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Homework Help: Exponents that have units (how to deal with them)

  1. Aug 2, 2010 #1
    1. The problem statement, all variables and given/known data
    So I'm going over the simple yearly interest formula A=P(1+r)^t, where 't' is in years, 'P' is in dollars and is the size of the initial investment, 'A' is in dollars and is the total interest earned in a given period of years, and 'r' is the rate at which the investment grows.

    I've done countless problems involving this and more complicated interest formulas. What I don't understand: How do I account for the variable 't', specifically its units, in the final answer, given that I'm looking for the total interest earned, which is in dollars? Where does the unit 'years' go?

    2. Relevant equations
    Interest Accumulated = A = P(1+r)^t

    3. The attempt at a solution
    Ex: Find the interest earned on a $750 dollar principal invested for 5 years at an interest rate of 9%.


    A=(750dollars)(1.09)^(5years)=(1153.97dollars) <-- My current understanding is that I can just consider the final answer as being in dollars because the units for the exponent are only there as a guide through the problem. On top of that I don't consider 'r' as having any units because it's just a decimal number used to modify P. Both of these explanations are arbitrary and I've learned those don't usually work in math. Can someone give me some guidance?
     
  2. jcsd
  3. Aug 2, 2010 #2

    Mark44

    Staff: Mentor

    I'm pretty sure there are no units associated with t. Instead, it is the number of years. The multiplier (1 + r)^t is a quantity without units. This has to be true, because it wouldn't make sense to have an equation where you have <dollars> on one side and <dollars>*<???> on the other side.
     
  4. Aug 2, 2010 #3

    berkeman

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    Staff: Mentor

    Another way to handle it would be to make the exponent t/1 year, and keep the units of t at years. That way the exponent is unitless, which it has to be.
     
  5. Aug 2, 2010 #4
    What about a situation where I need to solve for t? Do I say it is 'the number of years' needed to accumulate that interest or that it is a number whose units is years?

    I'm a little confused by the distinction between 'number of years' and a value of t whose units is years. Does that make any sense?
     
  6. Aug 2, 2010 #5

    berkeman

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    Staff: Mentor

    Yes, it appears in your equation that the units of t are years. So when you solve for t, you could get 11.85 years, for example, and then convert that into years, months and days, if that format was easier for the consumer to understand and use.
     
  7. Aug 2, 2010 #6
    Mark44: I was referring to your distinction between a variable with units and something like 'number of years'.

    I've always been taught to see if 'the units make sense'. They do make sense when I drop them, but why do I do so?
     
  8. Aug 3, 2010 #7

    Mark44

    Staff: Mentor

    I still think that the expression (1 + r)^t is a unitless multiplier. The reason is that (1 + r) has no units, so neither will (1 + r)(1 + r) or (1 + r)(1 + r)(1 + r), which would be the multipliers if you calculated A = P(1 + r)^2 or A = P(1 + r)^3. All we're doing is squaring or cubing a unitless number, 1 + r, giving us something that is also unitless.

    If you are working problems of a physical nature, involving length and mass and such, it makes sense to keep track of units, but if you're doing problems involving financial calculations, with identical currency units on both sides of the equation, then it doesn't seem as important to me to keep track of units.

    BTW, your formula, A = P(1 + r)^t is for compound interest, with interest computed annually, not simple interest. The formula for simple interest is A = Prt.
     
  9. Aug 3, 2010 #8
    Is that to say that if there was another situation with a variable in the exponent that had specific units we would just discount the units? A situation not involving this particular formula?
     
  10. Aug 3, 2010 #9

    vela

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    Staff Emeritus
    Science Advisor
    Homework Helper
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    The exponent isn't really time but the number of periods. It's just that if the period is a year and t has units of years, the expression works out numerically with t in the exponent.

    It's probably easier to see in the case of interest compounded monthly. Say t has units of years while the interest rate r has units of 1/year. The balance after t years would be given by

    [tex]A = P \left(1 + \frac{r}{12/\textrm{year}}\right)^{(12/\textrm{year})t}[/tex]

    where the factor 12/year is the number of periods per year.
     
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