# How do you solve 64^(x-1)/16^(2x+2)=2^(x-2)?

Aug 25, 2016

$x = 4$

#### Explanation:

${64}^{x - 1} / {16}^{2 x + 2} = {2}^{x - 2}$

Well. We know that $64 = {2}^{6}$ and $16 = {2}^{4}$. Also we know that

${\left({a}^{b}\right)}^{c} = {a}^{b c}$

so

${64}^{x - 1} / {16}^{2 x + 2} = {2}^{x - 2} \equiv {\left({2}^{6}\right)}^{x - 1} / {\left({2}^{4}\right)}^{2 x + 2} = {2}^{x - 2}$

and also

${64}^{x - 1} / {16}^{2 x + 2} = {2}^{x - 2} \equiv {2}^{6 \left(x - 1\right)} / {2}^{4 \left(2 x + 2\right)} = {2}^{x - 2}$

Now, applying the ${\log}_{2}$ transformation to both sides and keeping in mind that

${\log}_{a} {a}^{b} = b$

$6 \left(x - 1\right) - 4 \left(2 x + 2\right) = x - 2$. Now, solving for $x$

$x = 4$

Aug 25, 2016

$x = - 4$

#### Explanation:

All the bases are powers of 2 - change them all to a base of 2.
Then we can use laws of indices.

$\frac{{\textcolor{red}{64}}^{x - 1}}{{\textcolor{b l u e}{16}}^{2 x + 2}} = {2}^{x - 2}$

$\frac{{\left(\textcolor{red}{{2}^{6}}\right)}^{x - 1}}{{\left(\textcolor{b l u e}{{2}^{4}}\right)}^{2 x + 2}} = {2}^{x - 2} \textcolor{w h i t e}{\times \times \times \times \times}$ Multiply the indices.

${2}^{\textcolor{m a \ge n t a}{6 x - 6}} / \left({2}^{\textcolor{m a \ge n t a}{8 x + 8}}\right) = {2}^{x - 2} \textcolor{w h i t e}{\times \times \times \times \times \times}$ Subtract the indices

${2}^{\textcolor{m a \ge n t a}{- 2 x - 14}} = {2}^{x - 2} \textcolor{w h i t e}{\times \times \times \times \times \times}$Equate the indices.

$- 2 x - 14 = x - 2 \textcolor{w h i t e}{\times \times \times \times \times}$ Solve for x

$- 14 + 2 = x + 2 x$

$- 12 = 3 x$

$x = - 4$