A message on the topic of converting expressions containing logarithms. Properties of logarithms and examples of their solutions

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 factorization of large numbers into prime factors.

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.

Let us reduce all logarithms to a common base 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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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 factorization of large numbers into prime factors.

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.

Let us reduce all logarithms to a common base 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.

Still have questions? Not sure how to find the value of a logarithmic expression?
To get help from a tutor -.
The first lesson is free!

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Lesson type: lesson of generalization and systematization of knowledge

Goals:

• update students' knowledge about logarithms and their properties as part of general repetition and preparation for the Unified State Exam;
• promote the development of students’ mental activity, skills in applying theoretical knowledge when performing exercises;
• promote the development of students’ personal qualities, self-control skills and self-assessment of their activities; cultivate hard work, patience, perseverance, and independence.

Equipment: computer, projector, presentation (Annex 1), cards with homework (you can attach a file with the assignment in the electronic diary).

During the classes

I. Organizational moment. Greetings, get ready for the lesson.

II. Discussion of homework.

III. State the topic and purpose of the lesson. Motivation.(Slide 1) Presentation.

We continue our general review of the mathematics course in preparation for the Unified State Exam. And today in lesson we will talk about logarithms and their properties.

Tasks for calculating logarithms and converting logarithmic expressions are necessarily present in control and measurement materials of both the basic and profile levels. Therefore, the goal of our lesson is to restore ideas about the meaning of the concept “logarithm” and update the skills of converting logarithmic expressions. Write down the topic of the lesson in your notebooks.

IV. Updating knowledge.

1. /Orally/ First, let's remember what is called a logarithm. (Slide 2)

(The logarithm of a positive number b to base a (where a > 0, a?1) is the exponent to which the number a must be raised to obtain the number b)

Log a b = n<->a n = b, (a> 0, a 1, b> 0)

So, “LOGARITHM” is an “EXPONSOR”!

(Slide 3) Then a n = b can be rewritten in the form = b – basic logarithmic identity.

If the base a = 10, then the logarithm is called decimal and is denoted lgb.

If a = e, then the logarithm is called natural and is denoted lnb.

2. /In writing/ (Slide 4) Fill in the blanks to get the correct equations:

Log? x + Log a ? =Log? (?y)

Log a? - Log? y = Log ? (x/?)

Log a x ? = pLog? (?)

Examination:

1; 1; a,y,x; x,a,a,y; p,a,x.

These are properties of logarithms. And another group of properties: (Slide 5)

Examination:

a,1,n,x; n,x,p,a; x,b,a,y; a,x,b; a,1,b.

V. Oral work

(Slide 6) No. 1. Calculate:

a B C D) ; d) .

Answers : a) 4; b) – 2; at 2; d) 7; e) 27.

(Slide 7) No. 2. Find X:

A) ; b) (Answers: a) 1/4; b) 9).

No. 3. Does it make sense to consider such a logarithm:

A) ; b) ; V) ? (No)

VI. Independent work in groups, strong students - consultants. (Slide 8)

No. 1. Calculate: .

#2: Simplify:

No. 3. Find the value of the expression if

No. 4. Simplify the expression:

No. 5. Calculate:

No. 7. Calculate:

No. 8. Calculate:

After completion, check and discussion using the prepared solution or using a document camera.

VII. Solving a task of increased complexity(strong student on the board, the rest in notebooks) (Slide 9)

Find the meaning of the expression:

VIII. Homework (on cards) is differentiated.(Slide 10)

No. 1. Calculate:

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 the 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 a complex function is given, then it is necessary to multiply the derivative of the 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 at a 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 is the difference between an irrational equation and a rational one? If the unknown variable is under the square root sign, 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 this 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, to the 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, an ordinary 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, it is necessary to carry out identical transformations until the set 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 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

Variable Replacement Method

Solving integrals of the second kind

Substitution of integration limits