Logarithmic Equation

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

Formula

Formula

For loga x,
x > 0
0 < a < 1, a > 1.

The solution of a logarithmic equation
should satisfy these conditions.

Example 1

Example

Solution

See log3 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.

Then solve the equation.

log3 x = 4

The exponent is 4.
The base is 3.

So x = 34.

Logarithmic Form

34
= 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 logx 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.

logx 64 = 3

The exponent is 3.
The base is x.

So x3 = 64.

64 = 43

x3 = 43

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 log5 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 (log3 (log5 x)) as a whole.

Then the given equation is
log2 [whole] = 0.

The exponent is 0.
The base is 2.

Then [whole] = 20.

So log3 (log5 x) = 20.

20 = 1

log3 (log5 x) = 1

Think (log5 x) as a whole.

Then log3 [whole] = 1.

The exponent is 1.
The base is 3.

Then [whole] = 31.

So log5 x = 31.

31 = 3

log5 x = 3

The exponent is 3.
The base is 5.

So x = 53.

53 = 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 log2 x.

x is in the log.

So x > 0.

Next, see log2 (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 log2.

So combine the left side terms.

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

Logarithm of a Product

log2 x(x - 1) = log2 12

The bases of the logs are the same:
2.

Then the numbers in the logs are the same.

So x(x - 1) = 12.

Move 12 to the left side.

Factor the left side x2 - 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

Solve the quadratic equation for each case.

Case 1: x - 4 = 0
Then 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.

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 log5 x.

x is in the log.

So x > 0.

Next, see log5 x2.

x2 is in the log.

So x2 > 0.

x2 > 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.

log5 x2 = 2 log5 x

There's (log5 x)2 term.

So you cannot combine the terms
like the last example.

Then, think (log5 x) as a variable
and factor [log5 x]2 x - 2[log5 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 ([log5 x] - 3)([log5 x] + 1) = 0.

Solve the quadratic equation for each case.

Case 1: log5 x - 3 = 0

Move -3 to the right side.

Then log5 x = 3.

log5 x = 3

The exponent is 3.
The base is 5.

Then x = 53.

53
= 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: log5 x + 1 = 0

Move +1 to the right side.

Then log5 x = -1.

log5 x = -1

The exponent is -1.
The base is 5.

Then x = 5-1.

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.