Question 878648
I'm assuming the problem is {{{log(8,(x+1))-log(8,(x))=2}}}



If so, then...



{{{log(8,(x+1))-log(8,(x))=2}}}



{{{log(8,((x+1)/x))=2}}}



{{{(x+1)/x=8^2}}}



{{{(x+1)/x=64}}}



{{{x+1=64x}}}



{{{1=64x-x}}}



{{{1=63x}}}



{{{1/63=x}}}



{{{x=1/63}}}