CAT 2003 | Question: 2-74

Question

If $\frac{1}{3} \log_3 \text{M} + 3 \log_3 \text{N} =1 + \log_{0.008} 5$, then

• A. $\text{M}^9 = \frac{9}{\text{N}}$
• B. $\text{N}^9 = \frac{9}{\text{M}}$
• C. $\text{M}^3 = \frac{3}{\text{N}}$
• D. $\text{N}^9 = \frac{3}{\text{M}}$

Answer 1

$\frac{1}{3} \log_3 M + 3 \log_3 N=1 + \log_{0.008} 5$

$\log_3 M^{\frac{1}{3}} + \log_3 N^{3}=1 + \log_{0.008} 5$

$\log_3 M^{\frac{1}{3}} N^{3}=1 + \log_{0.008} 5$

$\log_3 M^{\frac{1}{3}} N^{3}=1 + \frac{\log 5 }{\log 0.008}$

$\log_3 M^{\frac{1}{3}} N^{3}=1 + \frac{\log \frac{10}{2} }{\log \frac{8}{1000}}$

$\log_3 M^{\frac{1}{3}} N^{3}=1 + \frac{\log10 - \log 2 }{\log 8 - \log 1000}$

$\log_3 M^{\frac{1}{3}} N^{3}=1 + \frac{\log10 - \log 2 }{3(\log 2 - \log 10)}$

$\log_3 M^{\frac{1}{3}} N^{3}=1-\frac{1}{3}$

$\log_3 M^{\frac{1}{3}} N^{3}=\frac{2}{3}$

$M^{\frac{1}{3}} N^{3}=3^{\frac{2}{3}}$

$M* N^{9}=9$

$N^{9}=\frac{9}{M}$

 

Hence,Option(B)$N^{9}=\frac{9}{M}$ is the correct choice.