# Approximation of $\log(x)$ for very small $x$

To avoid the $$\log()$$ function, I am looking for a good approximation of $$\log(x)$$ for very small $$x$$ (e.g. order $$10^{-5}$$).

I think Taylor series expansion is useless because around these small $$x$$, the first order derivative approachs $$+\infty$$.

I did try this approximation $$\log_{10}(x) \approx 1 - \frac{1}{\sqrt{x}}$$ but still don't have satisfactory results.

Could anyone suggest some better approximations?

• $\frac{\ln x}{\ln 10}$ – gammatester Oct 27 '18 at 8:32
• Since $\log 1/x=-\log x$, an approximation for very small $x$ would be the same as an approximation for very large $x$. – Aaron Oct 27 '18 at 8:36
• @gammatester very funny. – AlexTP Oct 27 '18 at 8:41
• @Aaron thanks for quick comment. However, as I said, I am trying to avoid the $\log()$ function. Can you please tell me the good approximation of $\log(x)$ for very large $x$? – AlexTP Oct 27 '18 at 8:42
• Please see here: math.stackexchange.com/a/977657/471884. – TheSimpliFire Oct 27 '18 at 8:47

## 2 Answers

Since you seem to allow square roots, then the sequence of functions $$\, f_n(x) := 2^n(\sqrt[2^n]{x}-1)\,$$ give better and better results. In fact, $$\, f_n(x) \to \ln(x)\,$$ as $$\, n \to \infty\,$$ for all $$\,x>0.\,$$ Once you have $$\,\ln(x)\,$$ you can use $$\, \log_{10}(x) = \frac{\ln(x)}{\ln(10)}.$$

$$\ln x\sim\frac{1-x^{-x}}x\qquad{(x\to 0)}$$

or

$$\ln x\sim \frac{x^x-1}x$$

Since $$x^x\approx 1$$ for small $$x$$,

$$\ln x=\frac1x\ln x^x=\frac1x\ln(1+(x^x-1))\sim\frac{x^x-1}x$$