Answer

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Hint: - Use the properties of logarithms.

To convert the given term of logarithm in simpler form, we proceed as follows:

$

\Rightarrow \log {\left( {\sqrt {{a^2}{b^3}} } \right)^6} \\

\Rightarrow 6\log \left( {\sqrt {{a^2}{b^3}} } \right){\text{ }}\left[ {\because \log {m^n} = n\log m} \right] \\

\Rightarrow 6\log {\left( {{a^2}{b^3}} \right)^{\dfrac{1}{2}}} \\

\Rightarrow \dfrac{6}{2}\log \left( {{a^2}{b^3}} \right) \\

\Rightarrow 3\log \left( {{a^2}{b^3}} \right) \\

\Rightarrow 3\left( {\log \left( {{a^2}} \right) + \log \left( {{b^3}} \right)} \right){\text{ }}\left[ {\because \log mn = \log m + \log n} \right] \\

\Rightarrow 3\left( {2\log a + 3\log b} \right){\text{ }}\left[ {\because \log {m^n} = n\log m} \right] \\

\Rightarrow 6\log a + 9\log b \\

$

Note:- Different properties of logarithms have been used in the above question in order to simplify the terms, also all of these formulae used have been mentioned along with for better understanding and these formulae must be remembered. Also the logarithm is the inverse function to exponentiation which has a very big advantage while solving large problems.

To convert the given term of logarithm in simpler form, we proceed as follows:

$

\Rightarrow \log {\left( {\sqrt {{a^2}{b^3}} } \right)^6} \\

\Rightarrow 6\log \left( {\sqrt {{a^2}{b^3}} } \right){\text{ }}\left[ {\because \log {m^n} = n\log m} \right] \\

\Rightarrow 6\log {\left( {{a^2}{b^3}} \right)^{\dfrac{1}{2}}} \\

\Rightarrow \dfrac{6}{2}\log \left( {{a^2}{b^3}} \right) \\

\Rightarrow 3\log \left( {{a^2}{b^3}} \right) \\

\Rightarrow 3\left( {\log \left( {{a^2}} \right) + \log \left( {{b^3}} \right)} \right){\text{ }}\left[ {\because \log mn = \log m + \log n} \right] \\

\Rightarrow 3\left( {2\log a + 3\log b} \right){\text{ }}\left[ {\because \log {m^n} = n\log m} \right] \\

\Rightarrow 6\log a + 9\log b \\

$

Note:- Different properties of logarithms have been used in the above question in order to simplify the terms, also all of these formulae used have been mentioned along with for better understanding and these formulae must be remembered. Also the logarithm is the inverse function to exponentiation which has a very big advantage while solving large problems.

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