4^x+1=(1/8)^x

Solution for 4^x+1=(1/8)^x equation:

4^x+1=(1/8)^x

We move all terms to the left:

4^x+1-((1/8)^x)=0


Domain of the equation: 8)^x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

4^x-((+1/8)^x)+1=0

We multiply all the terms by the denominator


4^x*8)^x)-((+1*8)^x)+1=0

We add all the numbers together, and all the variables

4^x*8)^x)-(8^x)+1=0

We add all the numbers together, and all the variables

4^x*8)^x)-8^x+1=0

Wy multiply elements

32x^2=0

a = 32; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·32·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{64}=0$