32^-x=(1/4)^(12-8x)

Solution for 32^-x=(1/4)^(12-8x) equation:

32^-x=(1/4)^(12-8x)

We move all terms to the left:

32^-x-((1/4)^(12-8x))=0


Domain of the equation: 4)^(12-8x))!=0

x∈R


We add all the numbers together, and all the variables

-x-((+1/4)^(-8x+12))+32^=0

We add all the numbers together, and all the variables

-1x-((+1/4)^(-8x+12))=0

We multiply all the terms by the denominator


-1x*4)^(-8x+1+12))-((=0

We add all the numbers together, and all the variables

-1x*4)^(-8x+13))-((=0

We add all the numbers together, and all the variables

-8x-1x*4)^(=0

Wy multiply elements

-4x^2-8x=0

a = -4; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-4)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*-4}=\frac{0}{-8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-4}=\frac{16}{-8}
=-2
$