Find the value of \(2log{2\over 5}\) + \(3log{25\over 8}\) – \(log{625\over 128}\).

Solution :

\(2log{2\over 5}\) + \(3log{25\over 8}\) + \(log{128\over 625}\)

= \(log{2^2\over 5^2}\) + \(log({5^2\over 2^3})^3\) + \(log{2^7\over 5^4}\)

= \(log({2^2\over 5^2}{5^6\over 2^9}{2^7\over 5^4})\) = log 1 = 0

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