Solve 4^x + 2^x + 1 = 80

4^x+2^x+1 = 80

To solve this
we put 4^x = (2^2)^x = (2^x)^2.

So we put 2^x = t and
rewite the equation:

t^2+t+1 =
80

t^2 +t +1-80 = 0

 t^2 +t
-79 = 0.

t1 = {-1 +sqrt[1-4*1*(-79)}/2 =
{-1+sqrtsqrt317}/2

t2 =
{-(1+sqrt(317)}/2

 So 2^x = (sqrt317
)-1}/2

 x =  log
{[(sqrt317)-1]/2}/log2

x =  3.070775181. is the real
solution.

If we take t2 which is negative,  the slution 
would not be real