Question

$\sqrt[?]{3 ^{2x+1} } = \frac{9}{ \sqrt[6]{3} }$
Помогите решить​

Answer 1

$\sqrt[7]{3^{2x+1} }=\frac{9}{\sqrt[6]{3} }\\\\3^{\frac{2x+1}{7} }=\frac{3^{2}}{3^{\frac{1}{6}}}\\\\3^{\frac{2x+1}{7} }=3^{\frac{11}{6} }\\\\\frac{2x+1}{7}=\frac{11}{6}\\\\6(2x+1)=11*7\\\\12x+6=77\\\\12x=71\\\\\boxed{x=5\frac{11}{12} }$