Exponential Equation

How to solve an exponential equation: 5 examples and their solutions.

Example 1

Example

Solution

To solve an exponential equation,
make the bases of both sides the same.

The base of the left side 23x - 1 is 2.

So change the base of the right side to 2:
4 = 22.

The bases of both sides are the same.
Then the exponents of both sides are the same.

So 3x - 1 = 2.

Solve 3x - 1 = 2.

Move -1 to the right side.

Divide both sides by 3.
Then x = 1.

So x = 1.

Example 2

Example

Solution

The base of the left side 3x - 5 is 3.
So change the base of the right side to 3.

9 = 32

(32)4 = 32⋅4 = 38

Power of a Power

3x - 5 = 38

The bases of both sides are the same.
Then the exponents of both sides are the same.

So x - 5 = 8.

Solve x - 5 = 8.

Move -5 to the right side.
Then x = 13.

So x = 13.

Example 3

Example

Solution

The base of the left side 74x + 8 is 7.
So change the base of the right side to 7.

1 = 70

Zero Exponent

The bases of both sides are the same.
Then the exponents of both sides are the same.

So 4x + 8 = 0.

Solve 4x + 8 = 0.

Move +8 to the right side.

Divide both sides by 4.
Then x = -2.

So x = -2.

Example 4

Example

Solution

The bases of the numbers are 125, 5, and 1/25.
So change the bases of the numbers to 5.

125 = 53

(1/25)x = 25-x

Negative Exponent

53⋅53 = 53 + x

Product of Powers

25-x = (52)-x

(52)-x = 5-2x

53 + x = 5-2x

The bases of both sides are the same.
Then the exponents of both sides are the same.

So 3 + x = -2x.

Solve 3 + x = -2x.

Move 3 to the right side.
Move -2x to the left side.

Then, x + 2x, 3x is equal to -3.

Divide both sides by 3.
Then x = -1.

So x = -1.

Example 5

Example

Solution

The given equation has 3 terms,
not 2 terms.

So you can't directly compare the exponents
like the previous examples.

To solve this equation,
first move the right side terms to the left side.

4x = (22)x = 22x

22x = (2x)2

Think 2x as a variable
and factor (2x)2 x - 3⋅2x - 4 = 0.

Find a pair of numbers
whose product is the constant term -4
and whose sum is the coefficient of the middle term -3.

-4⋅1 = -4
-4 + 1 = -3

So (2x - 4)(2x + 1) = 0.

Factor a Quadratic Trinomial

Solve the quadratic equation for each case.

Case 1: 2x - 4 = 0

Move -4 to the right side.
Then 2x = 4.

Change the base of 4 to 2.
4 = 22

2x = 22

So x = 2.

This is the answer for case 1.

Case 2: 2x + 1 = 0

Move +1 to the right side.
Then 2x = -1.

See 2x = -1.

The left side 2x is always plus.
(The graph of y = ax is always above the x-axis.)

But the right side -1 is minus.

So this equation has no solution.

This is the answer for case 2.

For case 1, x = 2.
For case 2, there's no solution.

So the solution of (2x - 4)(2x + 1) = 0 is
x = 2.

So x = 2 is the answer.