Discussion

Explanation:

Given  $\log \frac{16}{27}-\log \frac{81}{64}+\log \frac{4}{3}$= log 16 – log 27 – (log 64 - log 81) + log 4 – log3

= 2log4 – 3log3 – 3log4 + 4log3 + log4 – log3 = 0

Alternately,

$\log \frac{16}{27}-\log \frac{64}{81}+\log \frac{4}{3}=\log (\frac{16}{27}\times \frac{64}{81}\times \frac{4}{3})$=

= log 1 = 0

Hence, option (a).

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