Solve for x:
(4^x+2^x-6)=0
4^x + 2^x - 6 = 0 factor as
(2^x + 3) ( 2^x - 2) = 0
Setting the first factor to 0 and solving produces a non-real solution
For the second factor we have
2^x - 2 = 0
And it's clear that x = 1