CAT 2003 Slot 2 — QA Question 24

$\frac{1}{3}{\log }_{3}M+3{\log }_{3}N+1+{\log }_{0.008}5$...(i)

Consider the R.H.S of (i):

$\begin{array}{rl}& 1+{\log }_{0.008}5\\ & =1+\frac{\log 5}{\log 0.008}\\ & =1+\frac{\log \left(\frac{10}{2}\right)}{\log \left(\frac{8}{1000}\right)}\\ & =1+\frac{\log 10-\log 2}{\log 8-\log 1000}=1+\frac{\log 10-\log 2}{3\log 2-3\log 10}\end{array}$

$=1+\frac{\log 10-\log 2}{-3(\log 10-\log 2)}=1-\frac{1}{3}=\frac{2}{3}$ ...(ii)

Consider the L.H.S of (i):

$\begin{array}{rl}& \frac{1}{3}{\log }_{3}M+3{\log }_{3}N\end{array}$

$={\log }_3{M}^{1/3}+{\log }_3{N}^3={\log }_3\left(M^{1/3}{N}^3\right)$ ....(iii)

Equating equations (ii) and (iii), we get,

$\begin{array}{rl}& {\log }_3\left(M^{1/3}{N}^3\right)=\frac{2}{3}\\ & \therefore M^{1/3}{N}^3=3^{2/3}\end{array}$

Cubing both sides, we get,

M × N⁹ = 9

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

Hence, option (b).