4^x+2^2x=1/8

Solution for 4^x+2^2x=1/8 equation:

4^x+2^2x=1/8

We move all terms to the left:

4^x+2^2x-(1/8)=0

We add all the numbers together, and all the variables

4^x+2^2x-(+1/8)=0

We get rid of parentheses

4^x+2^2x-1/8=0

We multiply all the terms by the denominator


4^x*8+2^2x*8-1=0

Wy multiply elements

16x^2+32x-1=0

a = 16; b = 32; c = -1;
Δ = b2-4ac
Δ = 322-4·16·(-1)
Δ = 1088
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:

$\sqrt{\Delta}=\sqrt{1088}=\sqrt{64*17}=\sqrt{64}*\sqrt{17}=8\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{17}}{2*16}=\frac{-32-8\sqrt{17}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{17}}{2*16}=\frac{-32+8\sqrt{17}}{32}
$