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
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\Leftrightarrow \log_2 ^2t+3(1+\log_2t)-13=0
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\it Notam\ \ \log_2t = y
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Ecuatia\ \ devine:
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y^2+3(1+y)-13=0 \Leftrightarrow y^2+3y-10=0
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.
[/tex]
Dupa rezolvare, obtinem:
[tex]y_1 = -5,\ \ y_2=2[/tex]
Revenind asupra notatiei, rezulta:
[tex]\log_2t=-5 \Longrightarrow t=2^{-5} \Longrightarrow t =\dfrac{1}{2^5}
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\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}
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\it 2x = 4 \Longrightarrow x = 2
[/tex]
Deci, ecuatia data are doua solutii:
[tex]\it x_1=\dfrac{1}{64}\ \ si\ \ x_2 = 2.[/tex]
