Explore: Exponential Equations Solutions

Exponential Equations Solutions

All of the Exponent Rules apply to real number exponents .  
To solve problems with exponents, the bases must be equal . 
When an exponent is negative, the reciprocal of the base is taken.

Note

Taking the reciprocal of the base is also helpful when working with equations with bases that need to be manipulated before solving.

Using technology to check the solution to exponential equations is recommended because the values are often very large or small.

Note

It may be helpful to complete the More to Explore to review the first few multiples of each base.

Example 1

Solve.

${(\frac{1}{32})}^{x + 5} = 8^{x - 3}$

Plan

Write the equation with a common base

Set the exponential expressions equal to one another

Solve

Check work using technology

Implement

$\begin{array}{rcl} (\frac{1}{32}) & = & 2^{- 5} 8 = 2^{3} \\ 2^{- 5 (x + 5)} & = & 2^{3 (x - 3)} \\ - 5 (x + 5) & = & 3 (x - 3) \\ - 5 x - 25 & = & 3 x - 9 \\ - 16 & = & 8 x \\ x & = & - 2 \end{array}$

$Check {(\frac{1}{32})}^{- 2 + 5} = 8^{- 2 - 3} {(\frac{1}{32})}^{3} = 8^{- 5} 2^{- 5 (3)} = 2^{3 (- 5)} ✓$

Note

Remember a negative exponent represents the reciprocal of the base.

Example 2

Solve.

${(\frac{64}{25})}^{2 x + 1} = {(\frac{125}{512})}^{1 - 3 x}$

$\begin{array}{rcl} \frac{64}{25} = {(\frac{8}{5})}^{2} \frac{125}{512} = {(\frac{5}{8})}^{3} = {(\frac{8}{5})}^{- 3} \\ {(\frac{8}{5})}^{2 (2 x + 1)} & = & {(\frac{8}{5})}^{- 3 (1 - 3 x)} \\ 2 (2 x + 1) & = & - 3 (1 - 3 x) \\ 4 x + 2 & = & - 3 + 9 x \\ 5 & = & 5 x \\ x & = & 1 \end{array}$

Note

Remember to maintain the power rule when rewriting the number(s) raised to a single power. The exponent needs to be the same for both bases.