Question

The value of ${\log }_9\left({\mathrm{antilog}}_{\frac{1}{243}}\left(\frac{-3}{5}\right)\right)$ equals

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $\frac{4}{3}$
• D. $\frac{3}{4}$

Answer 1

Answer: (A)

$\mathrm{anti}{\log }_{\frac{1}{243}}\left(\frac{-3}{5}\right)={\left(\frac{1}{243}\right)}^{\frac{-3}{5}}={\left(3^{-5}\right)}^{\times \frac{-3}{5}}=27$
$\therefore {\log }_{9}27={\log }_{3^2}(3{)}^3=\frac{3}{2}$