Question

How do you evaluate ${\log _{81}}9$?

Answer 1

Hint: 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 : ${\log _{81}}9$
Here consider that when the given expression ${\log _{81}}9$ is compared to standard notation of logarithms which is ${\log _b}x$, here :
$ \Rightarrow x = 9$
The base $b$, is given below:
$ \Rightarrow b = 81$
Now consider the value of the base $b$ which is equal to 81, this number is a perfect square.
The number 81 is a whole square of the number 9, which is given below:
$ \Rightarrow {9^2} = 81$
Now substituting this in the logarithmic expression of ${\log _{81}}9$, as shown below:
$ \Rightarrow {\log _{{9^2}}}9$
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 ${\log _{{9^2}}}9$, as given below:
$ \Rightarrow {\log _{{9^2}}}9 = \dfrac{1}{2}{\log _9}9$
We know that the value of the logarithmic function of the base is the same as the number is equal to 1.
Here ${\log _9}9 = 1$
$ \Rightarrow {\log _{{9^2}}}9 = \dfrac{1}{2}$
$\therefore {\log _{81}}9 = \dfrac{1}{2}$

The value of the expression ${\log _{81}}9$ is equal to $\dfrac{1}{2}$.

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 $b = {\log _e}a$.