How do you solve Log_6(x+8) + log_6 (x-8)=2?

Oct 22, 2015

$x = 10$ is the (unique) answer. See explanation below.

Explanation:

First, use the fact that ${\log}_{b} \left(A\right) + {\log}_{b} \left(B\right) = {\log}_{b} \left(A B\right)$ (when $A , B > 0$) to rewrite the equation as ${\log}_{6} \left(\left(x + 8\right) \left(x - 8\right)\right) = 2$, or ${\log}_{6} \left({x}^{2} - 64\right) = 2$.

Now rewrite this last equation in exponential form as $36 = {6}^{2} = {x}^{2} - 64$. Therefore, ${x}^{2} = 100$ and $x = \pm \sqrt{100} = \pm 10$.

Sometimes the solution of logarithmic equations gives extraneous (fictitious) roots, so these should be checked in the original equation (with close attention paid to the domain). In fact, $x = - 10$ is an extraneous solution since ${\log}_{6} \left(z\right)$ is only defined for $z > 0$.

What about $x = 10$? Upon substitution, the left-hand side of the original equation becomes ${\log}_{6} \left(18\right) + {\log}_{6} \left(2\right) = {\log}_{6} \left(36\right) = 2$, so it works.