Question

How do you solve \[{{\log }_{10}}0.01\]?

Answer 1

Hint: A common logarithm is the logarithm of base 10. To get the logarithm of a number n, find the number x that when the base is raised to that power, the resulting value is n. The properties of logarithm required for evaluating are \[{{\log }_{b}}{{a}^{n}}=n{{\log }_{b}}a\].If bases are equal, then we can use \[{{\log }_{b}}{{b}^{n}}=n\].

Complete step by step answer:
As per the given question, we have to evaluate the given logarithmic expression. We can easily simplify the given logarithmic expression by converting decimal to exponent form. Here, we have the logarithmic expression \[{{\log }_{10}}0.01\].
In the given expression, let \[a=0.01\]. Since a is a decimal. We need to convert it into fractions. So, in order to get rid of the decimal we multiply the decimal with 100 and divide with 100. Thus, we can rewrite the expression as
\[\Rightarrow 0.01\times {100}/{100}\;=\dfrac{1}{100}\]
We can write 100 as \[10\times 10\]. Then
\[\Rightarrow \dfrac{1}{100}=\dfrac{1}{10\times 10}=\dfrac{1}{{{10}^{2}}}\]
From the properties of exponents,\[\dfrac{1}{{{a}^{n}}}={{a}^{-n}}\].
Using the above property we can rewrite a as
\[\begin{align}
  & \Rightarrow \dfrac{1}{100}=0.01={{10}^{-2}} \\
 & \Rightarrow a=0.01={{10}^{-2}} \\
\end{align}\]
On substituting \[a\] value we get the expression as \[{{\log }_{10}}{{10}^{-2}}\].
Now we can use the property \[{{\log }_{b}}{{b}^{n}}=n\] as the base is 10 and the number is 10 raised to certain powers. Using the property \[{{\log }_{b}}{{b}^{n}}=n\], we get
\[\Rightarrow {{\log }_{10}}{{10}^{-2}}=-2\]
\[\therefore {{\log }_{10}}0.01=-2\] is the required answer.

Note:
We can find the value of \[{{\log }_{10}}0.01\] directly by writing \[0.01\] as \[{{10}^{-2}}\] and using the property \[{{\log }_{b}}{{b}^{n}}=n\],we get -2 as the answer. We should remember that we are dealing with a logarithmic function which doesn’t exist for \[x\underline{<}0\]. In order to solve these types of questions, we need to have enough knowledge on properties of logarithms. We should avoid calculation mistakes to get the desired results.