Đáp án:
$x\in\{1,\dfrac{1}{6}\}$
Giải thích các bước giải:
$\log_22x.\log_33x=1$
$\rightarrow \dfrac{\log_32x}{\log_32}.\log_33x=1$
$\rightarrow \dfrac{\log_3x+\log_32}{\log_32}.(\log_3x+\log_33)=1$
$\rightarrow \dfrac{\log_3x+\log_32}{\log_32}.(\log_3x+1)=1$
Đặt $\log_3x=t$
$\rightarrow \dfrac{(log_32+t)(1+t)}{\log_32}=1$
$\rightarrow t^2+(log_32+1)t=0$
$+)t=0\rightarrow \log_3x=0\rightarrow x=1$
$+)t=-log_32-1\rightarrow x=3^{-log_23-1}=\dfrac{1}{6}$
