Question

How do you evaluate ${\log _6}\left( {216} \right)$?

Answer 1

This problem deals with logarithms. This problem is rather very easy and very simple, though it seems to be complex. In mathematics logarithms are an inverse function of exponentiation. Which means that the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x.
If given ${\log _b}x = a$, then $x$ is given by :
$\Rightarrow x = {b^a}$
In order to solve this problem basic formulas of logarithms are used here, such as:
$\Rightarrow \log a - \log b = \log \left( {\dfrac{a}{b}} \right)$
$\Rightarrow {\log _{{a^n}}}{b^m} = \dfrac{m}{n}{\log _a}b$
$\Rightarrow {\log _a}a = 1$

Complete step-by-step answer:
Here we are given with the logarithmic expression which is : ${\log _6}\left( {216} \right)$
Here consider that when the given expression ${\log _6}\left( {216} \right)$ is compared to standard notation of logarithms which is ${\log _b}x$, here :
$\Rightarrow x = 216$
The base $b$, is given below:
$\Rightarrow b = 6$
Now consider the value of $x$ which is equal to 216, this number is the perfect cube.
The number 216 is a whole cube of the number 6, which is given below:
$\Rightarrow {6^3} = 216$
Now substituting this in the logarithmic expression of ${\log _6}\left( {216} \right)$, as shown below:
$\Rightarrow {\log _6}{6^3}$
We know that if a logarithmic function is expressed in the form of ${\log _{{a^n}}}{b^m}$, then the value of the expression is equal to $\dfrac{m}{n}{\log _a}b$.
So now, applying the same formula to the expression ${\log _6}{6^3}$, as given below:
$\Rightarrow {\log _6}{6^3} = 3{\log _6}6$
We know that the value of the logarithmic function of the base is the same as the number is equal to 1.
Here ${\log _6}6 = 1$
$\Rightarrow {\log _6}{6^3} = 3$
$\therefore {\log _6}\left( {216} \right) = 3$
Final answer: The value of the expression ${\log _6}\left( {216} \right)$ is equal to 3.

Note:
Please note that here while solving this problem, here basic logarithmic formulas are used, few others are mentioned here. There are more important logarithmic basic formulas such as:
$\Rightarrow {\log _{10}}\left( {ab} \right) = {\log _{10}}a + {\log _{10}}b$
$\Rightarrow {\log _{10}}\left( {\dfrac{a}{b}} \right) = {\log _{10}}a - {\log _{10}}b$
$\Rightarrow$If ${\log _e}a = b$, then $a = {e^b}$
Hence $a = {e^{{{\log }_e}a}}$, since $b = {\log _e}a$.