9^x=(1/27)^(4x+2)

Solution for 9^x=(1/27)^(4x+2) equation:

9^x=(1/27)^(4x+2)

We move all terms to the left:

9^x-((1/27)^(4x+2))=0


Domain of the equation: 27)^(4x+2))!=0

x∈R


We add all the numbers together, and all the variables

9^x-((+1/27)^(4x+2))=0

We multiply all the terms by the denominator


9^x*27)^(4x+1+2))-((=0

We add all the numbers together, and all the variables

9^x*27)^(4x+3))-((=0

We add all the numbers together, and all the variables

9^x*27)^(4x=0

Wy multiply elements

243x^2=0

a = 243; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·243·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{486}=0$