8^-3x+1=(1/16)^2x

Solution for 8^-3x+1=(1/16)^2x equation:

8^-3x+1=(1/16)^2x

We move all terms to the left:

8^-3x+1-((1/16)^2x)=0


Domain of the equation: 16)^2x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-3x-((+1/16)^2x)+1+8^=0

We add all the numbers together, and all the variables

-3x-((+1/16)^2x)=0

We multiply all the terms by the denominator


-3x*16)^2x)-((+1=0

Wy multiply elements

-48x^2+1=0

a = -48; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-48)·1
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*-48}=\frac{0-8\sqrt{3}}{-96}
=-\frac{8\sqrt{3}}{-96}
=-\frac{\sqrt{3}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-48}=\frac{0+8\sqrt{3}}{-96}
=\frac{8\sqrt{3}}{-96}
=\frac{\sqrt{3}}{-12}
$