Collapse the logarithms into one:
[tex]\log_3(x+6)+\log_3(x-6)=\log_3((x+6)(x-6))=4[/tex]
Then write both sides as powers of 3:
[tex]3^{\log_3((x+6)(x-6))}=3^4[/tex]
[tex](x+6)(x-6)=81[/tex]
Simplify the left side and solve for [tex]x[/tex]:
[tex]x^2-36=81[/tex]
[tex]x^2=117[/tex]
[tex]x=\pm\sqrt{117}=\pm3\sqrt{13}[/tex]
But we're not done yet. Notice that both [tex]x+6[/tex] and [tex]x-6[/tex] are negative when [tex]x=-3\sqrt{13}[/tex]. We can't take the logarithm of a negative number (the result isn't real-valued, anyway), so we throw this solution out, and we're left with just
[tex]x=3\sqrt{13}[/tex]