Is $(a/b)-1$ approximately equal to $\log_e (a/b)$

Question

I was reading an article where in one of the steps we were trying to calculate the daily return. It said

Return = (a / b) – 1

It then said, this equation can be approximated to:

Return = Log e (a/b)

Could someone explain a proof around how these are equal? Why $\log_e$ (and not $\log$ base of another value)?

Answer 1

This is explained by the Taylor's theorem, expanding to the first order:

$$\log(1+x)\approx \log(1+x)_{x=0}+\left(\log(1+x)\right)'_{x=0}x=\frac x{1+0}=x.$$

For logarithms in other bases, it suffices to apply the conversion factor. (The natural logarithm is used because no factor is required by the derivative.)

With $x:=\dfrac ab-1$,

$$\log\left(\frac ab\right)\approx \frac ab-1.$$

The closer to $1$ the ratio, the better the approximation.

In fact, you are replacing the curve by its tangent:

Answer 2

Take it in the other direction $$\log \left(\frac{a}{b}\right)=\log \left(1+\frac{a-b}{b}\right)\approx \frac{a-b}{b}=\frac{a}{b}-1$$

Answer 3

Set $x:=\log_e (a/b)$, $a/b>0$;

Then

$(a/b) =e^x= 1+x+x^2/2! +...;$

$(a/b) =$

$1+\log_e (a/b) + O((\log_e (a/b))^2)$.