Question 450913
{{{4^x - 2^x - 12 = 0}}}
<==> {{{2^(2x) - 2^x - 12 = 0}}}
<==> {{{(2^x - 4)(2^x + 3) = 0}}}
<==> {{{2^x - 4 = 0}}}, or {{{2^x + 3 = 0}}}

The second equation has no real solutions, and so, 

{{{2^x - 4 = 0}}}, or {{{2^x = 4}}}, or x = 2.