Question 256085
    log base 9 of (x-6) + log base 9 of (x+2) =1
.
{{{    log(9,(x-6)) + log(9,(x+2)) =1}}}
.
Applying "log rules":
.
{{{    log(9,(x-6)(x+2)) =1}}}
.
{{{    (x-6)(x+2) = 9^1}}}
.
{{{    x^2+2x-6x-12 = 9}}}
.
{{{    x^2-4x-12 = 9}}}
.
{{{    x^2-4x-21 = 0}}}
.
{{{    (x-7)(x+3) = 0}}}
.
x = {7, -3}
But, plugging it back into the original equation shows that -3 is an "extraneous" solution -- throw it out.  Leaving:
x = 7