# How do you solve log_8(11-6x)=log_8(1-x)?

Sep 4, 2016

$\left\{\emptyset\right\}$

#### Explanation:

${\log}_{8} \left(11 - 6 x\right) - {\log}_{8} \left(1 - x\right) = 0$

You need use the apply the rule ${\log}_{a} n - {\log}_{a} m = {\log}_{a} \left(\frac{n}{m}\right)$.

${\log}_{8} \left(\frac{11 - 6 x}{1 - x}\right) = 0$

$\frac{11 - 6 x}{1 - x} = {8}^{0}$

$11 - 6 x = 1$

$- 6 x = - 10$

$x = \frac{5}{3}$

However, checking in the original equation, you will realize that this solution is extraneous, because if in ${\log}_{b} \left(n\right)$, $n \le 0$, the equation will be undefined.

Hence, the solution set is $\left\{\emptyset\right\}$.

Hopefully this helps!