64^-n=(1/8)^2n

Solution for 64^-n=(1/8)^2n equation:

64^-n=(1/8)^2n

We move all terms to the left:

64^-n-((1/8)^2n)=0


Domain of the equation: 8)^2n)!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

-n-((+1/8)^2n)+64^=0

We add all the numbers together, and all the variables

-1n-((+1/8)^2n)=0

We multiply all the terms by the denominator


-1n*8)^2n)-((+1=0

Wy multiply elements

-8n^2+1=0

a = -8; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-8)·1
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*-8}=\frac{0-4\sqrt{2}}{-16}
=-\frac{4\sqrt{2}}{-16}
=-\frac{\sqrt{2}}{-4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*-8}=\frac{0+4\sqrt{2}}{-16}
=\frac{4\sqrt{2}}{-16}
=\frac{\sqrt{2}}{-4}
$