1/9^x=3-^x/3*81^x

Solution for 1/9^x=3-^x/3*81^x equation:

1/9^x=3-^x/3*81^x

We move all terms to the left:

1/9^x-(3-^x/3*81^x)=0


Domain of the equation: 9^x!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 3*81^x)!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

1/9^x+^x/3*81^x-3=0

We calculate fractions


9x^2/17496x^2+1944x/17496x^2-3=0

We multiply all the terms by the denominator


9x^2+1944x-3*17496x^2=0

Wy multiply elements

9x^2-52488x^2+1944x=0

We add all the numbers together, and all the variables

-52479x^2+1944x=0

a = -52479; b = 1944; c = 0;
Δ = b2-4ac
Δ = 19442-4·(-52479)·0
Δ = 3779136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3779136}=1944$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1944)-1944}{2*-52479}=\frac{-3888}{-104958}
=216/5831
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1944)+1944}{2*-52479}=\frac{0}{-104958}
=0
$