If $p^{3}=q^{4}=r^{5}=s^{6}$, then the value of $\log_{s}\left ( pqr \right )$ is equal to

- $16/5$
- $1$
- $24/5$
- $47/10$

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## 1 Answer

Let say $, \;p^{3} = q^{4} = r^{5} = s^{6} = k$

Now,

- $ p^{3} = k \Rightarrow p = k^{\frac{1}{3}} $
- $ q^{4} = k \Rightarrow q = k^{\frac{1}{4}} $
- $ r^{5} = k \Rightarrow r = k^{\frac{1}{5}} $
- $ s^{6} = k \Rightarrow s = k^{\frac{1}{6}} $

Now, $ \log_{s} (pqr) = \log_{k^{\frac{1}{6}}} \left( k^{\frac{1}{3}} \cdot k^{\frac{1}{4}} \cdot k^{\frac{1}{5}} \right) $

$ \quad = \log_{k^{\frac{1}{6}}} \left[ k^{\left( \frac{1}{3} + \frac{1}{4} + \frac{1}{5}\right)} \right] $

$ \quad = \log_{k^{\frac{1}{6}}} \left[ k^{\left(\frac{ 20+15+12}{60}\right)} \right] $

$ \quad = \log_{k^{\frac{1}{6}}} \left( k^{\frac{47}{60}} \right) $

$ \quad = \frac{47}{60} \; \log_{k^{\frac{1}{6}}} k \quad [ \because \log_{b} a^{x} = x\log_{b} a ] $

$ \quad = \frac{47}{60} \times 6 \log_{k} k \quad \left[ \because \log_{b^{x}} a = \frac{1}{x} \log_{b} a \right] $

$ \quad = \frac{47}{10} \quad [ \because \log_{a} a = 1 ] $

Correct Answer $ : \text{D}$