Question

Solve
4^y -7(2^y) = 8

Answer 1

4^y = (2^2)^y = (2^y)^2
so, we have a quadratic in 2^y:
(2^y)^2 - 7*2^y - 8 = 0
(2^y-8)(2^y+1) = 0
y=3
2^y is never negative, so 2^y+1=0 has no real solution.