Question

How do you evaluate \[{\log _6}\left( {\dfrac{1}{{36}}} \right)\]?

Answer 1

Hint: Here in this question, we see the word log then the function is logarithmic function so to solve this question we use the logarithmic functions properties. The logarithmic function is having a base value 6. accordingly we apply properties and obtain the required result.

Complete step-by-step answer:
The logarithmic function is an inverse of exponential function. Here we have to solve the above logarithmic function. This logarithmic function contains 6 as a base. Hence by applying the properties of logarithmic functions we solve this function.
Now consider the given function \[{\log _6}\left( {\dfrac{1}{{36}}} \right)\]
By using the property \[\log \left( {\dfrac{a}{b}} \right) = \log a - \log b\], using this property the above function is written as
 \[ \Rightarrow {\log _6}\left( {\dfrac{1}{{36}}} \right) = {\log _6}1 - {\log _6}36\]
In the second term of RHS of the above equation we have 36. The number 36 can be written as a square of 6. So we have
 \[ \Rightarrow {\log _6}\left( {\dfrac{1}{{36}}} \right) = {\log _6}1 - {\log _6}{6^2}\]
By using the property \[\log {a^m} = m\log a\], using this property the above function is written as
 \[ \Rightarrow {\log _6}\left( {\dfrac{1}{{36}}} \right) = {\log _6}1 - 2{\log _6}6\]
We know the value of \[{\log _6}1 = 0\] and \[{\log _6}6 = 1\]. So by substituting we get
 \[ \Rightarrow {\log _6}\left( {\dfrac{1}{{36}}} \right) = 0 - 2(1)\]
On simplification we get
 \[ \Rightarrow {\log _6}\left( {\dfrac{1}{{36}}} \right) = - 2\]
We can also solve the above logarithmic function by using another method.
Now consider the given function \[{\log _6}\left( {\dfrac{1}{{36}}} \right)\]
The number 36 can be written as a square of 6. So we have
 \[ \Rightarrow {\log _6}\left( {\dfrac{1}{{36}}} \right) = {\log _6}\left( {\dfrac{1}{{{6^2}}}} \right)\]
The exponential number which is in the denominator we can write the exponential number in the numerator by changing the sign of the exponent.
So we have
 \[ \Rightarrow {\log _6}\left( {\dfrac{1}{{36}}} \right) = {\log _6}{6^{ - 2}}\]
By using the property \[\log {a^m} = m\log a\], using this property the above function is written as
 \[ \Rightarrow {\log _6}\left( {\dfrac{1}{{36}}} \right) = - 2{\log _6}6\]
We know the value of \[{\log _6}6 = 1\]. So by substituting we get
 \[ \Rightarrow {\log _6}\left( {\dfrac{1}{{36}}} \right) = - 2(1)\]
On simplification we get
 \[ \Rightarrow {\log _6}\left( {\dfrac{1}{{36}}} \right) = - 2\]
So, the correct answer is “- 2”.

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.