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I have a natural log equation from excel and I need to plot it on 3D plane at a particular z(for example, z=1) enter image description here

I would expect a vertical plane, but what I am getting is a horizontal plane.

This is another equation I tried to plot.

Plot3D[{y = 0.1448 Log[x] + 2.4618}, {x, 0, 50}, {y, 0, 4.5}, ColorFunction -> "Rainbow"] enter image description here

I am not sure how to include the z in the equation and draw the plane. I have seven sets of natural log equations with corresponsing z values. It will be helpful if someone could help me with it!

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2 Answers 2

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eqs = {
   {x, 0.1448 Log[x] + 2.4618, 1},
   {x, 0.4125 Log[x] + 1.7234, 2},
   {x, 0.7234 Log[x] + 3.6123, 3}
   };

ParametricPlot3D[Evaluate[eqs], {x, 0.1, 3}, AxesLabel -> {"x", "y", "z"}]

enter image description here

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Perhaps, you meant something like this?

Mathematica graphics

One can grab and rotate the image to the desired viewpoint.

From what you said, I guess that you wanted the $y$ axis to be vertical. I selected and assigned a viewpoint to present a "reasonable" view. Of course, your opinion may differ.

Plot3D[{y = 0.1448*Log[x] + 2.4618}, {x, 0, 50}, {y, 0, 4.5}, 
  ColorFunction -> "Rainbow", ViewVertical -> {0, 1, 0}, 
  ViewPoint -> {60, 3, -35}]
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  • $\begingroup$ Hi, thanks for working it out. What I meant was that I need that curve to be at a particular z (like z = 0.5). The maximum limit of z-axis should be 1(that's how far the data goes). But when I plotted for different natural log equations in the same graph, the z axiz went until 4. I am not sure how to give in the z axis value. Let me know if you have any idea about it. Thanks again. $\endgroup$
    – Seya
    Mar 24 at 1:54

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