Question

How do you evaluate \[{\log _{216}}6\] ?

Answer 1

Hint: Here in this question we are going to solve this function. The function is of the form of a logarithmic function. The logarithmic function is represented as \[{\log _b}a\] , where b is a base and a is a number either it is a fraction or a whole number. By using the properties of logarithmic functions, we are going to simplify the question.

Complete step by step solution:
The logarithmic function is an inverse of exponential function. Here we have to solve the above logarithmic function. This logarithmic function contains 216 as a base. Hence by applying the properties of logarithmic functions we solve this function.
Now consider the given function \[{\log _{216}}6\]
When the logarithmic function is defined as \[{\log _b}a\] , by the property of logarithmic function it is written as \[\dfrac{{\log a}}{{\log b}}\]
Therefore the given function is written as
 \[ \Rightarrow \dfrac{{\log 6}}{{\log 216}}\]
Since it is just log so it is a common logarithm.
Let we factorise the number 216.
Therefore the number 216 is written as \[6 \times 6 \times 6\] . The number 6 is multiplied thrice. Therefore the number is written in the form of exponential. So it is written as \[{6^3}\]
So the above function is written as
 \[ \Rightarrow \dfrac{{\log 6}}{{\log {6^3}}}\]
In the denominator of the above function we have the log function is in the form \[\log {a^n}\] , so we have property \[\log {a^n} = n\log a\] . By applying the property we have
 \[ \Rightarrow \dfrac{{\log 6}}{{3\log 6}}\]
In the numerator and the denominator we have log6 so we can cancel it. So we have
 \[ \Rightarrow \dfrac{1}{3}\]
Hence we have evaluated the function and obtained an answer.
Therefore \[{\log _{216}}6 = \dfrac{1}{3}\]
So, the correct answer is “$\dfrac{1}{3}$”.

Note: The logarithmic functions have many properties. These properties are based on the exponential number and on the arithmetic operation like addition, subtraction, multiplication and division. So by using these properties we can solve the logarithmic properties. We have different values of logarithms for different base values.