Question

log₂² (2x) + 3log₂ (4x)= 13
cum se rezolva?! ca nu inteleg :(

Answer 1

Vom transforma ecuatia, dupa cum urmeaza:

[tex]\log_2^2 2x+3\log_24x-13=0 [/tex]

Notam 2y = t,  t > 0

[tex]\log_2^2 t+3\log_2 2t-13=0 \Leftrightarrow \log_2 ^2 t+3(\log_22+\log_2t)-13=0 \\\;\\ \Leftrightarrow \log_2 ^2t+3(1+\log_2t)-13=0 \\\;\\ \it Notam\ \ \log_2t = y \\\;\\ Ecuatia\ \ devine: \\\;\\ y^2+3(1+y)-13=0 \Leftrightarrow y^2+3y-10=0 \\ . [/tex]

Dupa rezolvare, obtinem:  

$y_1 = -5,\ \ y_2=2$

Revenind asupra notatiei, rezulta:

[tex]\log_2t=-5 \Longrightarrow t=2^{-5} \Longrightarrow t =\dfrac{1}{2^5} \\\;\\ \log_2t=2 \Longrightarrow t = 2^2 \Longrightarrow t=4[/tex]

Dar, t=2x, iar de aici obtinem:

[tex]\it 2x =\dfrac{1}{2^5} \Longrightarrow x = \dfrac{1}{2^6}\Longrightarrow x=\dfrac{1}{64} \\\;\\ \it 2x = 4 \Longrightarrow x = 2 [/tex]

Deci, ecuatia data are doua solutii:

$\it x_1=\dfrac{1}{64}\ \ si\ \ x_2 = 2.$