Question

Resolva em R as equacoes :

4^x+1 - 2^x+1 - 56=0

Answer 1

4^(x+1) - 2^(x+1) - 56=0

Repare que 4=2^2.Logo,4^(x+1)=(2^2)^(x+1)= (2^(x+1))^2

(2^(x+1))^2 -2^(x+1) - 56=0

Tome y=2^(x+1).Logo:

y^2-y-56=0

Δ=1+224=225

Sendo y1 e y2 as raízes:

y1=(1+15)/2 = 8

y2=(1-15)/2=-7

Descobrindo x:

2^(x+1)=8 => x=2

2^(x+1) = -7 (não existe x real que satisfaça isso)

Logo,x=2