<s>[INST] Nicolaus Mercator's extension of logarithmic power series from $(-1,1) \to \mathbb{R}$ I was reading this Quora answer in which the expansion of $\log(1+x)$ was extended in such a way that one could approximate any real number's natural logarithm quickly. The following was written in the answer as an identity:
$$ \ln(Z) - \ln(z)= 2 \left[ Q+ \frac{1}{3} Q^3... O(Q^5) \right] \tag{1}$$
With $Q = \frac{Z-z}{Z+z}$
How do we derive $(1)$ starting from :  $\ln(1+x) = x- \frac{x^2}{2!} +O(x^3)$? In the post it is mentioned that is done by $x= \frac{X-1}{X+1}$.. I suppose you put $X= \frac{Z}{z}$ but I can't see how the even powers on left side cancel and how one will get $\ln(Z) - \ln(z)$ on LHS.

It maybe noted that for finding $\ln(2), \ln(3)$ just by clever application of the formula.
$\ln(2)$
$$ \ln(2) - \ln(1) \approx  2 \left[ \frac{1}{3} \right] = .66$$
$\ln (3)$
$$ \ln(3) = \ln\left(2(1.5) \right)
= \ln(2) + \ln(1.5) = .66 + 2  \left[  \frac{1.5-1}{1.5+1}\right]=1.06$$
$\ln(4)$
$\ln(4) = 2 \ln(2) \approx  1.32$
$\ln(5)$
$$ \ln(5) - \ln(4) \approx 2 \frac{1}{9} \approx  .\overline{22}$$
Hence, $\ln(5) \approx 1.54$
And so on with more clever manipulation.
 [/INST] Hint:
For $|x|\lt 1$,
$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+O(x^5)$
and $\ln(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4+O(x^5)$
Subtract these to get: $\ln(1+x)-\ln(1-x)=2(x+\frac{x^3}3+O(x^5))$
Put $x=Q$
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