(1/9^x)+(1/3^x)=6

Solution for (1/9^x)+(1/3^x)=6 equation:

(1/9^x)+(1/3^x)=6

We move all terms to the left:

(1/9^x)+(1/3^x)-(6)=0


Domain of the equation: 9^x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 3^x)!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

1/9^x+1/3^x-6=0

We calculate fractions


3x/27x^2+9x/27x^2-6=0

We multiply all the terms by the denominator


3x+9x-6*27x^2=0

We add all the numbers together, and all the variables

12x-6*27x^2=0

Wy multiply elements

-162x^2+12x=0

a = -162; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-162)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-162}=\frac{-24}{-324}
=2/27
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-162}=\frac{0}{-324}
=0
$