# Question #07bdd

Sep 29, 2017

$- \frac{2}{3}$

#### Explanation:

Notice we can express ${\left(\frac{1}{512}\right)}^{x}$ as:

${\left(\frac{1}{8} ^ 3\right)}^{x}$

${\left(\frac{1}{8} ^ 3\right)}^{x} \implies {1}^{x} / {8}^{3 x} \implies \frac{1}{{8}^{3 x}} \implies {8}^{- 3 x}$

$64 \implies {8}^{2}$

So we have:

${8}^{- 3 x} = {8}^{2}$

If bases are equal, then exponents are equal.

So:

$- 3 x = 2 \implies x = - \frac{2}{3}$

Exponential equations are generally best solved using logarithms. In this case it was relatively easy to express the values using a common base. In the vast number of cases this is either not possible or would take an excessive amount of time to find them.

This is the problem above solved using logarithms.

Taking logs of both sides.( You can use logarithms to any base for this).

$\ln {\left(\frac{1}{512}\right)}^{x} \implies x \ln \left(\frac{1}{512}\right) \implies x \left(\ln 1 - \ln 512\right) \implies x \left(- \ln 512\right)$

So we have:

$x \left(- \ln 512\right) = \ln 64$

$x = \ln \frac{64}{- \ln 512} \implies x = - .6 \overline{6} = - \frac{2}{3}$