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# Given $\log 2$ and $\log 3$ , find the value of $\log \left( \sqrt{48}\times {{108}^{\dfrac{1}{4}}}\div \sqrt{6} \right)$ . Verified
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Hint: The given question is related to logarithms and its properties. Try to recall the properties of logarithms which are related to logarithm of product and division of two numbers and logarithms of numbers with exponents.

Before solving the problem , we must know about the properties of logarithms. The following properties will be used in solving the question :

$\log (a\times b)=\log (a)+\log (b)$
$\log \left( \dfrac{a}{b} \right)=\log (a)-\log (b)$
$\log {{(a)}^{b}}=b\times \log (a)$
Now , we know that $\sqrt[n]{a}$ can be written as ${{a}^{\dfrac{1}{n}}}$ , or we can say $\sqrt[n]{a}={{a}^{\dfrac{1}{n}}}$.
So , we can write $\sqrt{48}$ as ${{48}^{\dfrac{1}{3}}}$ and $\sqrt{6}$ as ${{6}^{\dfrac{1}{12}}}$ .
Now , we need to find the value of $\log \left( \sqrt{48}\times {{108}^{\dfrac{1}{4}}}\div \sqrt{6} \right)$.
We know , $\sqrt{48}={{48}^{\dfrac{1}{3}}}$ and $\sqrt{6}={{6}^{\dfrac{1}{12}}}$. So , $\log \left( \sqrt{48}\times {{108}^{\dfrac{1}{4}}}\div \sqrt{6} \right)=\log \left( {{48}^{\dfrac{1}{3}}}\times {{108}^{\dfrac{1}{4}}}\div {{6}^{\dfrac{1}{12}}} \right)$.
Now , we know $\log (a\times b)=\log (a)+\log (b)$ and $\log \left( \dfrac{a}{b} \right)=\log (a)-\log (b)$.
So, $\log \left( {{48}^{\dfrac{1}{3}}}\times {{108}^{\dfrac{1}{4}}}\div {{6}^{\dfrac{1}{12}}} \right)=\log \left( {{48}^{\dfrac{1}{3}}} \right)+\log \left( {{108}^{\dfrac{1}{4}}} \right)-\log \left( {{6}^{\dfrac{1}{12}}} \right)$.
Now , we know $\log {{(a)}^{b}}=b\times \log (a)$.
So , $\log \left( {{48}^{\dfrac{1}{3}}} \right)+\log \left( {{108}^{\dfrac{1}{4}}} \right)-\log \left( {{6}^{\dfrac{1}{12}}} \right)=\dfrac{1}{3}\log (48)+\dfrac{1}{4}\log (108)-\dfrac{1}{12}\log (6)$
We know we can write $48$ as $2\times 2\times 2\times 2\times 3$ ; $108$ as $2\times 2\times 3\times 3\times 3$ and $6$ as $2\times 3$.
So , $\dfrac{1}{3}\log (48)+\dfrac{1}{4}\log (108)-\dfrac{1}{12}\log (6)=\dfrac{1}{3}\left( \log \left( 2\times 2\times 2\times 2\times 3 \right) \right)+\dfrac{1}{4}\left( \log \left( 2\times 2\times 3\times 3\times 3 \right) \right)-\dfrac{1}{12}\left( \log \left( 2\times 3 \right) \right)$
Now , we know $\log (a\times b)=\log (a)+\log (b)$.
So , we can write $\log \left( 2\times 2\times 2\times 2\times 3 \right)$ as $\log 2+\log 2+\log 2+\log 2+\log 3=4\log 2+\log 3$.
Again , $\log \left( 2\times 2\times 3\times 3\times 3 \right)=\log 2+\log 2+\log 3+\log 3+\log 3=2\log 2+3\log 3$.
And , $\log \left( 2\times 3 \right)=\log 2+\log 3$
Now , after calculating all these values , we can write $\dfrac{1}{3}\log (48)+\dfrac{1}{4}\log (108)-\dfrac{1}{12}\log (6)$ as $\dfrac{1}{3}\left( 4\log 2+\log 3 \right)+\dfrac{1}{4}\left( 2\log 2+3\log 3 \right)-\dfrac{1}{12}\left( \log 2+\log 3 \right)$
Now , we will open the brackets . On opening the brackets , we get
$\log \left( \sqrt{48}\times {{108}^{\dfrac{1}{4}}}\div \sqrt{6} \right)=\dfrac{4}{3}\log 2+\dfrac{1}{3}\log 3+\dfrac{2}{4}\log 2+\dfrac{3}{4}\log 3-\dfrac{1}{12}\log 2-\dfrac{1}{12}\log 3$
Now , we will write all the terms with $\log 2$together .
So , we get $\log \left( \sqrt{48}\times {{108}^{\dfrac{1}{4}}}\div \sqrt{6} \right)=\dfrac{4}{3}\log 2+\dfrac{2}{4}\log 2-\dfrac{1}{12}\log 2+\dfrac{3}{4}\log 3+\dfrac{1}{3}\log 3-\dfrac{1}{12}\log 3$.
Now , we will take $\log 2$ and $\log 3$ common.
So , we get $\log \left( \sqrt{48}\times {{108}^{\dfrac{1}{4}}}\div \sqrt{6} \right)=\log 2\left( \dfrac{4}{3}+\dfrac{2}{4}-\dfrac{1}{12} \right)+\log 3\left( \dfrac{3}{4}+\dfrac{1}{3}-\dfrac{1}{12} \right)$
Now , we will take the LCM of the denominators and solve the fractions in the brackets.
To find the LCM , we will factorize the denominators.
\begin{align} & 3=3\times 1 \\ & 4=4\times 1 \\ & 12=4\times 3\times 1 \\ \end{align}
So , the LCM of the denominators is $4\times 3=12$.
So , $\left( \dfrac{4}{3}+\dfrac{2}{4}-\dfrac{1}{12} \right)=\dfrac{16+6-1}{12}=\dfrac{21}{12}=\dfrac{7}{4}$ and $\left( \dfrac{3}{4}+\dfrac{1}{3}-\dfrac{1}{12} \right)=\dfrac{9+4-1}{12}=\dfrac{12}{12}=1$
So , we can write $\log \left( \sqrt{48}\times {{108}^{\dfrac{1}{4}}}\div \sqrt{6} \right)$ as $\dfrac{7}{4}\log 2+\log 3$ .
Hence , the value of $\log \left( \sqrt{48}\times {{108}^{\dfrac{1}{4}}}\div \sqrt{6} \right)$ is $\dfrac{7}{4}\log 2+\log 3$.

Note: Students generally get confused between $\log \left( \dfrac{a}{b} \right)$ and $\dfrac{\log a}{\log b}$ . Both are not the same. $\log \left( \dfrac{a}{b} \right)=\log (a)-\log (b)$ which is not equal to $\dfrac{\log a}{\log b}$. Such confusion should be avoided and the formulas should be remembered. They are helpful in solving various problems related to logarithms.
Last updated date: 27th Sep 2023
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