The error of the natural logarithm Can anyone explain why the error for  $\ln (x)$ (where for $x$ we have $x\pm\Delta x$) is simply said to be $\frac{\Delta x}{x}$? I would very much appreciate a somewhat rigorous rationalization of this step. Additionally, is this the case for other logarithms (e.g. $\log_2(x)$), or how would that be done? 
 A: Simple error analysis assumes that the error of a function $\Delta f(x)$ by a given error $\Delta x$ of the input argument is approximately
$$
\Delta f(x) \approx \frac{\text{d}f(x)}{\text{d}x}\cdot\Delta x
$$
The mathematical reasoning behind this is the Taylor series and the character of $\frac{\text{d}f(x)}{\text{d}x}$ describing how the function $f(x)$ changes when its input argument changes a little bit. In fact this assumption makes only sense if $\Delta x \ll x$ (see Emilio Pisanty's answer for details on this) and if your function isnt too nonlinear at the specific point (in which case the presentation of a result in the form $f(x) \pm \Delta f(x)$ wouldnt make sense anyway). Note that sometimes $\left| \frac{\text{d}f(x)}{\text{d}x}\right|$ is used to avoid getting negative erros. 
Since
$$
\frac{\text{d}\ln(x)}{\text{d}x} = \frac{1}{x}
$$
the error would be
$$
\Delta \ln(x) \approx \frac{\Delta x}{x}
$$
For arbitraty logarithms we can use the change of the logarithm base:
$$
\log_b x = \frac{\ln x}{\ln b}\\
(\ln x = \log_\text{e} x)
$$
to obtain
$$
\Delta \log_b x \approx \frac{\Delta x}{x \cdot \ln b}
$$
A: While appropriate in many important contexts, LeFitz's answer can fail in one important situation, and can lead you astray, for example, when plotting graphs in logarithmic scale.
More specifically, LeFit'zs answer is only valid for situations where the error $\Delta x$ of the argument $x$ you're feeding to the logarithm is much smaller than $x$ itself:
$$
\text{if}\quad \Delta x\ll x\quad\text{then}\quad \Delta\ln(x)\approx\frac{\Delta x}{x}.
$$
However, if this condition fails then the result also fails. The reason for this is that the logarithm becomes increasingly nonlinear as its argument approaches zero; at some point, the nonlinearities can no longer be ignored.
One immediately noticeable effect of this is that error bars in a log plot become asymmetric, particularly for data that slope downwards towards zero. For example:

 (Image source)
This asymmetry in the error bars of $y=\ln(x)$ can occur even if the error in $x$ is symmetric. Consider, for example, a case where $x=1$ and $\Delta x=1/2$. Here you'll observe a value of 
$$y=\ln(x+\Delta x)=\ln(3/2)\approx+0.40$$ with the same probability as 
$$y=\ln(x-\Delta x)=\ln(1/2)\approx-0.69,$$
although their distances to the central value of $y=\ln(x)=0$ are different by about 70%. In a more radical example, if $\Delta x$ is equal to $x$ (and don't even think about it being even bigger), the error bar should go all the way to minus infinity, as there is a chance that the measured variable be negative. (Unless, of course, you're doing your statistics wrong, and assuming e.g. a symmetric distribution of errors in a situation where that doesn't even make sense.)
In more general terms, when this thing starts to happen then you have stumbled out of the gaussian statistics that underpin most of the standard formulas. In such cases there are often established methods to deal with specific situations, but you should watch your step and consult your resident statistician when in doubt.
If you just want a rough-and-ready error bars, though, one fairly trusty method is to draw them in between $y_\pm=\ln(x\pm\Delta x)$. If you know that there is some specific probability of $x$ being in the interval $[x-\Delta x,x+\Delta x]$, then obviously $y$ will be in $[y_-,y_+]$ with that same probability.
A: We can just see that the error bar in the logarithm of the data $y=\log x$ can be approximated to $\Delta x /x$ to the first order. If the error in the data $x$ is symmetric, then the upper and lower value of the data within 1$\sigma$ are
$$ x_{\mathrm{upper}}=x+\Delta x/2$$
and,
$$ x_{\mathrm{lower}}=x-\Delta x/2,$$
thus the error in y should be,
$$\Delta y = \log(x_{\mathrm{upper}}) - \log(x_{\mathrm{lower}})=\log(\frac{x+\Delta x/2}{x-\Delta x/2})= \log(1+\Delta x/2x) - \log(1-\Delta x/2x)  \approx \Delta x/x$$
In the last step, Taylor expansion is used to find the error in the $\log (x)$ in the first order in $\Delta x/x$. The expansion can only be cut off to the first term if $\Delta x/(2x) <<1$.
