Question

How do you calculate $$\log 0.1$$?

Answer 1

Hint: In the above question we are given a logarithmic term that is $$\log 0.1$$. While approaching such kind of questions we should know about the logarithmic terms that is the logarithm is the inverse function of the exponentiation. That means the logarithm of a given number $$x$$ is the exponent to which another number, the base $$b$$, must be raised to produce that number $$x$$. Now the given logarithmic term that is $$\log 0.1$$ can also be written as $${\log }_{10}0.1$$ and can be solved further to get the desired answer.

Complete step-by-step answer:
Here we are given a logarithmic term (the logarithmic term is the inverse function of the exponentiation) that is $$\log 0.1$$ .
Firstly for calculating the given term that is $$\log 0.1$$ this expression can be written as- $${\log }_{10}0.1$$ now it can be solved further
Here the question one can ask is by how many times does the base $$10$$ needs to be raised to get the answer as $$0.1$$ that is the governing condition comes out to be –
$${10}^x=0.1={\displaystyle \frac{1}{10}}$$
Now further solving it we get –
$${10}^x={10}^{-1}$$
Now using the formula that if bases are same powers can be compared that is –
$$\begin{gathered} t^a=t^b \\ \Rightarrow a=b \\ \Rightarrow a-b=0 \end{gathered}$$
Now using this formula in above part we get-
$$x=-1$$
So the resultant answer is equal to $$-1$$
Hence the value of the equation asked in the question that is $$\log 0.1$$ comes out to be $$-1$$ .

Note: in such kind of questions care should be taken while making the equations from the given logarithmic term as it governs the whole question and also the calculation should be done with the utter concentration and precision as a silly mistake there can make the whole answer wrong .