Question

Sa se calculeze: $3log in baza 2 (tot logaritmul la patrat)^{2} din (x+1)-2log in baa 2 din (x+1)^2+1=0$

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Răspuns:

Explicație pas cu pas:

$3log_{2}^{2}(x+1)-2log_{2}(x+1)^{2}+1=0\\C.E.~~x+1>0,~=>~x>-1\\3log_{2}^{2}(x+1)-2log_{2}(x+1)^{2}+1=0~<=>~3log_{2}^{2}(x+1)-2*2log_{2}(x+1)+1=0~<=>~3log_{2}^{2}(x+1)-4log_{2}(x+1)+1=0~$

Notăm  log2(x+1)=t, obtinem,  3t²-4t+1=0, Δ=(-4)²-4·3·1=16-12=4

Deci t1=(4-2)/(2·3)=1/3, sau t1=(4+2)/6=1.

$Deci, log_{2}(x+1)=\frac{1}{3},~=>~x+1=2^{\frac{1}{3} },~=>x=\sqrt[3]{2}-1 ~>~-1,\\sau~ log_{2}(x+1)=1,~=>~x+1=2,~=>~x=1~>~-1\\Raspuns:~\sqrt[3]{2}-1; 1.$