Lesson Solving logarithmic equations

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Solving logarithmic equations

Logarithmic equations contain the unknown variable under the logarithmic function.

The simplest logarithmic equation has the form

log%28a%2Cf%28x%29%29+=+log%28a%2Cg%28x%29%29

log%28a%2Cf%28x%29%29+=+log%28a%2Cg%28x%29%29

,

where

is a real positive number not equal to 1.

To solve the equation log%28a%2Cf%28x%29%29+=+log%28a%2Cg%28x%29%29

, you need
1) to solve the equation f%28x%29+=+g%28x%29

;
2) to check found roots for conditions f%28x%29%3E0

.
Only those roots that satisfy these conditions, are the solutions of the original equations.
Other roots are not solutions of the original equations.

There are two major methods for solving logarithmic equations:
1) A method of transforming the original equation to the form log%28a%2Cf%28x%29%29+=+log%28a%2Cg%28x%29%29

and then replacing it by the equation f%28x%29+=+g%28x%29

, and
2) A method that introduces a new variable.

Examples below show how to use these methods.

Example 1

Solve the equation log%285%2C%28x%5E2-7x%2B11%29%29+=+log%285%2C%285-2x%29%29

.

Solution
First, pass from the given equation to the equation
x%5E2-7x%2B11+=+5-2x

.
Then, to simplify this equation move terms from the right side to the left side with the opposite signs and collect common terms. As a result, you get the quadratic equation
x%5E2+-+5x+%2B+6+=+0

.
You can solve this equation by applying the quadratic formula (see the lesson Introduction into Quadratic Equations)
x+=+%285%2B-sqrt%285%5E2-4%2A1%2A6%29%29%2F2+=+%285%2B-sqrt%2825-24%29%29%2F2+=+%285%2B-sqrt%281%29%29%2F2

x+=+%285%2B-sqrt%285%5E2-4%2A1%2A6%29%29%2F2+=+%285%2B-sqrt%2825-24%29%29%2F2+=+%285%2B-sqrt%281%29%29%2F2

.
Two roots are x%5B1%5D+=+3

and

.
Now check these two roots for conditions x%5E2-7x%2B11+%3E+0

and

.
Value

do not satisfies these conditions, while value x=2

does.

Answer: x=2

.

Exercise 1
Solve yourself the equation log%283%2C%28x%5E2-3x-5%29%29+=+log%283%2C%287-2x%29%29

Example 2

Solve the equation log%2810%2C%28x%2B6%29%29+%2B+log%2810%2C%282x%2B3%29%29+=+log%2810%2C%287%2B2x%29%29

.
Solution
Using logarithm properties (sum of logarithms is equal to the logarithm of the product; see for example the lesson Properties of the logarithm),
you can transform the original equation into the form
log%2810%2C%28%28x%2B6%29%2A%282x%2B3%29%29%29+=+log%2810%2C%287%2B2x%29%29

log%2810%2C%28%28x%2B6%29%2A%282x%2B3%29%29%29+=+log%2810%2C%287%2B2x%29%29

,
Then consider the equation
%28x%2B6%29%2A%282x%2B3%29+=+7%2B2x

.

Step by step simplify this equation. Open brackets in the left side, move terms from the right side to the left side with the opposite signs and collect common terms:
2x%5E2%2B12x%2B3x%2B18+=+7%2B2x

.

Eventually, you get the quadratic equation, which you can solve using quadratic formula (see the lesson Introduction into Quadratic Equations)
x+=+%28-13%2B-sqrt%2813%5E2-4%2A2%2A11%29%29%2F%282%2A2%29+=+%28-13%2B-sqrt%28169-88%29%29%2F4+=+%28-13%2B-sqrt%2881%29%29%2F4