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

4^x+2^x+1 = 80. To find x .

solution:

4^x = (2^x )^2

So  the given equation is equivalent to:

(2^x)^2+2^x+1-80 = 0. Or

y^2 + y -79 = 0. or

y1 = [-1+sqrt(1+4*79)]/2 or y2 = [-1-sqrt(1+4*79)]/2.

 Or taking the positive root,

2^x = 8.402246907

x log2  =  log(8.402246907) , or

x = 8.402246907 / log2 =  3.070775181. The root is imaginary.