9^-3x=(1/27)^x+3

Solution for 9^-3x=(1/27)^x+3 equation:

9^-3x=(1/27)^x+3

We move all terms to the left:

9^-3x-((1/27)^x+3)=0


Domain of the equation: 27)^x+3)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-3x-((+1/27)^x+3)+9^=0

We add all the numbers together, and all the variables

-3x-((+1/27)^x+3)=0

We multiply all the terms by the denominator


-3x*27)^x+3)-((+1=0

Wy multiply elements

-81x^2+1=0

a = -81; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-81)·1
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*-81}=\frac{-18}{-162}
=1/9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*-81}=\frac{18}{-162}
=-1/9
$