Question

Pam-pam-pam Va rog frumos!

Answer 1

1)

$\it 3^{log^2_3x} = (3^{log_3x})^{log_3 x} =x^{log_3x}$

Ecuația devine:


[tex]\it x^{log_3x} +x^{log_3x} =162 \Rightarrow 2x^{log_3x}=162|_{:2}\Rightarrow x^{log_3x}= 81 \Rightarrow \\\;\\ \Rightarrow x^{log_3x}= 3 ^4 \Rightarrow log_3 x^{log_3 x} =log_3 3^4 \Rightarrow (log_3 x)(log_3 x) =4log_3 3 \\\;\\ \Rightarrow (log^2_3 x) =4 \Rightarrow log_3 x = \pm2 [/tex]

[tex]\it log_3 x = -2 \Rightarrow x = 3^{-2} \Rightarrow x = \dfrac{1}{3^2} \Rightarrow x = \dfrac{1}{9} \\\;\\ \\\;\\ log^2_3 x =2 \Rightarrow x = 3^2 \Rightarrow x = 9[/tex]

Soluțiile ecuației sunt:

[tex]\it x_1 = \dfrac{1}{9}, \ \ \ \ x_2 = 9 \\\;\\ \\\;\\ x_1\cdot x_2 = \dfrac{1}{9}\cdot9= 1[/tex]