# How do you solve 8^(2x-4)=16^(2x+4)?

Oct 24, 2016

I got: $x = - 14$

#### Explanation:

Let us take the log in base $2$ of both sides and use a property of logs to move the exponents in front of the logs. You should get:
${\log}_{2} {\left(8\right)}^{2 x - 4} = {\log}_{2} {\left(16\right)}^{2 x + 4}$
$\left(2 x - 4\right) {\log}_{2} \left(8\right) = \left(2 x + 4\right) {\log}_{2} \left(16\right)$
we now use the definition of log to solve the two logs and get:
$\left(2 x - 4\right) \cdot 3 = \left(2 x + 4\right) \cdot 4$
rearrange and solve for $x$:
$6 x - 12 = 8 x + 16$
$2 x = - 28$
$x = - \frac{28}{2} = - 14$

Oct 24, 2016

$x = - 14$

#### Explanation:

Some exponential equations need to be solved using logs, others can be solved by making the bases or the indices the same.

This is an example of the latter.

Both 8 and 16 are powers of 2, so we can write this equation in terms of a common base on each side.

color(red)(8 = 2^3) and color(blue)(16 = 2^4)color(red)(

${\textcolor{red}{8}}^{2 x - 4} = {\textcolor{b l u e}{16}}^{2 x + 4}$

${\textcolor{red}{\left({2}^{3}\right)}}^{2 x - 4} = {\textcolor{b l u e}{\left({2}^{4}\right)}}^{2 x + 4} \text{ } \leftarrow$ multiply the indices.

${2}^{6 x - 12} = {2}^{8 x + 16} \text{ } \leftarrow$ if ${x}^{m} = {x}^{n} \Leftrightarrow m = n$

$\therefore 6 x - 12 = 8 x + 16$

$- 12 - 16 = 8 x - 6 x$

$- 28 = 2 x$

$- 14 = x$