2/10^(2x)=10

Solution for 2/10^(2x)=10 equation:

2/10^(2x)=10

We move all terms to the left:

2/10^(2x)-(10)=0


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

x!=0/1

x!=0

x∈R


We multiply all the terms by the denominator


-10*10^2x+2=0

Wy multiply elements

-100x^2+2=0

a = -100; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-100)·2
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*-100}=\frac{0-20\sqrt{2}}{-200}
=-\frac{20\sqrt{2}}{-200}
=-\frac{\sqrt{2}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*-100}=\frac{0+20\sqrt{2}}{-200}
=\frac{20\sqrt{2}}{-200}
=\frac{\sqrt{2}}{-10}
$