Question

Prove that : log 3√56÷9 = 1/3( log 7 - 2log 3) + log 2

Answer 1

Step-by-step explanation:

To Prove that : log 3√56÷9 = 1/3 log 7 - 2log 3 + log 2

Take

LHS

$log( \frac{ \sqrt[3]{56} }{9} ) \\ \\ = log( \sqrt[3]{56} ) - log(9) \\ \\ \because log( \frac{ a}{b} ) = log(a) - log(b) \\ \\ and \: log(ab) = log(a) + log(b) \\ \\ so \\ \\ = log( \sqrt[3]{56} ) - log( {3}^{2} ) \\ \\ = log( {56}^{ \frac{1}{3} } ) - 2log( {3} ) \\ \\ \because \: log( {m}^{n} ) = n log(m) \\ \\ so \\ \\ = \frac{1}{3} log(56) - 2 log(3) \\ \\ = \frac{1}{3} log(7 \times 8) - 2 log(3) \\ \\ = \frac{1}{3} ( log(7) + log(8) ) - 2 log(3) \\ \\ = \frac{1}{3} ( log(7) + log( {2}^{3} ) ) - 2 log(3) \\ \\ = \frac{1}{3} ( log(7) + 3log( {2}) ) - 2 log(3) \\ \\ \frac{1}{3} log(7) + \frac{1}{3} \times 3log( {2}) ) - 2 log(3) \\ \\ = \frac{1}{3} log(7) - 2 log(3) + log(2)$

Hence proved

Answer 2

Answer:

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