How do you evaluate \[{{\log }_{9}}729\]?

Hint: Evaluating logarithmic equations we have to first check for the logarithm’s base. In this case we have the logarithm base as 9, so we will express the number given as a power of the base of the logarithm. Then using the logarithm formula, we will solve the equation.

Complete step-by-step solution:
Logarithm equation can be explained as an equation consisting of logarithm in the expression. To solve this expression, we will use the logarithm formula which is as follows:
Suppose we have, \[{{a}^{x}}=b\], then using logarithm to write this expression we have,
\[{{\log }_{a}}b=x\]
Here, we have ‘a’ as the logarithm base. We can say that, ‘x’ is the logarithm of ‘b’ to the base ‘a’.
Let us understand this property using an example.
For example – we know that the product of two written thrice is eight. It can be written as:
\[{{2}^{3}}=8\]
If we use logarithm in the above expression, we can rewrite it as:
\[{{\log }_{2}}8=3\]
So we have the base of the logarithm as 2, and since 8 can be written as two raised to the power 3. We get the following answer as 3.
\[\Rightarrow {{\log }_{2}}{{2}^{3}}=3{{\log }_{2}}2=3\], as \[{{\log }_{2}}2=1\]
According to the question we have, we have to solve for \[{{\log }_{9}}729\],
We can see here that the base of the logarithm is 9, so we will write the number 729 as a power of 9. It can be written as:
\[729=9\times 9\times 9={{9}^{3}}\]
Therefore, we have
 \[{{\log }_{9}}729\]
\[\Rightarrow {{\log }_{9}}{{9}^{3}}=3{{\log }_{9}}9=3\]
Therefore, \[{{\log }_{9}}729=3\]
We can confirm this as \[{{\log }_{9}}729=3\] would mean, \[{{9}^{3}}=729\] and which is correct.

Note: We can also solve the equation by taking,
\[{{\log }_{9}}729=y\]
As we know that, \[{{\log }_{a}}b=x\] can also be written as \[{{a}^{x}}=b\]. So, we have
\[{{9}^{y}}=729\]
Writing the above expression as a power of same bases, we can have
\[\Rightarrow {{({{3}^{2}})}^{y}}={{(3)}^{6}}\]
\[\Rightarrow {{(3)}^{2y}}={{(3)}^{6}}\]
Since, the bases are same, we can equate their powers, we get
\[\Rightarrow 2y=6\]
\[\Rightarrow y=3\]
Therefore, \[{{\log }_{9}}729=3\]