Question

Find the vale of the expression:  \(\rm \frac{log_{7}3 \times log_{2}7\times log_{3}2}{log_{5}\sqrt{15}}+\frac{1}{{log_{3}\sqrt{15}}} \)
1. 3
2. 1
3. 5
4. 2

Answer 1

Correct Answer - Option 4 : 2

Concept:

• \(\rm log_bm.log_ab = log_am\)
• \(\rm log_ba =\frac{1}{ log_ab}\)

Calculation:

Here we have to find the value of \(\rm \frac{log_{7}3 \times log_{2}7\times log_{3}2}{log_{5}\sqrt{15}}+\frac{1}{{log_{3}\sqrt{15}}} \)

⇒ \(\rm \rm \frac{log_{7}3 \times log_{2}7\times log_{3}2}{log_{5}\sqrt{15}}+\frac{1}{{log_{3}\sqrt{15}}} = \frac{log_{7}3 \times log_{3}7}{log_{5}\sqrt{15}}+\frac{1}{{log_{3}\sqrt{15}}}\)        ( ∴ \(\rm log_bm \times log_ab = log_am\))

⇒  \(\rm \frac{log_{7}3 \times log_{3}7}{log_{5}\sqrt{15}}+\frac{1}{{log_{3}\sqrt{15}}}= \frac{log_{3}3}{log_{5}\sqrt{15}}+\frac{1}{{log_{3}\sqrt{15}}}\)   ( ∴ \(\rm log_bm.log_ab = log_am\))

= \(\rm \frac{1}{log_{5}\sqrt{15}}+\frac{1}{{log_{3}\sqrt{15}}}\)

= \(\rm {log_{\sqrt{15}}5}+{{log_{\sqrt{15}}3}}\)            ( ∴ \(\rm log_ba =\frac{1}{ log_ab}\))

= \(\rm {log_{\sqrt{15}}15}\)

=  \(\rm {log_{\sqrt{15}}{(\sqrt{15})}^2}\)

= \(\rm 2 {log_{\sqrt{15}}{(\sqrt{15})}}\)

= 2

Hence, option D is correct.