Before you try to solve exponential equations, you must be quite comfortable using the rules and laws of exponents. An exponential equation is simply an equation in which a variable appears in the exponent.

**Examples** of exponential equations

- 2
^{x}= 4 - 2
^{2x}= 16 - 2
^{x + 1}= 256

**Steps** to solve exponential equations

There are many different kinds of exponential equations. First, let's focus on exponential equations that have a single term on both sides. These equations can be classified into two different types.

- Type #1) When the bases are of both terms are the same
- Type #2) When the bases are of the terms are different

**Solving** Exponential Equations of the same base (type #1)

**Example 4**Step 1

^{x + 1}= 4^{9}Ignore the bases, and simply set the exponents equal to each other

X + 1 = 9

Step 2Solve for the variable

X = 9 − 1

X = 8

**Solving** Exponential Equations of with unlike bases (type #2)

**Example Equation: 4**Step 1

^{3}=2^{x}Rewrite the bases as powers of a common base (Ignore the exponents and Answer the following question: *"4 and 2 are powers of what number? "* Once you have your answer, rewrite both of the bases as powers of that number)

They are both powers of 2

4 = 2²

Substitute the rewritten bases into original equation

4^{3} =2^{x}

(2^{2})^{3} =2^{x}

Simplify exponents

^{6}= 2

^{x}

Solve like an exponential equation of like bases

**Example** Problems: solving exponential equations

**Demonstration** Problem 2: Solving Exponential Equations

### Exponential Equations with Negative Exponents

It is possible to use the steps outlined on this page to solve exponential equations with negative exponents. Again, it is necessary that both of the bases are powers of a common base. See the example below

### Mixed **Practice** Problems

This is the second type of equation (exponents with unlike bases). So follow the steps.

Rewrite the bases as powers of a common base (Ignore the exponents and Answer the following question: *"9 and 81 are powers of what number? "* Once you have your answer, rewrite both of the bases as powers of that number)

They are both powers of 3 and of 9. You can use either base. I will use 3.

81 = 3^{4}

9 = 3^{2}

Substitute the rewritten bases into original equation

(3^{2})^{x} =3^{4}

Simplify exponents

3^{2}^{x} =3^{4}

Solve like an exponential equation of like bases

2x = 4

x = 2

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

Rewrite the bases as powers of a common base (Ignore the exponents and Answer the following question: *"$$ \frac{1}{4} $$ and 32 are powers of what number? "* Once you have your answer, rewrite both of the bases as powers of that number)

They are both powers of 2

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

Substitute the rewritten bases into original equation

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

Simplify exponents

$$ 2 ^{(-2 \cdot x)} = 2^5 \\ 2 ^{-2 x} = 2^5 $$

Solve like an exponential equation of like bases

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

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

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

Since these equations have different bases, followthe steps for unlike bases.

Rewrite the bases as powers of a common base (Ignore the exponents and Answer the following question: *"4 and 64 are powers of what number? "* Once you have your answer, rewrite both of the bases as powers of that number)

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

$$ 4 = 4^1 \\ 64 = 4^3 $$

Substitute the rewritten bases into original equation

$$ \bf{{4^{2x} +1 = 64}} \\ 4^{2x} = 4^3 $$

Simplify exponents

4^{2x} = 4^{3}

Solve like an exponential equation of like bases

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

Rewrite this equation so that it looks like the other ones we solved--In other words, isolate the exponential expression as follows: $$ \left( \frac{1}{9} \right)^x -3 \red{+3} =24\red{+3} \\ \bf{\left( \frac{1}{9} \right)^x=27 } $$

Since these equations have different bases, followthe steps for unlike bases

Rewrite the bases as powers of a common base

(Ignore the exponents and Answer the following question:*" $$ \frac{1}{9}$$ and 27 are powers of what number? "*Once you have your answer, rewrite both of the bases as powers of that number)

They are both powers of 3. $$ \frac{1}{9} = 3^{-2} \\ 27 = 3^3 $$

Substitute into original equation

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

Simplify exponents

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

Solve like an exponential equation of like bases

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

Rewrite this equation so that it looks like the other ones we solved--In other words, isolate the exponential expression as follows: $$ \left( \frac{1}{25} \right)^{(3x -4)} -1 \red{+1} = 124 \red{+1} \\ \left( \frac{1}{25} \right)^{(3x -4)} = 125 $$

Since these equations have different bases, followthe steps for unlike bases

Rewrite the bases as powers of a common base

(Ignore the exponents and Answer the following question:*"1/25 and 125 are powers of what number? "*Once you have your answer, rewrite both of the bases as powers of that number)

They are both powers of 5.

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

Substitute into original equation

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

Simplify exponents

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

Solve like an exponential equation of like bases

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