2^-2x=(1/32)^x-3

Solution for 2^-2x=(1/32)^x-3 equation:

2^-2x=(1/32)^x-3

We move all terms to the left:

2^-2x-((1/32)^x-3)=0


Domain of the equation: 32)^x-3)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-2x-((+1/32)^x-3)+2^=0

We add all the numbers together, and all the variables

-2x-((+1/32)^x-3)=0

We multiply all the terms by the denominator


-2x*32)^x-3)-((+1=0

Wy multiply elements

-64x^2+1=0

a = -64; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-64)·1
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-64}=\frac{-16}{-128}
=1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-64}=\frac{16}{-128}
=-1/8
$