Please Find the equation, Graph is provided.

  • Thread starter mri44
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  • #1
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1. Graph is provided. Please find the appropriate equation for it.



2. Equation should be in form of x = f (y)



3. Example: X =10^((Y-349.09)/(-48.78751)), This equation is not related to this graph.
 

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Answers and Replies

  • #2
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nice graph.
 
  • #3
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Please give me any reference. Is there any software that could determine the equation from the graph?
 
  • #4
SteamKing
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Funny how the axes don't use evenly spaced divisions. I wonder why?
 
  • #5
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ok, your plot is semi-log, with the x-axis being a log scale and the y-axis being a linear scale, and the graph looks like a straight line. This means that your general equation is:

[tex]
y = a + b \, \log{x}
[/tex]

where the [itex]\log[/itex] is in whatever base you like (I would say decade), but the coefficient [itex]b[/itex] depends on your choice for a base. Then, you can use the method of least squares to get the points.
 
  • #6
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Funny how the axes don't use evenly spaced divisions. I wonder why?

It is a semi log graph.
 
  • #7
NascentOxygen
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It is a semi Log graph.
No, it's not semi-log. It is log-log. Both axes are logarithmically scaled.

You determine the slope exactly as you do for your ordinary linear graphs, (viz., "rise" and "run" technique) except with the graph you have here you shouldn't try to interpolate between marked lines if you can avoid it. (Not until you know what you're doing, anyway.)

The marked scale lines are accurate, but if you try to draw in extra ones you will introduce considerably more inaccuracy than you'd think. (You have been warned! :smile: ) So try to limit yourself to using those lines that are already marked.

The need to write it as x=f(y) adds a bit of awkwardness, but that's a separate issue and you just need to keep your wits about you. :smile:
 
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  • #8
HallsofIvy
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Whatever the scale, that graph is a straight line so its equation is linear: y= ax+ b.
If it is in fact "log-log", and we call the vertical axis variable "V" and the horizontal axis variable "U", then the equation is log(V)= a(log(U))+ b. Take any two points on the graph and put their values into that equation. That gives two linear equations for a and b.

Taking the exponential of both sides, that equation becomes
[tex]e^{log(V)}= e^{alog(U)+ b}= e^{alog(U)}e^b= (e^b)(e^{log(U)})^a[/tex]
or
[tex]V= (e^b)U^a[/tex]
 
  • #9
Redbelly98
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Moderator's note:
Now that the OP has been given plenty of hints, let's please let him/her respond with an attempt at solving the problem before offering further help.
 
  • #10
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ok, your plot is semi-log, with the x-axis being a log scale and the y-axis being a linear scale, and the graph looks like a straight line. This means that your general equation is:

[tex]
y = a + b \, \log{x}
[/tex]

where the [itex]\log[/itex] is in whatever base you like (I would say decade), but the coefficient [itex]b[/itex] depends on your choice for a base. Then, you can use the method of least squares to get the points.

Thank you very much.

I got a result by this way.
 
  • #11
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Whatever the scale, that graph is a straight line so its equation is linear: y= ax+ b.
If it is in fact "log-log", and we call the vertical axis variable "V" and the horizontal axis variable "U", then the equation is log(V)= a(log(U))+ b. Take any two points on the graph and put their values into that equation. That gives two linear equations for a and b.

Taking the exponential of both sides, that equation becomes
[tex]e^{log(V)}= e^{alog(U)+ b}= e^{alog(U)}e^b= (e^b)(e^{log(U)})^a[/tex]
or
[tex]V= (e^b)U^a[/tex]

I will try it.

I will let you know.

Thank you very much for your help.
 

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