Question

logaritmando $\sqrt[3]{7}$ e base 49 

Answer 1

$\log_{49 } \sqrt[3]{7} =x$

$49^x= \sqrt[3]{7}$

$7^2^x=7^ \frac{1}{3}$

$2x= \frac{1}{3}$

$6x=1$

$\fbox{ x= \frac{1}{6} }$