8^-x-1=(1/64)^x

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

8^-x-1=(1/64)^x

We move all terms to the left:

8^-x-1-((1/64)^x)=0


Domain of the equation: 64)^x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-x-((+1/64)^x)-1+8^=0

We add all the numbers together, and all the variables

-1x-((+1/64)^x)=0

We multiply all the terms by the denominator


-1x*64)^x)-((+1=0

Wy multiply elements

-64x^2+1=0

a = -64; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-64)·1
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-64}=\frac{-16}{-128}
=1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-64}=\frac{16}{-128}
=-1/8
$