Question

Me explican cómo resolver este ejercicio, por favor :(

Answer 1

$27logc-8=0$
$27logc=8$
$logc= \frac{8}{27}$
*$log x^{b} =b(logx)$
$log \sqrt[3]{c^{2} } =logc^{ \frac{2}{3}} = \frac{2}{3} logc= \frac{2}{3}. \frac{8}{27} = \frac{16}{81}$