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- Thread starter mri44
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- #2

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nice graph.

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- #4

SteamKing

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Funny how the axes don't use evenly spaced divisions. I wonder why?

- #5

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[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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No, it's not semi-log. It is log-log.It is a semi Log graph.

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! ) 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.

Last edited:

- #8

HallsofIvy

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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

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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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[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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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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