32^-3x=(1/4)^-3x

Solution for 32^-3x=(1/4)^-3x equation:

32^-3x=(1/4)^-3x

We move all terms to the left:

32^-3x-((1/4)^-3x)=0


Domain of the equation: 4)^-3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-3x-((+1/4)^-3x)+32^=0

We add all the numbers together, and all the variables

-3x-((+1/4)^-3x)=0

We multiply all the terms by the denominator


-3x*4)^-3x)-((+1=0

Wy multiply elements

-12x^2+1=0

a = -12; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-12)·1
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-12}=\frac{0-4\sqrt{3}}{-24}
=-\frac{4\sqrt{3}}{-24}
=-\frac{\sqrt{3}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-12}=\frac{0+4\sqrt{3}}{-24}
=\frac{4\sqrt{3}}{-24}
=\frac{\sqrt{3}}{-6}
$