# How do you solve 2^(5x)/4^(3x)=8*16^(9x) ?

Apr 16, 2017

$x = - \frac{3}{37}$

#### Explanation:

Given:

${2}^{5 x} / {4}^{3 x} = 8 \cdot {16}^{9 x}$

This can be rewritten in terms of powers of $2$...

${2}^{5 x} / \left({\left({2}^{2}\right)}^{3 x}\right) = {2}^{3} \cdot {\left({2}^{4}\right)}^{9 x}$

Then using:

${\left({a}^{b}\right)}^{c} = {a}^{b c} \text{ }$ (when $a > 0$)

we can rewrite this as:

${2}^{- x} = {2}^{5 x - 6 x} = {2}^{5 x} / {2}^{6 x} = {2}^{3} \cdot {2}^{36 x} = {2}^{36 x + 3}$

Note that ${2}^{x}$ is a one-one function as a real function, so the exponents are equal for the real solution and we find:

$- x = 36 x + 3$

Hence:

$x = - \frac{3}{37}$