How do you write \[\log 0.001\] in exponential form?

Hint: First find the value of the given logarithmic expression \[\log 0.001\] by converting the argument of log, i.e., 0.001 in fractional form first. Now, in the next step we have to convert the obtained fractional form into the exponential form having base 10. Use the property of logarithm given as: - \[\log \left( {{a}^{m}} \right)=m\log a\] to simplify the expression. Now, use the identity \[{{\log }_{n}}n=1\] to get the value of log. Once the value of this logarithmic expression is found, use the formula: - if \[{{\log }_{m}}n=k\] then \[n={{m}^{k}}\] to get the required exponential form.

Complete step by step answer:
Here, we have been provided with the logarithmic expression: - \[\log 0.001\] and we have to convert it into the exponential form. But first, we need to find the value of this logarithmic expression.
Now, as we can see that the argument of log is in decimal form, 0.001, so on converting it into the fractional form, we get,
\[\Rightarrow \log 0.001=\log \left( \dfrac{1}{1000} \right)\]
The above expression can be written as: -
\[\Rightarrow \log 0.001=\log \left( \dfrac{1}{{{10}^{3}}} \right)\]
Using the identity, \[\dfrac{1}{{{a}^{m}}}={{a}^{-m}}\], we get,
\[\Rightarrow \log 0.001=\log \left( {{10}^{-3}} \right)\]
Using the formula: - \[\log {{a}^{m}}=m\log a\], we get,
\[\Rightarrow \log 0.001=-3\log 10\]
Since the given log is common log, i.e., log to the base 10, so we have,
\[\Rightarrow {{\log }_{10}}0.001=-3{{\log }_{10}}10\]
Using the formula: - \[{{\log }_{n}}n=1\], we get,
\[\Rightarrow {{\log }_{10}}0.001=-3\]
Now, we know that is \[{{\log }_{m}}n=k\] then \[n={{m}^{k}}\], so for the above relation, we have,
\[\Rightarrow 0.001={{10}^{-3}}\]
Hence, the above relation shows the exponential form and our answer.

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 ahs base e. Natural log is denoted by ln. You must remember the basic formulas of logarithmic like: - \[\log m+\log n=\log \left( mn \right)\], \[\log m-\log n=\log \left( \dfrac{m}{n} \right)\], \[\log {{m}^{n}}=n\log m\], \[{{\log }_{\left( {{a}^{b}} \right)}}m=\dfrac{1}{b}{{\log }_{a}}m\] 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.