1/9=27^2^x

Solution for 1/9=27^2^x equation:

1/9=27^2^x

We move all terms to the left:

1/9-(27^2^x)=0

We multiply all the terms by the denominator


-27^2^x*9+1=0

Wy multiply elements

-243x^2+1=0

a = -243; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-243)·1
Δ = 972
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:

$\sqrt{\Delta}=\sqrt{972}=\sqrt{324*3}=\sqrt{324}*\sqrt{3}=18\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{3}}{2*-243}=\frac{0-18\sqrt{3}}{-486}
=-\frac{18\sqrt{3}}{-486}
=-\frac{\sqrt{3}}{-27}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{3}}{2*-243}=\frac{0+18\sqrt{3}}{-486}
=\frac{18\sqrt{3}}{-486}
=\frac{\sqrt{3}}{-27}
$