Question

How do you find the exact value of \[{{\log }_{6}}\sqrt[3]{6}\] ?

Answer 1

Hint:The problem given above is related to logarithm and laws of indices. For that we will first express the log above in fraction form. Then we will simplify the radical form. After that the simplification of the fraction will give the exact answer.

Complete step by step answer:
Given that, \[{{\log }_{6}}\sqrt[3]{6}\]. Now as we know that,
\[{{\log }_{b}}a=\dfrac{\log a}{\log b}\]
Thus the log above can be written as,
\[{{\log }_{6}}\sqrt[3]{6}=\dfrac{\log \sqrt[3]{6}}{\log 6}\]
Now the radical so given is of the form\[\sqrt[3]{6}={{6}^{\dfrac{1}{3}}}\]
Thus the fraction becomes,
\[{{\log }_{6}}\sqrt[3]{6}=\dfrac{\log {{6}^{\dfrac{1}{3}}}}{\log 6}\]
Now the numerator we can say the power can be written as,
\[{{\log }_{6}}\sqrt[3]{6}=\dfrac{\dfrac{1}{3}\log 6}{\log 6}\]
Now the log in numerator and denominator both can be cancelled,
\[{{\log }_{6}}\sqrt[3]{6}=\dfrac{1}{3}\]

Therefore, the exact value of \[{{\log }_{6}}\sqrt[3]{6}\] is $\dfrac{1}{3}$.

Note:In order to solve this question we should only note that, \[{{\log }_{b}}a=\dfrac{\log a}{\log b}\] is the very basic and important simplification or formula we can say. Also note that the power of base can be taken as the coefficient of it, \[\log {{6}^{\dfrac{1}{3}}}=\dfrac{1}{3}\log 6\]. There are formulas applicable for both logarithms and indices.