125^-b=(1/5)^-b-3

Solution for 125^-b=(1/5)^-b-3 equation:

125^-b=(1/5)^-b-3

We move all terms to the left:

125^-b-((1/5)^-b-3)=0


Domain of the equation: 5)^-b-3)!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

-b-((+1/5)^-b-3)+125^=0

We add all the numbers together, and all the variables

-1b-((+1/5)^-b-3)=0

We multiply all the terms by the denominator


-1b*5)^-b-3)-((+1=0

Wy multiply elements

-5b^2+1=0

a = -5; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-5)·1
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*-5}=\frac{0-2\sqrt{5}}{-10}
=-\frac{2\sqrt{5}}{-10}
=-\frac{\sqrt{5}}{-5}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*-5}=\frac{0+2\sqrt{5}}{-10}
=\frac{2\sqrt{5}}{-10}
=\frac{\sqrt{5}}{-5}
$