Question 747802
{{{log(3,x+6)+log(3,x-6)-log(3,x)=2}}}....in base {{{10}}} we will have

{{{log(x+6)/log(3)+log(x-6)/log(3)-log(x)/log(3)=2}}}



{{{(log(x+6)+log(x-6)-log(x))/log(3)=2}}}

{{{log(x+6)+log(x-6)-log(x)=2log(3)}}}

{{{log((x+6)(x-6))-log(x)=log(3^2)}}}

{{{log(((x+6)(x-6))/x)=log(9)}}}

{{{((x+6)(x-6))/x=9}}}

{{{(x+6)(x-6)=9x}}}


{{{x^2-6x+6x-36=9x}}}

{{{x^2-36=9x}}}

{{{x^2-9x-36=0}}}..write {{{-9x}}} as {{{-12x+3x}}}

{{{x^2-12x+3x-36=0}}}....group

{{{(x^2-12x)+(3x-36)=0}}}

{{{x(x-12)+3(x-12)=0}}}

{{{(x+3)(x-12)=0}}}

solutions:

if {{{x+3=0}}} => {{{x=-3}}}

if {{{x-12=0}}} => {{{x=12}}}


recall: 

logarithm of {{{negative}}} number is {{{undefined}}}

so, your solution is only {{{highlight(x=12)}}}