# How do you solve 8^(x-1)=root3(16)?

##### 1 Answer
May 24, 2016

$x = \frac{13}{9}$ or $x = 1 \frac{4}{9}$

#### Explanation:

To solve ${8}^{x - 1} = \sqrt[3]{16}$

We begin by taking the $16$ out of the root

${8}^{x - 1} = {16}^{\frac{1}{3}}$

Next convert the $8$ and $16$ to the same base.

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

Now simplify the exponents

${2}^{3 x - 3} = {2}^{\frac{4}{3}}$

Since the base values are the same we can simply simplify the exponents algebraically.

$3 x - 3 = \frac{4}{3}$

$3 x \cancel{- 3} \cancel{+ 3} = \frac{4}{3} + 3$

$3 x = \frac{13}{3}$

$\cancel{3} x \left(\frac{1}{\cancel{3}}\right) = \frac{13}{3} \left(\frac{1}{3}\right)$

$x = \frac{13}{9}$ or $x = 1 \frac{4}{9}$