# How do you solve 8^(2-x) = 4^(3x)?

Dec 16, 2015

Express each side of the given equation with a base of $2$
then equating the exponents, solve for $x$ to get
$\textcolor{w h i t e}{\text{XXX}} x = \frac{2}{3}$

#### Explanation:

${8}^{a} = {2}^{3 a}$
${4}^{b} = {2}^{2 b}$

Therefore
$\textcolor{w h i t e}{\text{XXX}} {8}^{2 - x} = {4}^{3 x}$
is equivalent to
$\textcolor{w h i t e}{\text{XXX}} {2}^{6 - 3 x} = {4}^{6 x}$

Which implies
$\textcolor{w h i t e}{\text{XXX}} 6 - 3 x = 6 x$

$\textcolor{w h i t e}{\text{XXX}} 9 x = 6$

$\textcolor{w h i t e}{\text{XXX}} x = \frac{2}{3}$