1/2^(2n)=256

Solution for 1/2^(2n)=256 equation:

1/2^(2n)=256

We move all terms to the left:

1/2^(2n)-(256)=0


Domain of the equation: 2^2n!=0

n!=0/1

n!=0

n∈R


We multiply all the terms by the denominator


-256*2^2n+1=0

Wy multiply elements

-512n^2+1=0

a = -512; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-512)·1
Δ = 2048
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{2}}{2*-512}=\frac{0-32\sqrt{2}}{-1024}
=-\frac{32\sqrt{2}}{-1024}
=-\frac{\sqrt{2}}{-32}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{2}}{2*-512}=\frac{0+32\sqrt{2}}{-1024}
=\frac{32\sqrt{2}}{-1024}
=\frac{\sqrt{2}}{-32}
$