1000^-4x=(1/10)^2x

Solution for 1000^-4x=(1/10)^2x equation:

1000^-4x=(1/10)^2x

We move all terms to the left:

1000^-4x-((1/10)^2x)=0


Domain of the equation: 10)^2x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-4x-((+1/10)^2x)+1000^=0

We add all the numbers together, and all the variables

-4x-((+1/10)^2x)=0

We multiply all the terms by the denominator


-4x*10)^2x)-((+1=0

Wy multiply elements

-40x^2+1=0

a = -40; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-40)·1
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*-40}=\frac{0-4\sqrt{10}}{-80}
=-\frac{4\sqrt{10}}{-80}
=-\frac{\sqrt{10}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*-40}=\frac{0+4\sqrt{10}}{-80}
=\frac{4\sqrt{10}}{-80}
=\frac{\sqrt{10}}{-20}
$