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Basic: Properties of ln & Exponents

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm taking a math of investment and credit course, in this case, solving for i which is the interest rate.

    My question is: is it possible to "ln both sides" of an equation? Like in (b) below, I know I could say y*ln(1+i) = ln(3).

    2. Relevant equations

    (a) 5(1+i)^8 + 10(1+i)^4 = 3
    (b) (1+i)^y = 3

    3. The attempt at a solution
    I've already solved the actual question (a) by making the substitution x=(1+i)^4 and solving via quadratic equation. What I'm looking for is a more general method for solving these sort of questions.

    Something like "lning both sides":
    40ln(1+i) + 40ln(1+i) = ln3

    ln(ab) = ln(a) + ln(b) so I know that isn't correct. What is the correct form?

    Thanks in advance!
     
  2. jcsd
  3. Sep 29, 2008 #2
    Remember, if you take the log of all of one side, you have to take the log of all of the other side. For instance if

    c = a + b then log(c) = log(a+b) NOT log(c) = log(a) + log(b) If you plug in numbers for a, b and c you will clearly see this is so.
     
  4. Sep 29, 2008 #3
    So what is ln(a+b) equal to in terms of ln(a) & ln(b)?
     
  5. Sep 29, 2008 #4

    HallsofIvy

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    Yes, it is possible to "ln both sides of an equation"- that is just "doing the same thing to both sides of an equation". But it may not help! log can be used to "move" a variable down from a power so while it should work nicely for the second problem, it is not appropriate for the first problem. I'm not sure what you mean by making the "substitution x= (1+ i)^4". The first problem has no "unknown" and doesn't need to be "solved". I expect you are simply asked to do the calculation on the right side and show that you get 3. Multiply (1+ i) by it self to get (1+i)^2, then multiply that by itself to get (1+ i)^4, multiply that by itself to get (1+i)^8. Only three multiplications: not that hard.

    For the second problem, (1+ i)^y= 3, since y is an exponent, yes, go ahead and use logarithms: ln((1+i)^y)= 3 so y ln(1+ i)= ln(3), as you say, and y= ln(3)/ln(1+i). Now you are going to have a problem with the fact that you have complex numbers! The complex number x+ iy can be represented as the point (x, y) in the "complex" plane and then changed to "polar coordinates". x+ iy= r e^{i theta} where r and theta are the polar coordinates: r= sqrt{x^2+ y^2} and theta= arctan(y/x). In this case, 1+ i corresponds to (1, 1) so r= sqrt(2) and theta =pi/4: 1+ i can be written as sqrt(3)e^(i pi/4) so ln(1+ i)= ln(sqrt(3))+ i pi/4= (1/2)ln(3)+ i pi/4.

    Warning: for complex numbers, ln is NOT a "single valued function": theta could also be written and pi/4+ 2pi= 9pi/4 so ln(1+ i)= (1/2)ln(3)+ 9pi/4. In fact ln(1+ i)= (1/2)ln(3)+ pi/4+ 2npi where n is any integer.

    there is no simple formula for ln(a+ b).
     
    Last edited: Sep 29, 2008
  6. Sep 29, 2008 #5
    Ouch, I wasn't expecting that, ugg!

    For (a) what I meant was to make 5(1+i)^8 + 10(1+i)^4 = 3 into a quadratic in x=(1+i)4 so: 5x2+10x - 3 = 0. The numbers are not correct, but the form is.

    Thanks for your reply.
     
  7. Sep 29, 2008 #6

    HallsofIvy

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    For what purpose? Exactly what does the problem ask you to do? You are not asked to solve an equation- there is no "unknown".
     
  8. Sep 29, 2008 #7
    I really don't understand why you keep saying there is no unknown. i is the rate of interest, which is the unknown variable.

    I don't know how I can make it plainer than that?
     
  9. Jun 12, 2011 #8
    --------------------------------------------------------------------------------
    I really don't understand why you keep saying there is no unknown. i is the rate of interest, which is the unknown variable.

    I don't know how I can make it plainer than that?
    _______________________________________________________________________

    Oooffff, there's the issue. You really should use r to represent rate of interest, as in math, i is used to represent sqrt(1)
     
  10. Jun 12, 2011 #9

    SteamKing

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    Let's take Eq. 2b. from the OP first:

    Find i, such that (1+i)^y = 3

    You can take the log of both sides (I prefer common logs for this):

    y * log (1+i) = log (3)

    That's as far as we can go, unless y or i is specified.

    For example, if you wanted to triple your investment in 20 years, what annual rate of interest would be required?

    Substituting, 20 * log (1+i) = log (3)

    log (1+i) = log (3) / 20 = 0.0239

    Converting the log equation by using x = 10^log(x):
    (1+i) = 10^0.0239 = 1.0565

    Therefore: i = 1.0565 - 1 = 0.0565 or 5.65% annual interest rate

    For Equation 2a from the OP:
    5(1+i)^8 + 10(1+i)^4 = 3

    we can't use log of both sides because the left side is a sum. If the interest rate i is the same in both terms, it would be possible to select a trial value of i and iterate until the equation is satisfied.
     
  11. Jun 12, 2011 #10
    This is wrong.

    i = sqrt(-1)
     
  12. Jun 12, 2011 #11

    Ray Vickson

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    There is something wrong with (a): you have "future values" of 5 and 10 adding up to only 3 (although as a purely mathematical question with no financial content it is OK). If you had said 5/(1+i)^8 + 10/(1+i)^4 = 5, that would make _financial_ sense. Anyway, your trick would not work for something like 5/(1+i)^3 + 8/(1+i)^7 + 3/(1+i)^8 = 10. To solve for i in that case would require *numerical* root-finding methods.

    As others have said, you can't usefully take logarithms of an equation like like c*r^a + d*r^b = f, because log(c*r^a + d*r^b) is not related in any simple, exact way to log(r^a) and log(r^b).

    RGV
     
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