Question 393005
 log(x+1)+log(x-1)=log 8  
==> log((x+1)(x-1)) = log 8
==> {{{log((x^2 - 1)) = log 8}}}
==> {{{x^2 - 1 = 8}}}, since the log function is one-to-one
<==> {{{x^2 - 9 = 0}}}, or (x-3)(x+3) = 0, or x = -3, 3.  Since -3 will not satisfy the original equation, the final answer is x  =3.