Question 29879
 log[4]x+log [4] (x+12) = 3   when adding, you are really multiplying
 log[4](x)(x+12) = 3 
 log [4] (x^2 + 12x) = 3
[4] is the base
3 is the power
x^2 + 12x is what they equal
{{{ 4^3 = x^2 + 12x }}} 
{{{ 64 = x^2 + 12x }}}
{{{ 0 = x^2 + 12x -64 }}} factor
{{{ 0 = (x+16)(x-4) }}}
{{{ 0 = x+16 }}} or {{{ 0 = x-4 }}}
{{{ -16 = x }}} or {{{ 4 = x }}}