Solve:

$\displaystyle log_8(x-3) - log_8 (3x + 11) = log_8 (\frac{1}{x+1})$

I have this:

$\displaystyle log_8(\frac {x-3}{3x + 11}) = log_8 (\frac{1}{x+1})$

$\displaystyle log_8(-2x -14) = log_8 (-x)$

But I'm not sure my second step is right...how would you solve this equation?

Answer: 7