Question 354511
log 2,(x+6) + log 2,(x-1) = 3
log 2,(x+6)(x-1) = 3
log 2,{{{(x^2+5x-6)}}} = 3.  This is the same as 
{{{x^2+5x-6 = 2^3}}},
{{{x^2+5x-6 = 8}}},
{{{x^2+5x-14 = 0}}},
{{{(x+7)(x-2) = 0}}}.
This gives {{{x = -7}}} or {{{x = 2}}}.  The first value won't satisfy the original equation, so the final answer is {{{x =2}}}.