243^2x=1/9

Solution for 243^2x=1/9 equation:

243^2x=1/9

We move all terms to the left:

243^2x-(1/9)=0

We add all the numbers together, and all the variables

243^2x-(+1/9)=0

We get rid of parentheses

243^2x-1/9=0

We multiply all the terms by the denominator


243^2x*9-1=0

Wy multiply elements

2187x^2-1=0

a = 2187; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·2187·(-1)
Δ = 8748
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:

$\sqrt{\Delta}=\sqrt{8748}=\sqrt{2916*3}=\sqrt{2916}*\sqrt{3}=54\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-54\sqrt{3}}{2*2187}=\frac{0-54\sqrt{3}}{4374}
=-\frac{54\sqrt{3}}{4374}
=-\frac{\sqrt{3}}{81}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+54\sqrt{3}}{2*2187}=\frac{0+54\sqrt{3}}{4374}
=\frac{54\sqrt{3}}{4374}
=\frac{\sqrt{3}}{81}
$