[latex] 3^{log_{x}3} = x^{9} [/latex]3^(logx 3(основание х) = x^9-x=x^9 получилось

ОДЗ: x>0 и х неравен 1 [latex] 3^{log_x 3}=3^{log_3 {x^9}} \\ log_x 3=log_3 {x^9} \\ \frac {1}{log_3 x}= 9 log_3 x \\ t=log_3 x \\ 1/t=9t \\ t^2=1/9 \\ t_1=1/3 <=> log_3 x=1/3 \\ x=3^{1/3} \\ t_2=-1/3 <=> log_3 x=-1/3 \\ x=3^{-1/3} [/latex] Ответ: [latex]3^{1/3}; 3^{-1/3}.[/latex]