# Exponential Equation

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

## Example2^{3x - 1} = 4

To solve an exponential equation,

make the bases of both sides the same.

The base of the left side 2^{3x - 1} is 2.

So change the base of the right side to 2:

4 = 2^{2}.

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.

## Example3^{x - 5} = 9^{4}

The base of the left side 3^{x - 5} is 3.

So change the base of the right side to 3.

9 = 3^{2}

(3^{2})^{4} = 3^{2⋅4} = 3^{8}

Power of a Power

3^{x - 5} = 3^{8}

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.

## Example7^{4x + 8} = 1

The base of the left side 7^{4x + 8} is 7.

So change the base of the right side to 7.

1 = 7^{0}

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.

## Example125⋅5^{x} = (1/25)^{x}

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

So change the bases of the numbers to 5.

125 = 5^{3}

(1/25)^{x} = 25^{-x}

Negative Exponent

5^{3}⋅5^{3} = 5^{3 + x}

Product of Powers

25^{-x} = (5^{2})^{-x}

(5^{2})^{-x} = 5^{-2x}

5^{3 + 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.

## Example4^{x} = 3⋅2^{x} + 4

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.

4^{x} = (2^{2})^{x} = 2^{2x}

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

Think 2^{x} as a variable

and factor (2^{x})^{2} x - 3⋅2^{x} - 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 (2^{x} - 4)(2^{x} + 1) = 0.

Factor a Quadratic Trinomial

Move -4 to the right side.

Then 2^{x} = 4.

Change the base of 4 to 2.

4 = 2^{2}

2^{x} = 2^{2}

So x = 2.

This is the answer for case 1.

Case 2: 2^{x} + 1 = 0

Move +1 to the right side.

Then 2^{x} = -1.

See 2^{x} = -1.

The left side 2^{x} is always plus.

(The graph of y = a^{x} 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 (2^{x} - 4)(2^{x} + 1) = 0 is

x = 2.

So x = 2 is the answer.