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

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4^x+2^x+1=80

4^x+2^x-79=0

Let us substitute 4 with 2^2

(2^x)^2 + 2^x-79=0

Assume that 2^x=y

y^2+y-79=0

y1=[-1+sqrt(1+4(79)]/2= (-1+sqrt(317))/2= 8

y2=-1-sqrt(317)= -9

Now to find x values:

y1=2^x1=8 ==> x1=3

y2=2^x2 =-9 impossible since 2^x is a positive value.

Then the function has one solution x= 3

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.

First, we'll make the substitution 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

We'll apply the quadratic formula:

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

2^x=t1

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

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

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

The only solution is x=3 (approx.)