9^x+3^x+1+3^1-x+1/9^x=8

Solution for 9^x+3^x+1+3^1-x+1/9^x=8 equation:

9^x+3^x+1+3^1-x+1/9^x=8

We move all terms to the left:

9^x+3^x+1+3^1-x+1/9^x-(8)=0


Domain of the equation: 9^x!=0

x!=0/1

x!=0

x∈R


determiningTheFunctionDomain
9^x+3^x-x+1/9^x+1-8+3^1=0

We add all the numbers together, and all the variables

-1x+9^x+3^x+1/9^x-4=0

We multiply all the terms by the denominator


-1x*9^x+9^x*9^x+3^x*9^x-4*9^x+1=0

Wy multiply elements

-9x^2+81x^2+27x^2-36x+1=0

We add all the numbers together, and all the variables

99x^2-36x+1=0

a = 99; b = -36; c = +1;
Δ = b2-4ac
Δ = -362-4·99·1
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-30}{2*99}=\frac{6}{198}
=1/33
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+30}{2*99}=\frac{66}{198}
=1/3
$