Question 238700
{{{log(8,x) + log(8,(x-12)) = 2}}}
{{{log(8,x(x-12)) = 2}}}
{{{x(x-12) = 8^2}}}
{{{x(x-12) = 64}}}
{{{x^2-12x = 64}}}
{{{x^2-12x-64 = 0}}}
{{{(x-16)(x+4) = 0}}}
.
x = {-4, 16}
We can throw out the -4 because it won't work therefore:
x = 16