# How do you solve log_2 x - log_8 x = 4?

Apr 7, 2016

You must first put in the same base. This can be started out by using the log rule ${\log}_{a} n = \log \frac{n}{\log} a$

#### Explanation:

$\log \frac{x}{\log} 2 - \log \frac{x}{\log} 8 = 4$

Now, rewrite the terms in the denominator by using the rule $\log {a}^{n} = n \log a$

$\log \frac{x}{1 \log 2} - \log \frac{x}{\log {2}^{3}} = 4$

$\log \frac{x}{1 \log 2} - \log \frac{x}{3 \log 2} = 4$

Place on an equal denominator.

$\frac{3 \log x}{3 \log 2} - \log \frac{x}{3 \log 2} = 4$

${\log}_{8} \left({x}^{3}\right) - {\log}_{8} \left(x\right) = 4$

Now, you must use the rule ${\log}_{a} n - {\log}_{a} m = {\log}_{a} \left(\frac{n}{m}\right)$.

${\log}_{8} \left({x}^{3} / x\right) = 4$

Convert to exponential form:

${x}^{2} = {8}^{4}$

${x}^{2} = 4096$

$x = 64$

Checking the solution in the equation, we find that it works. Thus our solution set is $\left\{x = 64\right\}$

Hopefully this helps!