3^4-x=(1/27)^x

Solution for 3^4-x=(1/27)^x equation:

3^4-x=(1/27)^x

We move all terms to the left:

3^4-x-((1/27)^x)=0


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

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-x-((+1/27)^x)+3^4=0

We add all the numbers together, and all the variables

-1x-((+1/27)^x)+81=0

We multiply all the terms by the denominator


-1x*27)^x)-((+81*27)^x)+1=0

We add all the numbers together, and all the variables

-1x*27)^x)-(2187^x)+1=0

We add all the numbers together, and all the variables

-1x*27)^x)-2187^x+1=0

Wy multiply elements

-27x^2=0

a = -27; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-27)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{-54}=0$