4^x+1/2^x+2=128

Solution for 4^x+1/2^x+2=128 equation:

4^x+1/2^x+2=128

We move all terms to the left:

4^x+1/2^x+2-(128)=0


Domain of the equation: 2^x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

4^x+1/2^x-126=0

We multiply all the terms by the denominator


4^x*2^x-126*2^x+1=0

Wy multiply elements

8x^2-252x+1=0

a = 8; b = -252; c = +1;
Δ = b2-4ac
Δ = -2522-4·8·1
Δ = 63472
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{63472}=\sqrt{16*3967}=\sqrt{16}*\sqrt{3967}=4\sqrt{3967}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-252)-4\sqrt{3967}}{2*8}=\frac{252-4\sqrt{3967}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-252)+4\sqrt{3967}}{2*8}=\frac{252+4\sqrt{3967}}{16}
$