1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Plotting to Get a Straight Line

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data

    State what you would plot to get a straight line if experimental (x,y) data are to be
    correlated by the following relations, and what the slopes and intercepts would be in
    terms of the relation parameters. If you could equally well use two different kinds of plots
    (e.g. rectangular or semilog), state what you could plot in each case [the solution to part
    (a) is given as an example]

    a) y^2=a*e^(-b/x)
    solution: construct a semilog plot of y^2 vs 1/x or a plot of ln(y^2) vs 1/x on rectangular coordinates, slope = -b, intercept =lna.

    c) 1/ln(y −3) = (1+ a*sqrt( x))/b

    e) y = exp(a sqrt(x) +b)

    g) y=(ax+b/x)^(-1)

    3. The attempt at a solution

    I'm really confused about this. I tried looking in the book, but I couldn't find anything. I feel like the answer should be pretty easy, but I'm just not getting it. If you could help me with one of the questions it should definitely help me solve the other ones. Thanks in advance!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 14, 2013 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Do you understand how to arrive at that answer you have for (a)?
     
  4. Sep 14, 2013 #3
    The answer for a was given. I was looking at it so that I could get an idea as to how to complete the problem, but I don't see how they got it.
     
  5. Sep 14, 2013 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    y^2=a*e^(-b/x)

    Taking the natural log of both sides gives:

    log (y^2) = log ( a*e^(-b/x) )

    The RHS can be simplified using rules associated with logs, viz., log (x*y) = log(x) + log(y)

    = log (a) + .......

    Can you fill in the blank?
     
  6. Sep 14, 2013 #5
    log(e^(-b/x))
     
  7. Sep 15, 2013 #6

    NascentOxygen

    User Avatar

    Staff: Mentor

    This can be simplified. Remember the log we are using is loge
     
  8. Sep 15, 2013 #7
    So it would be just log(-b/x)
     
  9. Sep 15, 2013 #8

    NascentOxygen

    User Avatar

    Staff: Mentor

    When you reach an impasse in algebra, it can be enlightening if you substitute numbers and play with the expression. When you think you have found the simplified form, test with numbers to see whether it equals the expression you started out with.

    Perhaps first try to simplify log10(10^7)
     
  10. Sep 15, 2013 #9

    jtbell

    User Avatar

    Staff: Mentor

    You can go further than that.
     
  11. Sep 15, 2013 #10

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    No, you made a mistake.

    Log(ez) = z
     
  12. Sep 15, 2013 #11
    In part a, you need to start out by taking the natural logarithm (i.e., ln) of both sides of the equation. Can you show that you get:
    [tex]2\ln{y}=\ln{a}-\frac{b}{x}[/tex]
    In these exercises, you need to manipulate the equations mathematically so that you can plot a straight line between one parameter involving only y and/or x and another parameter involving only y and/or x. By plotting the data in this way, you can determine the (presumably unknown) parameters a and b.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted