1/4^x=2^x+5

Solution for 1/4^x=2^x+5 equation:

1/4^x=2^x+5

We move all terms to the left:

1/4^x-(2^x+5)=0


Domain of the equation: 4^x!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

1/4^x-2^x-5=0

We multiply all the terms by the denominator


-2^x*4^x-5*4^x+1=0

Wy multiply elements

-8x^2-20x+1=0

a = -8; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·(-8)·1
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12\sqrt{3}}{2*-8}=\frac{20-12\sqrt{3}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12\sqrt{3}}{2*-8}=\frac{20+12\sqrt{3}}{-16}
$