# Logarithmic Equation

How to solve the logarithmic equation: formula, 5 examples, and their solutions.

## Formula

### Formula

For log_{a} x,

x > 0

0 < a < 1, a > 1.

The solution of a logarithmic equation

should satisfy these conditions.

## Example 1

### Example

### Solution

See log_{3} x.

x is in the log.

So x > 0.

Draw x > 0

on a number line.

Color the region.

So x should be in this colored region.

Before solving the equation,

write and draw the condition like this.

3^{4}

= 3⋅3⋅3⋅3

= 9⋅9

= 81

Power

x = 81

See if x = 81 is in the colored region.

Draw x = 81 on the number line.

x = 81 is in the colored region.

So x = 81 is the root.

So x = 81.

## Example 2

### Example

### Solution

See log_{x} 64.

x is the base.

So 0 < x < 1, x > 1.

Draw 0 < x < 1, x > 1

on a number line.

Color the region.

So x should be in this colored region.

Next, solve the equation.

log_{x} 64 = 3

The exponent is 3.

The base is x.

So x^{3} = 64.

64 = 4^{3}

x^{3} = 4^{3}

Cube root both sides.

Then x = 4.

x = 4

See if x = 4 is in the colored region.

Draw x = 4 on the number line.

x = 4 is in the colored region.

So x = 4 is the root.

So x = 4.

## Example 3

### Example

### Solution

See log_{5} x.

x is in the log.

So x > 0.

Draw x > 0

on a number line.

Color the region.

So x should be in this colored region.

Next, solve the equation.

Think (log_{3} (log_{5} x)) as a whole.

Then the given equation is

log_{2} [whole] = 0.

The exponent is 0.

The base is 2.

Then [whole] = 2^{0}.

So log_{3} (log_{5} x) = 2^{0}.

2^{0} = 1

log_{3} (log_{5} x) = 1

Think (log_{5} x) as a whole.

Then log_{3} [whole] = 1.

The exponent is 1.

The base is 3.

Then [whole] = 3^{1}.

So log_{5} x = 3^{1}.

3^{1} = 3

log_{5} x = 3

The exponent is 3.

The base is 5.

So x = 5^{3}.

5^{3} = 125

x = 125

See if x = 125 is in the colored region.

Draw x = 125 on the number line.

x = 125 is in the colored region.

So x = 125 is the root.

So x = 125.

## Example 4

### Example

### Solution

See log_{2} x.

x is in the log.

So x > 0.

Next, see log_{2} (x - 1).

(x - 1) is in the log.

So x - 1 > 0.

Move -1 to the right side.

Then x > 1.

x > 0

x > 1

Draw these two inequalities

on a number line.

Color the intersecting region.

The colored region is x > 1.

So x should be in this colored region.

Next, solve the equation.

Every term has log_{2}.

So combine the left side terms.

log_{2} x + log_{2} (x - 1) = log_{2} x(x - 1).

Logarithm of a Product

log_{2} x(x - 1) = log_{2} 12

The bases of the logs are the same:

2.

Then the numbers in the logs are the same.

So x(x - 1) = 12.

x(x - 1) = x^{2} - x

Multiply a Monomial and a Polynomial

Move 12 to the left side.

Factor the left side x^{2} - x - 12.

Find a pair of numbers

whose product is the constant term -12

and whose sum is the coefficient of the middle term -1.

-4⋅3 = -12

-4 + 3 = -1

So (x - 4)(x + 3) = 0.

Factor a Quadratic Trinomial

See if x = 4 is in the colored region.

Draw x = 4 on the number line.

x = 4 is in the colored region.

So x = 4 is the root.

Case 2: x + 3 = 0

Then x = -3.

See if x = -3 is in the colored region.

Draw x = -3 on the number line.

x = -3 is not in the colored region.

So x = -3 is not the root.

Case 1: x = 4

Case 2: No root.

So x = 4.

So x = 4.

## Example 5

### Example

### Solution

See log_{5} x.

x is in the log.

So x > 0.

Next, see log_{5} x^{2}.

x^{2} is in the log.

So x^{2} > 0.

x^{2} > 0 means

x ≠ 0.

x > 0

x ≠ 0

Draw these two

on a number line.

Color the intersecting region.

The colored region is x > 0.

So x should be in this colored region.

Next, solve the equation.

log_{5} x^{2} = 2 log_{5} x

There's (log_{5} x)^{2} term.

So you cannot combine the terms

like the last example.

Then, think (log_{5} x) as a variable

and factor [log_{5} x]^{2} x - 2[log_{5} x] - 3 = 0.

Find a pair of numbers

whose product is the constant term -3

and whose sum is the coefficient of the middle term -2.

-3⋅1 = -3

-3 + 1 = -2

So ([log_{5} x] - 3)([log_{5} x] + 1) = 0.

Solve the quadratic equation for each case.

Case 1: log_{5} x - 3 = 0

Move -3 to the right side.

Then log_{5} x = 3.

log_{5} x = 3

The exponent is 3.

The base is 5.

Then x = 5^{3}.

5^{3}

= 5⋅5⋅5

= 25⋅5

= 125

x = 125

See if x = 125 is in the colored region.

Draw x = 125 on the number line.

x = 125 is in the colored region.

So x = 125 is the root

for case 1.

Case 2: log_{5} x + 1 = 0

Move +1 to the right side.

Then log_{5} x = -1.

log_{5} x = -1

The exponent is -1.

The base is 5.

Then x = 5^{-1}.

5^{-1} = 1/5

Negative Exponent

x = 1/5

See if x = 1/5 is in the colored region.

Draw x = 1/5 on the number line.

x = 1/5 is in the colored region.

So x = 1/5 is the root

for case 2.

Case 1: x = 125

Case 2: x = 1/5

So x = 1/5, 125.

So x = 1/5, 125.