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

Question

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?

Answer 1

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

or

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

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

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

Answer 2

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 $\, \log_{10}(x) = \frac{\ln(x)}{\ln(10)}.$

Answer 3

Working in this another question I found that one could use this approximation: $$\ln(x) \approx \frac{\left(x^{\frac{1}{\ln(2)}}-1\right)}{x^{\frac{1}{\ln(2)}}}\ln(x+1)$$

From this you could use the Taylor's expansion of $\ln(x+1)$ at $x=0$. But from the plot looks it works only for $x>0.4$ and I don't think it going to match the exact values as the order increases (with the exact functions it fails), but at least works better than the Taylor's expansion of $\ln(x)$ at $x=1$ near $x=0$.

(see the plot here).