Question:
If p3 = q4 = r5 = s6, then the value of logs(pqr) is equal to
- 16/5
- 1
- 24/5
- 47/10
Correct Answer: 4
Let $p^{3}=q^{4}=r^{5}=s^{6}=k$
$p=k^{1 / 3}, q=k^{1 / 4}, r=k^{1 / 5}, s=k^{1 / 6}$
$\text{pqr}={{\text{k}}^{\left( \frac{20+15+12}{60} \right)}}={{\text{k}}^{\frac{47}{60}}}$
${{\log }_{\text{s}}}(\text{pqr})={{\log }_{{{\text{k}}^{\frac{1}{6}}}}}{{\text{k}}^{\frac{47}{60}}}$
$=\left(\frac{47}{60} \times 6\right) \log _{\mathrm{k}} \mathrm{k}$
$=\frac{47}{10}$
Let $p^{3}=q^{4}=r^{5}=s^{6}=k$
$p=k^{1 / 3}, q=k^{1 / 4}, r=k^{1 / 5}, s=k^{1 / 6}$
$\text{pqr}={{\text{k}}^{\left( \frac{20+15+12}{60} \right)}}={{\text{k}}^{\frac{47}{60}}}$
${{\log }_{\text{s}}}(\text{pqr})={{\log }_{{{\text{k}}^{\frac{1}{6}}}}}{{\text{k}}^{\frac{47}{60}}}$
$=\left(\frac{47}{60} \times 6\right) \log _{\mathrm{k}} \mathrm{k}$
$=\frac{47}{10}$
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