4^-3x=(1/8)^-x

Solution for 4^-3x=(1/8)^-x equation:

4^-3x=(1/8)^-x

We move all terms to the left:

4^-3x-((1/8)^-x)=0


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

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-3x-((+1/8)^-x)+4^=0

We add all the numbers together, and all the variables

-3x-((+1/8)^-x)=0

We multiply all the terms by the denominator


-3x*8)^-x)-((+1=0

Wy multiply elements

-24x^2+1=0

a = -24; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-24)·1
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-24}=\frac{0-4\sqrt{6}}{-48}
=-\frac{4\sqrt{6}}{-48}
=-\frac{\sqrt{6}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-24}=\frac{0+4\sqrt{6}}{-48}
=\frac{4\sqrt{6}}{-48}
=\frac{\sqrt{6}}{-12}
$