Evaluate ${\log _9}\left( {\dfrac{1}{{729}}} \right)$ ?

Hint: First use the identity of logarithmic function which states that ${\log _b}a = \dfrac{{\log a}}{{\log b}}$ and then use other identities to cancel out log 9 and thus obtain the answer.

Complete step by step answer:
Given that we need to evaluate ${\log _9}\left( {\dfrac{1}{{729}}} \right)$.
Since we know that ${\log _b}a = \dfrac{{\log a}}{{\log b}}$ for any a and b.
Replacing a by $\dfrac{1}{{729}}$ and b by 9, we will then obtain the following equation:-
$ \Rightarrow {\log _9}\left( {\dfrac{1}{{729}}} \right) = \dfrac{{\log \left( {\dfrac{1}{{729}}} \right)}}{{\log 9}}$ ………………(1)
Now, we also know that $\log {a^{ - 1}} = - \log a$.
Replacing a by 729, we will get: $\log \left( {\dfrac{1}{{729}}} \right) = \log {(729)^{ - 1}} = - \log (729)$
Putting this in equation number (1), we will then obtain the following equation:-
$ \Rightarrow {\log _9}\left( {\dfrac{1}{{729}}} \right) = \dfrac{{ - \log 729}}{{\log 9}}$ ………………(2)
Now, we also know that $\log {a^n} = n\log a$.
Replacing a by 9 and n by 3, we will get the expression given by: $ - \log 729 = - \log {9^3} = - 3\log 9$
Putting this in equation number (2), we will then obtain the following equation:-
$ \Rightarrow {\log _9}\left( {\dfrac{1}{{729}}} \right) = \dfrac{{ - 3\log 9}}{{\log 9}}$
Crossing off log 9 from both numerator and denominator in the above equation to get the following equation:-
$ \Rightarrow {\log _9}\left( {\dfrac{1}{{729}}} \right) = - 3$

Note: Here we crossed – off log 9 from both the numerator and denominator but we could do that only because it is not equal to zero that is definite. We can never cancel any such possible thing which can be zero.
There is no value of any logarithmic function given to us, so it is kind of evident that we will somehow eliminate the logarithmic function and thus get the required answer. It is also of note that 729 is the cube of 9 and in the denominator we already had a log of 9. These both things get common of both numerator and denominator and thus we could cancel log 9 from it.
Remember following formulas:-
${\log _b}a = \dfrac{{\log a}}{{\log b}}$
$\log {a^{ - 1}} = - \log a$
$\log {a^n} = n\log a$