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  4. The value of log 9( operatornameantilog(1/243)((-3/5))) equals

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

Continuity and Differentiability

Solution:

$ \operatorname{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$
$\therefore \log _9 27=\log _{3^2}(3)^3=\frac{3}{2}$