Question

How do you evaluate $\log (\dfrac{1}{{100}})$?

Answer 1

To evaluate $\log (\dfrac{1}{{100}})$
As we all know that the function used in this question is a logarithmic function and it has certain properties. So in this question, we will use one of its properties that is ${\log _a}{b^x}$. Then we can write it as ${\log _a}{b^x} = x{\log _a}b$ and then by substituting the value of ${\log _{10}}10 = 1$ we will evaluate the value of whole expression. But before using this property we satisfy its conditions which make it defined in its domain. Using this way we will try to solve this question.

Complete step by step solution:
To find the value of $\log (\dfrac{1}{{100}})$, we will use one of the properties of logarithmic
${\log _a}{b^x} = x{\log _a}b$
We also know that the value of ${\log _{10}}10 = 1$
So, we will use both the above statement to find the value of ${\log _{10}}\left( {\dfrac{1}{{100}}} \right)$.
We can write ${\log _{10}}\left( {\dfrac{1}{{100}}} \right)$ as ${\log _{10}}{10^{ - 2}}$
By using logarithmic property,
$\Rightarrow {\log _{10}}{10^{ - 2}} = - 2{\log _{10}}10$
And we know that ${\log _{10}}10 = 1$
So, substituting the above value we get,

\Rightarrow - 2{\log _{10}}10 = - 2 \times 1\\\ {\rm{ }} = - 2 \end{array}$$ **So, our answer is $$ - 2$$.** **Note:** While using this function we need to take care that base value must be greater than 0 and must not be equal to 1 that is $${\log _a}b,{\rm{ }}a > 0{\rm{ }}\& {\rm{ }}a \ne 1$$. In this case, the log form and index form are interchangeable. we also have to take care that b also should be greater than zero. If these conditions are satisfied then only we will be able to solve the logarithmic function.