Question

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

Pls. solve this question.. n step by step.. Pls!

Answer 1

4^x+2^x+1 = 80

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

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 sOlution  would not be real