Presentation for a mathematics lesson "solving logarithmic equations." Presentation on the topic "logarithmic equations" Presentation of exponential and logarithmic equations

Counting and calculations are the basis of order in the head

Johann Heinrich Pestalozzi

Find errors:

• log 3 24 – log 3 8 = 16
• log 3 15 + log 3 3 = log 3 5
• log 5 5 3 = 2
• log 2 16 2 = 8
• 3log 2 4 = log 2 (4*3)
• 3log 2 3 = log 2 27
• log 3 27 = 4
• log 2 2 3 = 8

Calculate:

• log 2 11 – log 2 44
• log 1/6 4 + log 1/6 9
• 2log 5 25 +3log 2 64

Find x:

• log 3 x = 4
• log 3 (7x-9) = log 3 x

Peer review

True equalities

Calculate

-2

-2

22

Find x

Results of oral work:

“5” - 12-13 correct answers

“4” - 10-11 correct answers

“3” - 8-9 correct answers

“2” - 7 or less

Find x:

• log 3 x = 4
• log 3 (7x-9) = log 3 x

Definition

• An equation containing a variable under the logarithm sign or in the base of the logarithm is called logarithmic

For example, or

• If an equation contains a variable that is not under the logarithmic sign, then it will not be logarithmic.

For example,

Are not logarithmic

Are logarithmic

1. By definition of logarithm

The solution to the simplest logarithmic equation is based on applying the definition of logarithm and solving the equivalent equation

Example 1

2. Potentization

By potentiation we mean the transition from an equality containing logarithms to an equality not containing them:

Having solved the resulting equality, you should check the roots,

because the use of potentiation formulas expands

domain of equation

Example 2

Solve the equation

Potentiating, we get:

Examination:

If

Answer

Example 2

Solve the equation

Potentiating, we get:

is the root of the original equation.

REMEMBER!

Logarithm and ODZ

together

are working

everywhere!

Sweet couple!

Two of a Kind!

HE

- LOGARITHM !

SHE

-

ODZ!

Two in one!

Two banks of one river!

We can't live

friend without

friend!

Close and inseparable!

3. Application of the properties of logarithms

Example 3

Solve the equation

0 Moving on to the variable x, we get: ; x = 4 satisfy the condition x 0, therefore, the roots of the original equation. "width="640"

4. Introduction of a new variable

Example 4

Solve the equation

Moving on to the variable x, we get:

; X = 4 satisfy the condition x 0 therefore

roots of the original equation.

Determine the method for solving the equations:

Applying

holy of logarithms

A-priory

Introduction

new variable

Potentiation

The nut of knowledge is very hard,

But don't you dare back down.

“Orbit” will help you chew it,

And pass the knowledge exam.

№ 1 Find the product of the roots of the equation

4) 1,21

3) 0 , 81

2) - 0,9

1) - 1,21

№ 2 Specify the interval to which the root of the equation

1) (- ∞;-2]

3)

2) [ - 2;1]

4) }