Question 579298
Your notation is confusing.  I'll assume you meant log base 6:
{{{log(6,x)+log(6,(x-9))=2}}}
{{{log(6,x(x-9))=2}}}
{{{x(x-9)=6^2}}}
{{{x^2-9x=36}}}
{{{x^2-9x-36=0}}}
{{{(x-12)(x+3)=0}}}
x = {-3, 12}
But, since you can't take the log of a negative number, the -3 is an extraneous answer -- throw it out leaving:
x = 12