1/9^x=3^x+1

Solution for 1/9^x=3^x+1 equation:

1/9^x=3^x+1

We move all terms to the left:

1/9^x-(3^x+1)=0


Domain of the equation: 9^x!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

1/9^x-3^x-1=0

We multiply all the terms by the denominator


-3^x*9^x-1*9^x+1=0

Wy multiply elements

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

a = -27; b = -9; c = +1;
Δ = b2-4ac
Δ = -92-4·(-27)·1
Δ = 189
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:

$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{21}}{2*-27}=\frac{9-3\sqrt{21}}{-54}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{21}}{2*-27}=\frac{9+3\sqrt{21}}{-54}
$