Răspuns:
Explicație pas cu pas:
mai intai observam ca [tex]log_{2} 4x = log_{2} 4 + log_{2} x = log_{2} 2^{2} + log_{2}x = 2log_{2} 2 + log_{2} x =2 + log_{2} x[/tex]
Asadar ecuatia devine:
[tex]log_{2} ^{2} x + 2+ log_{2} x =4[/tex] , sau
[tex]log_{2} ^{2} x + log_{2} x -2 = 0[/tex]
Daca notam
[tex]y= log_{2} x[/tex]
Avem ecuatia
[tex]y^{2} +y -2 =0[/tex]
[tex](y-1)(y+2)=0[/tex]
[tex]y=1 \\y=-2[/tex].
[tex]log_{2} x =1\\x=2^{1} \\x=2[/tex]
[tex]log_{2} x =-2\\x=2^{-2} \\x=\frac{1}{2^{2}}\\ \\x=\frac{1}{4}[/tex]
prin urmare, solutiile sunt 2 si 1/4.
