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people :)

i have this graph in Mathematica

Help change Log base in this LogLog graph

the function mumas[y] its defined by

mumas[y_] := 1/2 + (2 y + 2)/(2 y (y^2 + 4)^2)

My problem is, according the manual of Mathematica, Mathematica uses Log[y] to denote the function $\ln(y)$, the Logarithm base $e$ , so im afraid that LogLogPlot[mumas[y],...] is actually a graph $\ln\mu$ vs. $\ln y$, so my question is how can change the Log base so the graph is in the base I needed. I hope I was clear in specifying the problem.

Thanks in advance for any help

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    $\begingroup$ The difference between Log and Log10 is just a constant factor. This together with the fact that the axis still shows the original numbers (just scaled) means that it doesn't matter which logarithm is used - the plott will always look the same $\endgroup$ – Lukas Lang Sep 7 '17 at 22:22
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TL;DR

The base of the logarithm doesn't matter in log-plots or log-log-plots.

Explanation

As noted in my comment, there is no difference between the two since

$$\log_{10}x=\frac{\log x}{\log 10},$$

i.e. they differ by a factor, namely $\log 10$. This factor doesn't do anything as the axes are labeled using the original coordinates, not their logarithms. (The factor just scales both axes, but this is irrelevant)

To demonstrate:

μ[y_] := 1/2 + (2 y + 2)/(2 y (y^2 + 4)^2)
Plot[μ[y], {y, 0.05, 3}, Frame -> True, PlotRange -> All,
     ScalingFunctions -> {"Log", "Log"}]
Plot[μ[y], {y, 0.05, 3}, Frame -> True, PlotRange -> All,
     ScalingFunctions -> {"Log10", "Log10"}]

gives the following two plots:

log-log plot comparison

As you can see, everything is the same.

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    $\begingroup$ In short: you only need to worry about the base for a semilog plot. On a log-log plot, the base doesn't matter. $\endgroup$ – J. M. is away Sep 8 '17 at 0:25
  • $\begingroup$ @J.M. The base is also irrelevant for the semilog plot - only one axis is scaled, but since the aspect ratio of the plot is fixed, this doesn't do anything. (Also, thanks for prettifying my answer - is there any reason to reorder the options besides readability?) $\endgroup$ – Lukas Lang Sep 8 '17 at 8:27
  • $\begingroup$ Well, of course the options of a function can be in any order, by design. But it looks nice when they're alphabetized. $\endgroup$ – J. M. is away Sep 8 '17 at 8:32

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