Question 202790
Applying "log rules":
{{{log(8,(x-1))+1/3=log(8,(x+7))}}}
{{{log(8,(x-1)) - log(8,(x+7)) = -1/3}}}
{{{log(8,(x-1)/(x+7)) = -1/3}}}
{{{(x-1)/(x+7) = 8^(-1/3)}}}
{{{(x-1)/(x+7) = 1/8^(1/3)}}}
{{{(x-1)/(x+7) = 1/2}}}
{{{2(x-1) = (x+7)}}}
{{{2x-2 = x+7}}}
{{{x-2 = 7}}}
{{{x = 9}}}