How do you evaluate log 0.01?

Hint: Write the argument of log, i.e., 0.01 in the fractional form and then convert it into the exponential form having base 10. Use the property of logarithm given as: - \[\log \left( {{m}^{a}} \right)=a\log m\] to simplify the expression. Now, use the identity: - log n = 1, to get the answer. Here, ‘n’ denotes the same argument and base of the logarithmic function and n > 0, \[n\ne 1\].

Complete step by step answer:
Here, we have been provided with the logarithmic expression: - log 0.01 and we are asked to evaluate it. That means we need to find its value.
Now, the given logarithmic function is a common log expression. That means the base value of the log is 10. Let us assume the value of this expression as E. So, we have,
\[\Rightarrow E={{\log }_{10}}0.01\]
Converting the argument, i.e., 0.01, into fractional form, we get,
\[\Rightarrow E={{\log }_{10}}\left( \dfrac{1}{100} \right)\]
Now, converting this fractional form into exponential form, we get,
\[\Rightarrow E={{\log }_{10}}\left( \dfrac{1}{{{10}^{2}}} \right)\]
Using the identity: - \[\dfrac{1}{{{a}^{m}}}={{a}^{-m}}\], we have,
\[\Rightarrow E={{\log }_{10}}\left( {{10}^{-2}} \right)\]
Applying the formula of logarithm given as: - \[\log \left( {{m}^{a}} \right)=a\log m\], we get,
\[\Rightarrow E=\left( -2 \right)\times {{\log }_{10}}10\]
Now, since the base and argument of the logarithmic expression is the same and we know that \[{{\log }_{n}}n=1\], where n > 0, \[n\ne 1\].
So, we have,
\[\begin{align}
  & \Rightarrow E=\left( -2 \right)\times 1 \\
 & \Rightarrow E=\left( -2 \right) \\
\end{align}\]
Hence, (-2) is the simplified form of \[{{\log }_{10}}0.01\].

Note:
One must know the difference between common log and natural log to solve the above question. Common log has base 10 and natural log has base e. Natural log is denoted by ln. You must remember the basic formulas of logarithm like: - \[\log m+\log n=\log \left( mn \right)\], \[\log m-\log n=\log \dfrac{m}{n}\], \[\log {{m}^{n}}=n\log m\], \[{{\log }_{\left( {{a}^{b}} \right)}}m=\dfrac{1}{b}{{\log }_{a}}m\] etc. Note that it was very important to convert the given decimal number into exponential form, otherwise we would have not been able to solve the question easily.