49^x+2=(1/7)^11-x

Solution for 49^x+2=(1/7)^11-x equation:

49^x+2=(1/7)^11-x

We move all terms to the left:

49^x+2-((1/7)^11-x)=0


Domain of the equation: 7)^11-x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

49^x-((+1/7)^11-x)+2=0

We multiply all the terms by the denominator


49^x*7)^11-x)-((+2*7)^11-x)+1=0

We add all the numbers together, and all the variables

49^x*7)^11-x)-(14^11-x)+1=0

Wy multiply elements

343x^2=0

a = 343; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·343·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{686}=0$