Logarithmic Equations

Logarithmic Equations

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

Domain and Base of a Logarithm

The domain is greater than 0. The base satisfies 0 < (base) < 1, (base) > 1.

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 log3 x = 4

Solve the given equation. log_3 x = 4

Before solving the equation,
write the condition of the domain.

The domain of log3 x is
x.

So x > 0.

Solve the given equation.

Solve the log.
(base: 3)

Then x = 34.

Logarithmic form

34 = 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 logx 64 = 3

Solve the given equation. log_x 64 = 3

Before solving the equation,
write the condition of the base.

The domain of logx 64 is
x.

So 0 < x < 1, x > 1.

Solve the given equation.

Solve the log.
(base: x)

Then 64 = x3.

Switch both sides.

64 = 43

x3 = 43

So x = 4.

The exponent of x3 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 log2 (log3 (log5 x)) = 0

Solve the given equation. log_2 (log_3 (log_5 x)) = 0

Before solving the equation,
write the condition of the domain.

The domain of log5 x is
x.

So x > 0.

Solve the given equation.

Think log3 (log5 x) as the base
and solve the log.
(base: 2)

Then log3 (log5 x) = 20.

20 = 1

Zero exponent

Think log5 x as the base
and solve the log.
(base: 3)

Then log5 x = 31.

31 = 3

Solve the log.
(base: 5)

Then x = 53.

53 = 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 log2 x + log2 (x - 1) = log2 12

Solve the given equation. log_2 x + log_2 (x - 1) = log_2 12

Before solving the equation,
write the conditions of the domains.

First, the domain of log2 x is
x.

So x > 0.

Next, the domain of log2 (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.

log2 x + log2 (x - 1) = log2 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) = x2 - x

Common monomial factor

Move 12 to the left side.

Factor x2 - 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 (log5 x)2 - log5 x2 - 3 = 0

Solve the given equation. (log_5 x)^2 - log_5 x^2 - 3 = 0

Before solving the equation,
write the conditions of the domains.

First, the domain of log5 x is x.

So x > 0.

Next, the domain of log5 x2 is
x2.

So x2 > 0.

So x ≠ 0.

x > 0
x ≠ 0

The intersection of these two is
x > 0.

Solve the given equation.

log5 x2 = 2 log2 x

Logarithm of a power

Think log5 x as a variable.

Then (log5 x)2 - 2 log5 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:
(log5 x - 3)(log5 x + 1) = 0.

Solve the equation.

1) log5 x - 3 = 0
So log5 x = 3.

Solve the log.
(base: 5)

Then x = 53.

53 = 125

So, for case 1,
x = 125.

2) log5 x + 1 = 0
So log5 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.