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10-year zero coupon rate, how to calculate x?

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  1. Sep 17, 2012 #1
    Hi guys,

    I'm trying to sort out the following formula, but just can't find out how to solve for 'x'.
    Could you guys please help me out??

    http://imageshack.us/a/img198/4850/screenshot20120917at725.png [Broken]

    Uploaded with ImageShack.us

    And I have another one, is there a quick way to solve this one using the log function of a simpele Casio Fx82ms calculator?

    http://imageshack.us/a/img842/7760/screenshot20120917at731.png [Broken]

    Uploaded with ImageShack.us

    The 'x' here is the interest rate and the 'e' mentioned here is the mathematical constant (or 2.71828).

    I would really appreciate your help!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 17, 2012 #2

    Mark44

    Staff: Mentor

    Is this homework?
     
    Last edited by a moderator: May 6, 2017
  4. Sep 17, 2012 #3
    No it's not, I'm trying to find out more about financial engineering. But it has been a while since I studied math at highschool, haha.

    I put the formulas in an online scientific calculator and then uploaded this on Imageschack to make sure you would understand the formulas.

    Could anyone help me?
     
  5. Sep 17, 2012 #4
    These are equations. You can add, subtract, multiply, or divide things on both sides of the equations to try to isolate the expressions containing [itex]x[/itex] and eventually solve for it.

    For example, in the second equation, the obvious first thing to do is add 770 to both sides.
     
  6. Sep 17, 2012 #5
    Thanks, I know how to subtract/multiplay/divide things and simplified the formula to:
    http://imageshack.us/a/img209/4018/screenshot20120917at812.png [Broken]

    I still need to solve for 'x' now, how can I do that? The scientific calculator says the answer should be 0.0375 and I believe that's correct, but I still don't know how to get to that answer myself. I bought a 200usd book that is about financial engineering and has been put together by one of the professors of Kellogg School of Management, however it doesn't always explain the logic behind some of the formulas used, the book goes straight to the answers... haha, sorry to be a pain but any help is welcome!
     
    Last edited by a moderator: May 6, 2017
  7. Sep 17, 2012 #6
    You just need to undo the effect of raising to the 10th power. Raising to the one-tenth power would do that, just as squaring (raising to the 2nd power) is undone by a square root (raising to the 1/2 power).
     
  8. Sep 17, 2012 #7
    Hmm thanks but could please denote your words with a formula like I did? That makes it easier for me to interpret (as my native language is Dutch...)
     
  9. Sep 17, 2012 #8
    [tex](y^{10})^{1/10} = y[/tex]
     
  10. Sep 17, 2012 #9
    Great, this helps! Thanks a million.
     
  11. Sep 18, 2012 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]\frac{100}{(1+x)^{10}}= 69.20205[/tex]
    [tex]\frac{(1+x)^{10}}{100}= \frac{1}{69.20205}= 0.01445044[/tex]
    [tex](1+ x)^10= 1.4455044[/tex]
    [tex]1+ x= \sqrt[10]{1.4455044}[/tex]
    [tex]x= \sqrt[10]{1.4455044}- 1[/tex]
    You can do the 10th root (if your calculator doesn't have a "[itex]\sqrt[y]{x}[/itex]" key) using logarithms: [itex]\sqrt[10]{x}= e^{ln(x)/10}[/itex]

    [tex]-770+ 815e^{-x}= 0[/tex]
    [tex]815e^{-x}= 770[/tex]
    [tex]e^{-x}= \frac{770}{815}= 0.9447[/tex]
    [tex]-x= ln(0.9447)[/tex]
    [tex]x= -ln(0.9447)[/tex]
    (since 0.9447 is less than 1, ln(0.9447) is negative so x will be positive.)
     
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