Timeline for What is the reason why we use natural logarithm (ln) rather than log to base 10 in specifying function in econometrics?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 31, 2022 at 23:57 | comment | added | trogne | Can someone explain why "with a coefficient of 0.06, a difference of 1 in x corresponds to an approximate 6% difference in y" ? $0.06\ln\left(50\right)-0.06\ln\left(49\right) = 0.00121216243905$ , which is far from a 6% difference from $0.06\ln\left(50\right)$. | |
| Feb 8, 2022 at 9:36 | comment | added | Jonas Lindeløv | Expanding on @SextusEmpiricus. "small" x are definitely within [-0.5, 0.5] which yields 0.64 (an error of 0.14). See a plot of the relationship between x and the error here: stats.stackexchange.com/a/244237/17459. | |
| Oct 17, 2017 at 6:59 | comment | added | Sextus Empiricus | @cs0815 if you apply the Taylor expansion around the point b $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(b)}{n!}(x-b)^n$$ to the exponential function $f(x)=a^x$, with $f^{(n)}(x) = ln(a)^n a^x$ then you get for the first two terms: $$f(b+x) = f(b) + ln(a)f(b)x + \mathcal{O}(x^2)$$ and the $ln(a)$ term becomes 1 for $a=e$ such that you can use $f(b+x) \sim f(b) (1+x)$, which is however only true for small x. Also you can simply try it out exp(1.06)/exp(1) = 1.0618 and 10^1.06/10^1=1.1418154 | |
| Jul 23, 2017 at 18:02 | comment | added | cs0815 | would this not apply if we apply log10 to the dependent and the independent variable(s)? | |
| May 11, 2016 at 14:03 | comment | added | user603 | More generally, the exponential function is the only continuous function that is equal to its derivative. | |
| Aug 6, 2012 at 6:02 | comment | added | Neil G | +1: For concrete reason to prefer the natural logarithm. | |
| Aug 6, 2012 at 5:20 | history | answered | fmark | CC BY-SA 3.0 |