Intuition behind natural log returns in finance and how to calculate natural log without calculator?

Question

I got asked this question in an interview and the interviewer only gave a hint. I can't find a convincing answer behind it.

Regarding the intuition part, what I partially know are the properties of additivity and $\ln(1+x)\approx x$

The interviewer subsequently asked me to calculate $\ln(105/100)$ without using a scientific calculator. The answer that was given to me was $5/102.5$. I checked that the answers match only approximately. As of now, it appears to be related to mean value theorem.

The $\ln$ function is continuous on $[100,105]$ and differentiable on $(100,105)$. Hence, by MVT, there exists a $c\in (100,105)$ such that

$\ln105-\ln100=\dfrac{1}{c}(105-100)=\dfrac{5}{c}$

Now, $100<c<105$. If we take an average of these two numbers, we can arrive at the approximation of 5/102.5.

This works on other examples $\ln\dfrac{107}{105}\approx \dfrac{2}{106.5}$

But why do we take an average ? Is this related to the shape of the plot of $\ln(x)$ or that $\ln(a)$ gives the area under the curve $1/x$ from $1$ to $a$

I feel that the correct answer should tie these two parts.

Please help.

Answer 1

The estimation made is the following:

$$\log(1+x) \approx x-\frac{x^2}{2}$$

which is compared to

$$\frac{x}{1+x/2} \approx x-\frac{x^2}{2}$$

hence

$$\log\left(\frac{105}{100}\right) \approx \frac{5}{102.5} \; .$$

Answer 2

There is an alternative approach:

$~\displaystyle \ln(105/100) = \ln(105) - \ln(100).~$

If $~f(x) = \ln(x),~$ then $~f'(x) = \dfrac{1}{x}.$

This means that you are trying to estimate the change in the function $~f(x)~$ where:

• $~x~$ goes from $~100~$ to $~105.$
  This is a change in the value of $~x~$ of $~+5~$ units.
  
  This means that hypothetically, if the rate of change of $~f(x)~$ was a constant value $~C~$ (which it is not), then the difference between $~f(105)~$ and $~f(100)~$ would have to be $~5 \times C.$

• However, in the interval $~x \in [100,105],~$ you have that $~f'(x)~$ starts at $~\dfrac{1}{100},~$ and ends in $~\dfrac{1}{105}.~$
  
  Therefore, you know immediately that
  $\dfrac{5}{105} < \ln(105) - \ln(100) < \dfrac{5}{100}.$

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Therefore, without bothering to attempt to analyze the mean rate of change of a function that is concave (i.e. curved downward), a very reasonable estimate for $~\ln(105) - \ln(100)~$ is

$$5 \times \frac{1}{102.5} = \frac{10}{205} = \frac{1}{20.5},$$

which is just under $~0.05.$