# Solving Exponential Equations

To solve an exponential equation, we can rewrite both sides of the equation using the same base. Next, we equate the exponents and solve for the unknown variable.

**Example 1: **Solve the exponential equation below.

$${9^{3x – 8}} = {1 \over {81}}$$

Notice that both \(9\) and \(81\) are powers of \(3\).

$$\eqalign{

9 &= {3^2} \cr

81 &= {3^4} \cr} $$

This transforms our original equation into

$${\left( {{3^2}} \right)^{3x – 8}} = {1 \over {{3^4}}}$$

Distribute the exponents on the left using the Power of a Power Property.

$${3^{6x – 16}} = {1 \over {{3^4}}}$$

Take the reciprocal of the fraction on the right side but make sure to switch the sign of the exponent from positive to negative.

$${3^{6x – 16}} = {3^{ – 4}}$$

If the two expressions are equal, their exponents must also be equal. At this point, we set the exponents equal to each other and solve for the variable \(x\).

$$\eqalign{

6x – 16 &= – 4 \cr

6x &= – 4 + 16 \cr

6x &= 12 \cr

x &= 2 \cr} $$

The final answer is \(\boxed{x=2}\). We can check it with the original equation to verify.

$$\eqalign{

{9^{3x – 8}} &= {1 \over {81}} \cr

{9^{3\left( {\color{red}2} \right) – 8}} &= {1 \over {81}} \cr

{9^{6 – 8}} &= {1 \over {81}} \cr

{9^{ – 2}} &= {1 \over {81}} \cr

{1 \over {{9^2}}} &= {1 \over {81}} \cr

{1 \over {81}} &= {1 \over {81}} \cr} $$

**Example 2: **Solve the exponential equation below.

$$\large{{2^{2x – 3}} \cdot {4^{x + 6}} = {{{8^{1 – 2x}}} \over {16}}}$$

**STEP 1:** Express all bases as powers of \(2\).

$${2^{2x – 3}} \cdot {\left( {{2^2}} \right)^{x + 6}} = {{{{\left( {{2^3}} \right)}^{1 – 2x}}} \over {{2^4}}}$$

**STEP 2:** Apply the Power of a Power Property of Exponent.

$${2^{2x – 3}} \cdot {2^{2x + 12}} = {{{2^{3 – 6x}}} \over {{2^4}}}$$

**STEP 3:** Apply the Product Property of Exponent on the left side. That means, copy the common base then add the exponents.

$${2^{4x + 9}} = {{{2^{3 – 6x}}} \over {{2^4}}}$$

**STEP 4:** Apply the Quotient Property of Exponent on the right side. That means, copy the common base then subtract the exponents.

$${2^{4x + 9}} = {2^{ – 6x – 1}}$$

**STEP 5:** Now, set the exponents equal to each other, then solve for the variable \(x\).

$$\eqalign{

4x + 9 &= – 6x – 1 \cr

4x + 6x &= – 1 – 9 \cr

10x &= – 10 \cr

x &= – 1 \cr} $$

The final answer is \(\boxed{x=-1}\).