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How to solve exponential equations

Examples of Exponential Equations

$$ 2^{\red x} = 4 \\ 8^{\red{2x}} = 16 \\ \\ 16^{\red { x+1}} = 256 \\ \left( \frac{1}{2} \right)^{\red { x+1}} = 512 $$

As you might've noticed, an exponential equation is just a special type of equation. It's an equation that has exponents that are $$ \red{ variables}$$.

Steps to Solve

There are different kinds of exponential equations. We will focus on exponential equations that have a single term on both sides. These equations can be classified into 2 types.

Type #1: Same Bases like :

$$ 4^x = 4^9 $$.


Type #2: Different Bases like:

$$ 4^3 = 2^x $$.

$$ \left( \frac{1}{4} \right)^x = 32 $$ (Part II below)

Part I. Solving Exponential Equations with Same Base

Example 1

Solve: $$ 4^{x+1} = 4^9 $$

Step 1

Ignore the bases, and simply set the exponents equal to each other

$$ x + 1 = 9 $$

Step 2

Solve for the variable

$$ x = 9 - 1 \\ x = \fbox { 8 } $$

Check

We can verify that our answer is correct by substituting our value back into the original equation . .

$$ 4^{x+1} = 4^9 \\ 4^{\red{8}+1} = 4^9 $$
$$ 4^{\red{9} } = 4^9 $$

Exponential Equation Solver

Enter any exponential equation into the algebra solver below :

Example 2
Example 3

II. Solving Exponential Equations with un-like bases

What do they look like?

$$ \red 4^3 = \red 2^x $$
$$ \red 9^x = \red { 81 } $$
$$ \left( \red{\frac{1}{2}} \right)^{ x+1} = \red 4^3 $$
$$ \red 4^{2x} +1 = \red { 65 } $$

In each of these equations, the base is different. Our goal will be to rewrite both sides of the equation so that the base is the same.

Example 4

Solve: $$ 4^{3} = 2^x $$

Step 1

Forget about the exponents for a minute and focus on the bases:
Rewrite the bases as powers of a common base. Do this by asking yourself :

Answer: They are both powers of 2

Step 2

Rewrite equation so that both exponential expressions use the same base

$$ \red 4^{3} = 2^x \\ (\red {2^2})^{3} = 2^x $$

Step 3

Use exponents laws to simplify

$$ (\red {2^2})^{3} = 2^x \\ (2^\red {2 \cdot 3 }) = 2^x \\ (2^\red 6 ) = 2^x $$

Step 4

$$ (2^\red 6 ) = 2^x \\ x = \fbox{6} $$

Check Your work

Substitute $$\red 6 $$ into the original equation to verify our work.

$$ 4^{3} = 2^{\red 6} $$
$$ 64 = 64 $$

Example with Negative Exponent

Unlike bases often involve negative or fractional bases like the example below. We are going to treat these problems like any other exponential equation with different bases--by converting the bases to be the same.

Example 5

Practice Problems (un-like bases)

Problem 1

Solve the following exponential Equation: $$9^x = 81$$

Step 1

Forget about the exponents for a minute and focus on the bases:
Rewrite the bases as powers of a common base. Ask yourself :

You can use either 3 or 9. I will use 9.

$ \\ 81 = \red 9 ^{\blue 2} \\ 9 = \red 9 ^{\blue 1} \\ $

Step 2

Substitute the rewritten bases into original equation

$$ (\red 9^{\blue 1})^x = \red 9^{\blue 2} $$

Step 3

Use exponents laws to simplify

$$ (\red 9^{\blue 1})^x = \red 9^{\blue 2} \\ 9^{1 \cdot x } = 9 ^{2} \\ 9^{x } = 9 ^{2} $$

Step 4

$$ x = 2 $$

Problem 2

Solve the equation : $$ 4^{2x} +1 = 65 $$

Step 1

Rewrite this equation so that it looks like the other ones we solved. Isolate the exponential expression as follows:

$$ 4^{2x} +1 \red{-1} = 65\red{-1} \\ 4^{2x} = 64 $$


Forget about the exponents for a minute and focus on the bases:
Rewrite the bases as powers of a common base. Ask yourself :

They are both powers of 2 and of 4. You could use either base to solve this. I will use base 4

