4^x+4=(1/2)^x

Solution for 4^x+4=(1/2)^x equation:

4^x+4=(1/2)^x

We move all terms to the left:

4^x+4-((1/2)^x)=0


Domain of the equation: 2)^x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

4^x-((+1/2)^x)+4=0

We multiply all the terms by the denominator


4^x*2)^x)-((+4*2)^x)+1=0

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

4^x*2)^x)-8^x+1=0

Wy multiply elements

8x^2=0

a = 8; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·8·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{16}=0$