32^-2x+1=(1/64)^3x

Solution for 32^-2x+1=(1/64)^3x equation:

32^-2x+1=(1/64)^3x

We move all terms to the left:

32^-2x+1-((1/64)^3x)=0


Domain of the equation: 64)^3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-2x-((+1/64)^3x)+1+32^=0

We add all the numbers together, and all the variables

-2x-((+1/64)^3x)=0

We multiply all the terms by the denominator


-2x*64)^3x)-((+1=0

Wy multiply elements

-128x^2+1=0

a = -128; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-128)·1
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*-128}=\frac{0-16\sqrt{2}}{-256}
=-\frac{16\sqrt{2}}{-256}
=-\frac{\sqrt{2}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*-128}=\frac{0+16\sqrt{2}}{-256}
=\frac{16\sqrt{2}}{-256}
=\frac{\sqrt{2}}{-16}
$