Question

Evaluate $\log \left( {216\sqrt 6 } \right)$ to the base 6.

Answer 1

Hint: Here, we are asked to find the value of $\log \left( {216\sqrt 6 } \right)$ to the base 6.
Firstly, write $\log \left( {216\sqrt 6 } \right)$ to the base 6 as ${\log _6}\left( {216\sqrt 6 } \right)$ .
Then, simplify ${\log _6}\left( {216\sqrt 6 } \right)$ to the form of ${\log _6}{6^n}$ , by using the property ${a^m}{a^n} = {a^{m + n}}$ .
After that, use the property $\log {x^n} = n\log x$ and write ${\log _6}{6^n}$ in the form of $n{\log _6}6$ .
Finally, using the property ${\log _x}x = 1$ , find the required answer.

Complete step-by-step answer:
Here, we are asked to find the value of $\log \left( {216\sqrt 6 } \right)$ to the base 6.
We can also write $\log \left( {216\sqrt 6 } \right)$ to the base 6 as ${\log _6}\left( {216\sqrt 6 } \right)$ .
Now, we know that 216 is a cube of 6. $\therefore 216 = {6^3}$
 $\therefore {\log _6}\left( {216\sqrt 6 } \right) = {\log _6}\left( {{6^3}\sqrt 6 } \right)$
Also, applying the property ${a^m}{a^n} = {a^{m + n}}$ .
 $\therefore {\log _6}\left( {216\sqrt 6 } \right) = {\log _6}\left( {{6^{3 + \dfrac{1}{2}}}} \right) = {\log _6}{6^{\dfrac{7}{2}}}$
Now, we can also apply the property $\log {x^n} = n\log x$ and we get
 ${\log _6}\left( {216\sqrt 6 } \right) = \dfrac{7}{2}{\log _6}6$
Now, we know that ${\log _x}x = 1$ i.e. if the base of the logarithm function and the value are the same, then its value becomes 1.
 $\therefore {\log _6}\left( {216\sqrt 6 } \right) = \dfrac{7}{2} \times 1 = \dfrac{7}{2}$
Thus, the value of $\log \left( {216\sqrt 6 } \right)$ to the base 6 is $\dfrac{7}{2}$ .

Note: In these types of questions, we must remember all the properties and identities of logarithmic functions, to solve the question, it is not possible to do the question if we don’t know the properties of the logarithmic functions.