Question

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

Answer 1

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

Explanation:

First, use the fact that `log_{b}(A)+log_{b}(B)=log_{b}(AB)` (when `A,B>0`) to rewrite the equation as `log_{6}((x+8)(x-8))=2`, or `log_{6}(x^2-64)=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}(z)` is only defined for `z>0`.

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