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I have a calculation using the following pseudo-formula:

$ y = -ln(X/X_o)$

where both $X$ and $X_o$ have an associated error with them. I have propagated the error out simply using:

$ \delta y = y\sqrt{\frac{\delta X}{X} + \frac{\delta X_o}{X_o} } $

However, I'm not sure how to account for the use of the natural log - I feel like this is adequate since my general understanding is that the deviation $S_p$ in a measurement $p$ is found with:

$S_X = \frac{S_p}{p}$

My other thought is to calculate the error as:

$ \delta y = y\sqrt{\ln({\frac{\delta X}{X})} + \ln{(\frac{\delta X_o}{X_o} })} $

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The error from a logarithmic function can be estimated by a series of its derivatives. In a truely mathematical form, this should be

$ln(x+\Delta x) = ln(x)+\sum_{n=1}^{\infty} \frac{(-1)^n\Delta x}{x^n}$

but really we just use

$ln(x+\Delta x) = ln(x)+\frac{\Delta x}{x} $

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The "Guide to the expression of uncertainty in measurement" [ISO/IEC Guide 98-3] is very clear on this: If $Y=f(X_1, X_2, \ldots, X_k)$ then the variance is approximated by a Taylor series \begin{align} \sigma^2_y &\approx \sum_{i=1}^{k} \sum_{j=1}^k \frac{\partial f}{\partial x_i} \frac{\partial f}{\partial x_j} Cov[x_i, x_j] \\ %%% &=\sum_{i=1}^k \left(% \frac{\partial f}{\partial x_i} \right)^2 \sigma^2_{x_i} + 2 \sum_{i=1}^{k-1} \sum_{j=i+1}^k \frac{\partial f}{\partial x_i} \frac{\partial f}{\partial x_j} Cov[x_i, x_j] %+ \mathcal{O}(\sigma^4) \end{align} where $Cov[x_i, x_j] = \rho[x_i, x_j] \, \sigma_{x_i} \sigma_{x_j}$ is the covariance of the two variables $x_i, x_j$, and $\rho[x_i, x_j]$ is their correlation. So, if the two random variables are independent, then use \begin{align} y &= -\ln{(x/x_0)} = \ln{(x_0)}-\ln{(x)}\\ \Rightarrow \sigma_y^2 &\approx (1/x_0)^2 \sigma_{x_0}^2 - (1/x)^2 \sigma_{x}^2 = \left(\frac{\sigma_{x_0}}{x_0}\right)^2 + \left(\frac{\sigma_{x}}{x}\right)^2 \end{align}

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