[tex]\it log_3(9x-3)=log_3 ^2\left(x-\dfrac{1}{3}\right) \Leftrightarrow log_39\left(x-\dfrac{1}{3}\right) = log_3 ^2\left(x-\dfrac{1}{3}\right) \Leftrightarrow
\\ \\ \\
\Leftrightarrow log_3 9+log_3\left(x-\dfrac{1}{3}\right) =log_3 ^2\left(x-\dfrac{1}{3}\right) \Leftrightarrow
\\ \\ \\
\Leftrightarrow 2+log_3\left(x-\dfrac{1}{3}\right)=log_3 ^2\left(x-\dfrac{1}{3}\right) [/tex]
[tex]\it\ Notez\ \ t = log_3\left(x-\dfrac{1}{3}\right) [/tex]
Ecuația devine:
[tex]\it 2+t=t^2 \Leftrightarrow t^2-t-2=0 \Leftrightarrow t^2+t-2t-2=0\Leftrightarrow
\\ \\
\Leftrightarrow t(t+1) -2(t+1) =0 \Leftrightarrow (t+1)(t-2)=0 \Leftrightarrow t_1=-1,\ t_2=2[/tex]
Revenim asupra notației și obținem:
[tex]\it log_3\left(x-\dfrac{1}{3}\right) = -1 \Rightarrow x-\dfrac{1}{3}=3^{-1}\Rightarrow x-\dfrac{1}{3}= \dfrac{1}{3} \Rightarrow x_1=\dfrac{2}{3}
\\ \\ \\
log_3\left(x-\dfrac{1}{3}\right) = 2 \Rightarrow x-\dfrac{1}{3}=3^2 \Rightarrow x=9+\dfrac{1}{3} \Rightarrow x_2 = \dfrac{28}{3} [/tex]
Condiția de existență a ecuației este :
[tex]\it x-\dfrac{1}{3} \ \textgreater \ 0 \Rightarrow x\ \textgreater \ \dfrac{1}{3} \Rightarrow x\in \left(\dfrac{1}{3},\ \infty \right)[/tex]
[tex]\it x_1,\ x_2 \in\left(\dfrac{1}{3},\ \infty\right)[/tex]
Deci, mulțimea soluțiilor ecuației date este :
[tex]\it S=\left\{\dfrac{2}{3},\ \dfrac{28}{3}\right \}[/tex]
