Răspuns:
3
Explicație pas cu pas:
[tex]log_{2}(2^{x}+1)+log_{0,5}(4^{x}-55)=0,~=>~log_{2}(2^{x}+1)+log_{2^{-1}}(4^{x}-55)=0,~=>~log_{2}(2^{x}+1)+\frac{1}{-1} log_{2}(4^{x}-55)=0,~=>~log_{2}(2^{x}+1)-log_{2}(4^{x}-55)=0,~=>~log_{2}\dfrac{2^{x}+1}{4^{x}-55}=0,~=>~\dfrac{2^{x}+1}{4^{x}-55}=2^{0},~\dfrac{2^{x}+1}{4^{x}-55}=1,~=>4^{x}-55=2^{x}+1,~=>~4^{x}-2^{x}-56=0\\Fie~2^{x}=t>0,~=>~4^{x}=(2^{t})^{2}=t^{2}.~~Obtinem~ecuatia~t^{2}-t-56=0[/tex]
Δ=(-1)²-4·1·(-56)=1+224=225>0, deci t1=(1-15)/2=-7 care nu convine deoarece t>0
t2=(1+15)/2=8, care convine
Deci 2ˣ=8, ⇒x=3.
Verificarea o las pe seama ta... :))), dar convine.
