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Question Number 193253 by MATHEMATICSAM last updated on 08/Jun/23

$$\mathbf{Select}\mathbf{the}\mathbf{correct}\mathbf{option}\mathbf{with}$$$$\mathbf{explaination}:$$$$\mathrm{If}\frac{1}{3}{\log }_{3}M+{3\log }_{3}N=1+{\log }_{0.008}5\mathrm{then}$$$$\mathrm{a}.M^9=\frac{9}{N}$$$$\mathrm{b}.N^9=\frac{9}{M}$$$$\mathrm{c}.M^3=\frac{3}{N}$$$$\mathrm{d}.N^3=\frac{3}{M}$$

Answered by deleteduser1 last updated on 08/Jun/23

$${log}_3{\left({MN}^9\right)}^{\frac{1}{3}}=1+x$$$${log}_{0.008}5=x\Rightarrow {\left(5^{-3}\right)}^x=5\Rightarrow x=-\frac{1}{3}$$$$\Rightarrow 3^{\frac{2}{3}}={\left({MN}^9\right)}^{\frac{1}{3}}\Rightarrow 9={MN}^9\Rightarrow N^9=\frac{9}{M}$$

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