Помогите срочно1)3log3x+3logx3=102)3log7x-2logx7=1

Решите задачу:

$1) 3log_3(x)+3log_x(3)=10 \\ \\ 3(log_3(x)+log_x(3))=10 \\ \\ 3(log_3(x)+ \frac{log_3(3)}{log_3(x)})=10 \\ \\ 3(log_3(x)+ \frac{1}{log_3(x)}=10 \\ \\ 3log_3(x)+ \frac{3}{log_3(x)}=10 \\ \\ 3AMEHA:t=log_3(x) \\ \\ 3t+ \frac{3}{t} =10 \\ \\ t=3 \\ t= \frac{1}{3} \\ \\ log_3(x)=3 \\ log_3(x)= \frac{1}{3} \\ \\ x=3^{3}=27 \\ x=3^{ \frac{1}{3}}= \sqrt[3]{3}$

$2)3log_7(x)-2log_x(7)=1 \\ \\ 3log_7(x)-2* \frac{log_7(7)}{log_7(x)} =1 \\ \\ 3log_7(x)- \frac{2}{log_7(x)} =1 \\ \\ 3AMEHA:t=log_7(x) \\ \\ 3t- \frac{2}{t}=1 \\ \\ t=1 \\ t=- \frac{2}{3} \\ \\ log_7(x)=1 \\ log_7(x)=- \frac{2}{3} \\ \\ x=7^{1}=7 \\ x=7^{ -\frac{2}{3}}= \frac{ \sqrt[3]{7} }{7}$