Discussion

Explanation:

We know  ${\log }_{a}b=\frac{{\log }_{c}b}{{\log }_{c}a}$

Given,  ${\log }_{10}20\times {\log }_{20}50\times {\log }_{50}20\times {\log }_{20}30$

Converting all logs in base 10, we get:

= $\frac{\log 20}{\log 10}\times \frac{\log 50}{\log 20}\times \frac{\log 20}{\log 50}\times \frac{\log 30}{\log 20}=\frac{\log 30}{\log 10}={\log }_{10}30=\log 30$

Hence, option (d).

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