8^x+2=(1/32)^-5x

Solution for 8^x+2=(1/32)^-5x equation:

8^x+2=(1/32)^-5x

We move all terms to the left:

8^x+2-((1/32)^-5x)=0


Domain of the equation: 32)^-5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

8^x-((+1/32)^-5x)+2=0

We multiply all the terms by the denominator


8^x*32)^-5x)-((+2*32)^-5x)+1=0

We add all the numbers together, and all the variables

8^x*32)^-5x)-(64^-5x)+1=0

Wy multiply elements

256x^2=0

a = 256; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·256·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{512}=0$