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Logarithmic/Trig Questions

  1. May 23, 2006 #1

    Solve log x = 2 Cos x


    Answer is 1.48, 5.07

    What I have done so far:
    let y = log x
    y = 2 cos x
    i graphed this and found the x-intercepts to be 1.57 and 4.71
    not sure where to go from here...


    Determine the phase shift of the function f(x) = cos (1/3 x - pi)


    from the question is "pi" not the phase shift?? the answer says the phase shift is 3pi, but I am not sure this is right.
  2. jcsd
  3. May 24, 2006 #2
    Until someone wiser comes along with more advice...

    For the second question note that [tex] cos[\frac{x}{3}-\pi]= cos(\frac{1}{3}[x-3\pi])[/tex]:wink:

    For the first (and you may want to re-check your plots btw)...though I haven't yet done any work on *solving* these types of equations yet you can, by using Newtons Method, home in on the solutions. This however requires calculus and so it may be beyond the scope of what the question requires, but if you're interested...

    noting that log10x = (1/ln10)(lnx) you can differentiate log10x - 2cos x and then plug decent guesses into [itex] x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)} [/itex] and keep using your answer as the next guess.(I use log10x because this base fits the answer, ln x (I believe sometimes referred to as log x) doesn't)

    (additionally is the domain of x meant to be [0,2pi]?... because only if x > 100 is there no potential for 2cos x = log10x)
    Last edited: May 24, 2006
  4. May 25, 2006 #3
    For the first question, you need to find the x coordinates of the points of intersection of the two curves if you wish to solve the question graphically.
    It is only at these points that the value of the two functions become equal for the same value of x , which is then the solution .

    The second question is ambiguous.
    Usually, phase shift is defined with respect to some given function, which has the same period as the function in the problem .
    Please give the question in its entire form.

  5. May 25, 2006 #4
    Hi Arunbg... how could I solve question 1 non-graphically?? The domain for this question is 0 to 2pi. Also, that is the complete question for Question 2.
  6. May 25, 2006 #5


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    Staff: Mentor

    The second solution by the book is correct, but log(1.48) = 0.170 and 2cos(1.48) = 0.181. :uhh:
  7. May 25, 2006 #6
    p.s. i was able to get the correct answers graphing.
  8. May 25, 2006 #7


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    Staff: Mentor

    log(1.57) = 0.196, 2cos(1.57) = 0.002

    log(4.71) = 0.673, 2cos(4.71) = -0.005
  9. May 25, 2006 #8
    check your calculator mode.. it should be in radians.

    log(1.48) = .17 and 2Cos(1.48) = 0.18

    log(5.07) = .70 and 2Cos(5.07) = .70

    Therefore, the solutions are correct.
  10. May 25, 2006 #9


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    Staff: Mentor

    Yeah, I switched to radians for my calcs as well. I just don't consider 0.170 = 0.181

    But I just did a more careful check, and the answer is closer to 1.485. That reduces the error to almost nothing. I guess the book was happy with giving the answer to only 2 decimal places, even though that gives a 6% error. In the real world, a 6% error can be a very bad thing....
  11. May 26, 2006 #10
    Unless there is some alternate definition for phase shift, I don't think there is an answer to question 2.
    I can only safely say that the period of the function is [itex]3\pi[/itex] .
    Perhaps that is what the question is asking ?

  12. May 26, 2006 #11
    There are different interpretations of phase. Some authors (Halliday/Resnick) consider the argument of the sinusoidal function as the phase which means [itex](x/3 - \pi)[/itex] is the phase. Some consider the constant part of the argument that is not a function of the independent variable x as the phase, so then [itex]\pi[/itex] is the phase.

    A geometric interpretation is that the graph of [itex]\cos (x/3)[/itex] when shifted by [itex]3\pi[/itex] units along the positive x direction gives the graph of f(x). Your books seems to consider phase shift as the amount of displacement that must be given to the first function to construct f(x).

    The period is [itex]2\pi/(1/3) = 6\pi[/itex] and not [itex]3\pi[/itex] but that is not being asked here.
    Last edited: May 26, 2006
  13. May 26, 2006 #12
    This was what I was trying to allude to in my previous post .
    The function considered as the standard(from which phase diff. is calculated) here is probably cos(x/3), which is not given in the question.

    Thanks for the correction .
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