examples of problems and inequalities

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

Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, however, it can be applied to every mathematical inequality or logarithmic equation certain rules. 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 deciding logarithmic equations, we should 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. For solutions natural logarithms need to apply logarithmic identities or their properties. Let's look at examples of solving logarithmic problems of various types.

How to Use Logarithm Formulas: With Examples and Solutions

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

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

Assignments from the Unified State Exam

Logarithms are often found in entrance exams, especially 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.