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Homework Help: Natural log real life application

  1. Mar 23, 2006 #1
    this problem is an excerpt from an explanation of a time wieghted performance method. I feel that if I can follow this part the rest of it will make sense. now i know the answer is .1259 but i'm a little fuzzy on how exactly they got that, and their 'step by step' seems to miss some steps. it may be partially due to my rustiness with ln. It is as follows:

    3000 = 2500(1 + R)^(31/31) + 175(1 + R)^(15/31)

    3000 = 2500(1)ln(1+R) + 175(15/31)ln(1+R)

    here I thought if you applied the natural log you would have to do it to both sides of the equation, and why were the constants left out of it? but the explanation did not do so I will continue as such, please correct me if i'm wrong.

    3000 = [ln(1+R)][2500 + (175)(15/31)]

    1.160686427 = ln(1+R)

    this is about where i don't know the proper way to solve this. any explanations about using natural logs in these types of situations are much appreciated.
  2. jcsd
  3. Mar 23, 2006 #2


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    I think you've copied something down wrong:

    If ln(1+R)>1 then 1+R > e > 2.

    Thus the first term in the addition is larger than 5000 - so that's not a correct solution.
  4. Mar 23, 2006 #3
    the way they presented it was that they 'assumed' R=.1259 now I will type line by line:

    3000 = 2500(1+.1259)^(31/31) + 175(1 + .1259)^(15/31)
    3000 = 2500[ln(1+.1259) x (31/31)] + 175[ln(1 + .1259) x (15/31)]
    3000 = 2500[ln(.118554) x (31/31)] + 175[ln(.118554) x (15/31)]
    3000 = 2500ln(.118554) + 175ln(.057365)

    returning the logarithm to an exponet results in:

    3000 = 2500(1.1259) + 175(1.0590)
    3000 = 2814.67 + 185.332383
    3000 = 3000

    I have triple checked this one for typos....this is exactly how it reads. I'm very confused on two parts. the first term, the transition from line two to line three, I do not understand how (1 + .1259) turns into (.118554). and then in the second term, the transition from line three to line four, how come the(15/31) gets multiplied to the (.118554) I thought that the exponet would get multiplied to the 175. Could it be that I'm trying to learn from a faulty explanation? OR am I just not getting it yet?
  5. Mar 23, 2006 #4


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    Now I'm confused! You told us what the answer was, but you didn't tell us what the question was. I assume it was to find R, given that
    3000 = 2500(1 + R)^(31/31) + 175(1 + R)^(15/31)

    It is certainly NOT true that taking the logarithm of both sides of that
    gives 3000 = 2500(1)ln(1+R) + 175(15/31)ln(1+R). Even
    ln(175(1+R)^(15/31)) is not 175(15/31)ln(1+R). It is instead
    ln(175)+ (15/31)ln(1+R).

    I think what they were doing is showing that R= .1259 IS a solution to that equation, not solving it.

    To find (1+ .1259)^(15/31)= (1.1259)^(15/31), you can take logarithms:
    ln(1.259)= (15/31)ln(1.1259)= (15/31)0.11858. That "0.11855" is (approximately but not to the accuracy they show) ln(1.1259). If the book or paper you are getting this out of really has ln(0.11855) burn it now! As far as " I thought that the exponent would get multiplied to the 175" is concerned, addition is "associative". When you are multiplying abc, it doesn't matter whether you multiply ab together first or bc first.

    Looks to me like you have a really faulty explanation.
  6. Mar 24, 2006 #5
    Thank you for the helpful comments. I concur with the faulty explanation part, none the less I still want to figure it out. I should probably explain a bit more about what I’m trying to figure out. The formula is for computing the time weighted return of a portfolio. This formula is used in a computer system to track performance, the paper/explanation I have doesn’t say how to solve for R, it just kinda plugs in all the numbers in an example, then at the end …both sides of the equation equal each other. however I need to apply it by ‘hand’ to another set of data using this method, such that I can solve for ‘R’. The formula is as follows:

    [tex] MVE = MVB (1 + R)^\frac{d_{o}}{D} + CF_{1}(1 + R)^\frac{d_{1}}{D} + CF_{2}(1 + R)^\frac{d_{2}}{D} + …. + CF_{n}(1 + R)^\frac{d_{n}}{D}[/tex]

    Where MVE is the ending market value, MVB is the beginning market value, CF is a cashflow(+/-) that occurs in a time period which has D number of days, and that cashflow occurred on a day such that ‘d’ is the number of days until the end of the time period . (I hope that makes sense, to who ever reads this). obviously there are ‘n’ number of cashflows in the time period. There are other aspects of the formula, such as computing ‘net of fees’ and taking into account for accrued dividends etc…but what I really need to know is how to go about solving for R. with the previous comments help… I have assumed one cashflow for simplicity sake, and have worked the equation (hopefully correctly) down to the point where I have ln(1 + R) on one side and a constant on the other. This is the point where I’m not sure what to do to solve for R ( or rather I don’t remember what I’m allowed to do:

    [tex]MVE = MVB(1 + R)^\frac{d_{o}}{D} + CF(1+R)^\frac{d_{1}}{D}[/tex]
    [tex]\ln(MVE) = \ln(MVB) + \frac{d_o}{D}\ln(1 + R) + \ln(CF) + \frac{d_1}{D}\ln(1 + R)[/tex]
    [tex]\ln(MVE) - \ln(MVB) - \ln(CF) = \frac{d_o}{D}\ln(1 + R) + \frac{d_{1}}{D}\ln(1 + R)[/tex]
    [tex]\frac{\ln(MVE) - \ln(MVB) - \ln(CF)}{ \frac{d_{o}{D} + \frac{d_{1}{D}} = \ln(1+ R) [/tex]
    Last edited: Mar 24, 2006
  7. Mar 28, 2006 #6
    the last line in my previous post i had difficult editing until now it should read:

    [tex]\frac{\ln(MVE) - \ln(MVB) - \ln(CF)}{ \frac{d_{o}+d_{1}}{D}= \ln(1+ R) [/tex]

    I'm still having trouble editing the latex, but I do recognize that the denominator should have (do + d1)/D.

    how do I isolate 'R'??
    Last edited: Mar 28, 2006
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