# Logarithmic Equations

How to solve logarithmic equations: examples, and their solutions.

## Domain and Base of a Logarithm

For log_{[base]} [domain],

the domain (blue) and the base (brown)

satisfy the below conditions:

[domain] > 0

0 < [base] < 1, [base] > 1.

## Example 1: Solve log_{3} *x* = 4

Before solving the equation,

write the condition of the domain.

The domain of log_{3} *x* is*x*.

So *x* > 0.

Solve the given equation.

Solve the log.

(base: 3)

Then *x* = 3^{4}.

Logarithmic form

3^{4} = 81

So *x* = 81.

Draw [*x* > 0] on a number line.

(from the above condition)

And draw [*x* = 81] on the number line.

81 is in the colored region.

So *x* = 81.

## Example 2: Solve log_{x} 64 = 3

Before solving the equation,

write the condition of the base.

The domain of log_{x} 64 is*x*.

So 0 < *x* < 1, *x* > 1.

Solve the given equation.

Solve the log.

(base: *x*)

Then 64 = *x*^{3}.

Switch both sides.

64 = 4^{3}

*x*^{3} = 4^{3}

So *x* = 4.

The exponent of *x*^{3} is 3:

which is odd.

So you don't have to write ± sign.

Draw [0 < *x* < 1, *x* > 1] on a number line.

(from the above condition)

And draw [*x* = 4] on the number line.

4 is in the colored region.

So *x* = 4.

## Example 3: Solve log_{2} (log_{3} (log_{5} *x*)) = 0

Before solving the equation,

write the condition of the domain.

The domain of log_{5} *x* is*x*.

So *x* > 0.

Solve the given equation.

Think log_{3} (log_{5} *x*) as the base

and solve the log.

(base: 2)

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

2^{0} = 1

Zero exponent

Think log_{5} *x* as the base

and solve the log.

(base: 3)

Then log_{5} *x* = 3^{1}.

3^{1} = 3

Solve the log.

(base: 5)

Then *x* = 5^{3}.

5^{3} = 125

So *x* = 125.

Draw [*x* > 0] on a number line.

(from the above condition)

And draw [*x* = 125] on the number line.

125 is in the colored region.

So *x* = 125.

## Example 4: Solve log_{2} *x* + log_{2} (*x* - 1) = log_{2} 12

Before solving the equation,

write the conditions of the domains.

First, the domain of log_{2} *x* is*x*.

So *x* > 0.

Next, the domain of log_{2} (*x* - 1) is

(*x* - 1).

So *x* - 1 > 0.

So *x* > 1.

*x* > 0*x* > 1

Find the intersection of these two inequalities

by drawing the inequalities

on a number line.

Then *x* > 1.

Solve the given equation.

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

Logarithm of a product

The bases of the logs on both sides

are the same: 2.

Then *x*(*x* - 1) = 12.

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

Common monomial factor

Move 12 to the left side.

Factor *x*^{2} - *x* - 12.

Factor a quadratic trinomial

Find a pair of numbers

whose product is the constant term [-12]

and whose sum is the middle term's coefficient [-1].

The constant term is (-).

So the signs of the numbers are different:

one is (+), and the other is (-).

(-12, 1) and (-6, 2)

are not the right numbers.

[-12] = -4⋅3

-4 + 3 = [-1]

So -4 and 3 are the right numbers.

Use -4 and +3

to write the factored form:

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

Solve the equation.

Solving a quadratic equation by factoring

1) *x* - 4 = 0

So *x* = 4.

2) *x* + 3 = 0

So *x* = -3.

So *x* = 4, -3.

Draw [*x* > 1] on a number line.

(from the above condition)

And draw [*x* = 4, -3] on the number line.

4 is in the colored region.

But -3 is not in the colored region.

So -3 cannot be the answer.

So *x* = 4.

## Example 5: Solve (log_{5} *x*)^{2} - log_{5} *x*^{2} - 3 = 0

Before solving the equation,

write the conditions of the domains.

First, the domain of log_{5} *x* is *x*.

So *x* > 0.

Next, the domain of log_{5} *x*^{2} is*x*^{2}.

So *x*^{2} > 0.

So *x* ≠ 0.

*x* > 0*x* ≠ 0

The intersection of these two is*x* > 0.

Solve the given equation.

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

Logarithm of a power

Think log_{5} *x* as a variable.

Then (log_{5} *x*)^{2} - 2 log_{5} *x* - 3 = 0

is a quadratic equation.

Factor the left side.

Factor a quadratic trinomial

Find a pair of numbers

whose product is the constant term [-3]

and whose sum is the middle term's coefficient [-2].

The constant term is (-).

So the signs of the numbers are different:

one is (+), and the other is (-).

-1 and 3

are not the right numbers.

[-3] = -3⋅1

-3 + 1 = [-2]

So -3 and 1 are the right numbers.

Use -3 and +1

to write the factored form:

(log_{5} *x* - 3)(log_{5} *x* + 1) = 0.

Solve the equation.

1) log_{5} *x* - 3 = 0

So log_{5} *x* = 3.

Solve the log.

(base: 5)

Then *x* = 5^{3}.

5^{3} = 125

So, for case 1,*x* = 125.

2) log_{5} *x* + 1 = 0

So log_{5} *x* = -1.

Solve the log.

(base: 5)

Then *x* = 5^{-1}.

5^{-1} = 1/5

Negative exponent

So, for case 2,*x* = 1/5.

So *x* = 1/5, 125.

Draw [*x* > 0] on a number line.

(from the above condition)

And draw [*x* = 1/5, 125] on the number line.

Both 1/5 and 125 are in the colored region.

So *x* = 1/5, 125.