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Log-log scales

  • Thread starter nobahar
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  • #1
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Am I correct to say that a function that is linear on a log-log graph is also linear when it is plotted on a graph using regular scales on both axes, takes a square root form when plotted with the x-axis scaled regularly and the y-axis with a log scale, and exponential in appearance whn the x axis is log and the y axis is regular?
This is based on 'intuition', for want of a better word. How would I represent this mathematically?
 

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  • #2
tiny-tim
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Hi nobahar! :smile:
Am I correct to say that a function that is linear on a log-log graph is also linear when it is plotted on a graph using regular scales on both axes …
Nooo :redface: … if logy = Alogx + B, then y = … ? :wink:
 
  • #3
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Hi Tiny Tim!
Haven't had a response from you for a while!

[tex]y = x^a + 10^b[/tex]

[tex]10^b[/tex] is a constant, say, [tex]10^b = C[/tex].

Then [tex]y = x^a + C[/tex]

So it's exponential.

What confuses me is that if I had a y = mx + c function, then this would be a straight line. If I altered both axes so that they increase by the same increments, say 10^1, 10^2, 10^3, etc., then the graph should still appear straight, shouldn't it? After all, the m value hasn't changed, nor the c value, and the two axes have been altered by the same scale. I think I've misunderstood how a log-log scale works...
 
  • #4
tiny-tim
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Hi nobahar! :smile:

I've been swimming around in my little bowl, keeping an eye on the world! :wink:
[tex]y = x^a + C[/tex]
No, it's y = Cxa.
What confuses me is that if I had a y = mx + c function, then this would be a straight line. If I altered both axes so that they increase by the same increments, say 10^1, 10^2, 10^3, etc., then the graph should still appear straight, shouldn't it? After all, the m value hasn't changed, nor the c value, and the two axes have been altered by the same scale. I think I've misunderstood how a log-log scale works...
hmm …

y = mx would be a straight line (at 45º), but y = mx + c would be slightly curved.

You're just going to have to think about this some more. :smile:
 
  • #5
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Sorry!

To show that I've made an effort to understand why it is y = Cx^a:

[tex]\log{y} = a.\log{x} + b \left \left then \left \left y = 10^{a.\log{x} + b}[/tex]

[tex]y = 10^{a.\log{x}} * 10^b = (10^{\log{x}})^a * 10^b = x^a * 10^b[/tex]

and since 10^b is a constant

[tex]If \left \left 10^b = C, then \left \left y = x^a * C = Cx^a[/tex]

I think I understand what you mean by slightly curved: the +C component will make 'less and less of a contribution' to the appearance of the graph moving along the x-axis because the evenly spaced intervals actually increase by some factor, and the contirbution by the +C component is much less, erm, 'noticeable'.

I think I understand the idea of log-log scales now: If a function takes the form of a straight line on a log-log scale, it allows you to use the y = mx + c formula to deduce the equation of a line. I also failed to consider that the log scale would mean that each point was the log of the variable! It doesn't just make the scale more compact, it has to be taken into consideration for the value in the equation!
I'm slightly worried about my brain right now...

Thanks Tiny Tim!

Feel free to correct and/or improve my understanding.
 

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