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

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

Then

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

$(a/b) =$

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

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

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 $x:=dfrac ab-1$,

$$logleft(frac abright)approx frac ab-1.$$

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

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