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

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It is an exponential equation and we'll solvet it using substitution technique.

2^x=t

4^x=(2^2)^x=(2^x)^2=t^2

t^2 + t + 1 = 80

t^2 + t + 1 - 80 = 0

t^2 + t - 79 = 0

t1 = [-1+(1+1+316)^1/2]/2=16/2=approx. 8

2^x=t1

2^x=8 => 2^x=2^3 =>x=3

t2=(-1-17)/2=approx. -9

2^x>0 so it's impossible that 2^x=-9!

The only solution is x=3 (approx.)

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