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I have data in Mathematica that comes from y-log scale

Data = {{5.0, 23.87548081003781},
   {6.94392523364486, 0.511639358262082},
   {8.925233644859812, 0.23397526329810545},
   {10.962616822429906, 0.16190746961888203},
   {12.906542056074766, 0.17751810380557045},
   {14.925233644859812, 0.25653445869951874}};

These points should be connected by straight line on log scale like this

ListLogPlot[Data, Joined -> True]

I want to find interpolated function with straight lines on log plot(just like above code), however naive result gives me:

LogPlot[Interpolation[Data, InterpolationOrder -> 1][x], {x, 5, 14}]

which does not have straight lines on logPlot, it has straight lines on Plot.

How can I interpolate data on log plot?

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  • $\begingroup$ what if you use InterpolationOrder with ListLogPlot like ListLogPlot[Data, Joined -> True, InterpolationOrder -> 3] $\endgroup$ – Sumit Jul 9 '17 at 9:11
  • $\begingroup$ Dear Sumit, thank you for the reply, this gives me a smoother plot, but I want an interpolated function which would look like ListLogPlot[Data, Joined -> True] $\endgroup$ – Wint Jul 9 '17 at 9:28
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    $\begingroup$ It'll be due to the difference between interpolating in log space or linear space. This'll get you what you want: if = Exp@*Interpolation[{#1, Log[#2]} & @@@ Data, InterpolationOrder -> 1] then LogPlot[if[x], {x, 5, 14}] $\endgroup$ – Quantum_Oli Jul 9 '17 at 9:47
  • $\begingroup$ Quantum_Oli thank you very much!! Please post as an answer and I mark it as solved. :) $\endgroup$ – Wint Jul 9 '17 at 10:02
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[Just noticed this was @Quantum_Oil's idea in a comment above. Probably why I didn't answer before.]

Often one interpolates to avoid transcendental functions, but the OP's objective cannot be achieved with polynomial interpolation. So I assume something like the following, which reproduces ListLogPlot[Data, Joined -> True], is desired:

ClearAll[logIF];
logIF[x_] = 
  Exp[Interpolation[MapAt[Log, Data, {All, 2}], InterpolationOrder -> 1][x]];
LogPlot[logIF[x], {x, 5, 14}]

Mathematica graphics

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The first point of your Data is an extreme outlier which we eliminate with

data = Rest @ Data;

We also need

xmax = Max @ data[[All, 1]]

14.9252

Show[
 ListLogPlot[data],
 LogPlot[Evaluate @ NonlinearModelFit[data, Exp[a + b x], {a, b}, x][x], {x, 0, xmax}]]

enter image description here

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  • 2
    $\begingroup$ Technically a fit is not an interpolation. An interpolation reproduces every data point. $\endgroup$ – Markus Roellig Jul 10 '17 at 10:30
  • $\begingroup$ OP wants a "straight line on log scale" which, you're right, is a fit :) $\endgroup$ – eldo Jul 10 '17 at 10:40

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