$ \\ 64 = \red 4 ^{\blue 3} \\ 4 = \red 4 ^{\blue 1} \\ $

Step 2

Substitute the rewritten bases into original equation

$$ 4^{2x} = 64 \\ \red 4^{\blue{ 2x }} = \red 4^{\blue 3 } $$

Step 3

Use exponents laws to simplify

$$ \red 4^{\blue{ 2x }} = \red 4^{\blue 3 } $$

Not much to do this time :)

Step 4

$$ 2x = 3 \\ x = \frac{3}{2} $$

Problem 3

Solve the exponential Equation : $$ \left( \frac{1}{4} \right)^x = 32 $$

Step 1

Since these equations have different bases, follow the steps for unlike bases


Forget about the exponents for a minute and focus on the bases:
Rewrite the bases as powers of a common base. Ask yourself :

They are both powers of 2

$ \\ 32 = \red 2 ^{\blue 5} \\ \frac 1 4 = \red 2 ^{\blue {-2}} \\ $

Step 2

Rewrite as a negative exponent and substitute the rewritten bases into original equation

$$ \left( \frac{1}{4} \right)^x = 32 \\ \left( \frac{1}{2^2} \right)^x = 32 \\ \left(\red 2 ^{\blue{-2}} \right)^x = \red 2^{\blue 5} $$

Step 3

Use exponents laws to simplify

$$ \left(\red 2 ^{\blue{-2}} \right)^x = \red 2^{\blue 5} \\ 2 ^{-2 \cdot x} = 2^5 \\ 2 ^{-2x} = 2^5 $$

Step 4

$$ 2 ^{-2x} = 2^5 \\ -2x = 5 \\ \frac{-2x}{-2} = \frac{5}{-2} \\ x = -\frac{5}{2} $$

Problem 4

Solve this exponential equation: $$ \left( \frac{1}{9} \right)^x-3 = 24 $$

Step 1

Rewrite this equation so that it looks like the other ones we solved. Isolate the exponential expression as follows:

$$ \left( \frac{1}{9} \right)^x -3 \red{+3} =24\red{+3} \\ \left( \frac{1}{9} \right)^x=27 $$


Forget about the exponents for a minute and focus on the bases:
Rewrite the bases as powers of a common base. Ask yourself :

They are both powers of 3.

$ \\ \frac 1 9 = \red 3 ^{\blue {-2}} \\ 27 = \red 3 ^{\blue 3} \\ $

Step 2

Rewrite as a negative exponent and substitute into original equation

$$ \left( \frac{1}{9} \right)^x=27 \\ \left( \red{3^{-2}}\right)^x=\red{3^3 } $$

Step 3

Use exponents laws to simplify

$$ 3^\red{{-2 \cdot x}} = 3^3 \\ 3^\red{{-2x}} = 3^3 $$

Step 4

$$ -2x = 3 \\ x = \frac{3}{-2} \\ x = -\frac{3}{2} $$

Problem 5

Solve this exponential equation: $$ \left( \frac{1}{25} \right)^{(3x -4)} -1 = 124 $$

Step 1

Rewrite this equation so that it looks like the other ones we solved--In other words, isolate the exponential expression as follows:

$$ \left( \frac{1}{25} \right)^{(3x -4)} -1 \red{+1} = 124 \red{+1} \\ \left( \frac{1}{25} \right)^{(3x -4)} = 125 $$


Forget about the exponents for a minute and focus on the bases:
Rewrite the bases as powers of a common base. Ask yourself :

They are both powers of 5.

$ \\ \frac { 1 } { 25 } = \red 5 ^{\blue {-2}} \\ 125 = \red 5 ^{\blue 3} \\ $

Step 2

Rewrite as a negative exponent and substitute into original equation

$$ \left( \frac{1}{25} \right)^{(3x -4)} = 125 \\ \left( \red{5^{-2}} \right)^{(3x -4)} = \red{5^3} $$

Step 3

Use exponents laws to simplify

$$ 5^\red{{-2 \cdot (3x -4)}} = 5^3 \\ 5^\red{{(-6x + 8)}} = 5^3 $$

Step 4

$$ -6x + 8 =3 \\ -6x = -5 \\ x = \frac{-5}{-6} \\ x = \frac{5}{6} $$


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