7^x=(1/49)^(x+3)

Solution for 7^x=(1/49)^(x+3) equation:

7^x=(1/49)^(x+3)

We move all terms to the left:

7^x-((1/49)^(x+3))=0


Domain of the equation: 49)^(x+3))!=0

x∈R


We add all the numbers together, and all the variables

7^x-((+1/49)^(x+3))=0

We multiply all the terms by the denominator


7^x*49)^(x+1+3))-((=0

We add all the numbers together, and all the variables

7^x*49)^(x+4))-((=0

We add all the numbers together, and all the variables

7^x*49)^(x=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:
$x=\frac{-b}{2a}=\frac{0}{686}=0$