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Homework Help: Logarithm question using compound interest formula

  1. Jan 18, 2017 #1
    1. The problem statement, all variables and given/known data
    A colony of ants will grow by 12% per month. If the colony originally contains 2000 ants how long will it take for the colony to double in size?

    Answer - 6.12 months
    2. Relevant equations
    A = P(1+r/n)nt

    3. The attempt at a solution
    r = 12% = 0.12
    n = 12
    P = 2000
    A = 4000
    t = ?

    A = P(1+r/n)nt
    4000 = 2000(1+0.12/12)12t
    2 = (1.01)12t
    log2 = log(1.01)12t
    log2 = 12t log(1.01)
    log2/12log(1.01) = t
    t = 5.8 months

    I don't know what I'm doing wrong. I've followed another compound interest question that is very similar to this one but my answer is still 0.3 months out. Can anyone point out what I'm doing wrong?
    Thanks for your help
  2. jcsd
  3. Jan 18, 2017 #2


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    Homework Helper

    You seem to be mixing up the formula.
    Your base for the exponent should be 1 + the percentage growth for each time increment...in this case months.
    If you were given a yearly growth rate and asked to figure out monthly growth, then n would be 12, but in this case, n is 1, and you can let t represent months.
    This gives you
    ## A = P(1 + .12)^t. ##
  4. Jan 18, 2017 #3
    Yes that was definitely my problem. Physics Forum for the win again! Thanks a lot for your help!
  5. Jan 18, 2017 #4


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    Science Advisor
    Gold Member

    I'd actually suggest that you are solving for n, not t. Why? you have periodic model that is fixed at months, and you want to know how many iterations of your model occur before population size doubles. Furthermore, having t in the exponent is troubling / not interpretable from a units perspective. Exponents are always dimensionless. n iterations is per se dimensionless. Time always has units. So if you want the equation to be transparent/ dimensionally correct, the below is more appropriate

    ## A = P(1 + .12)^n.##
  6. Jan 18, 2017 #5

    Ray Vickson

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    Homework Helper

    Just to clarify: the formula you wrote initially would be appropriate to a mortgage situation with an annual interest rate (or growth rate) of 12%, but compounded monthly. Mortgages are typically computed that way, so an annual rate of 12% is regarded as a rate of 1% per month in a normal mortgage contract.

    Mathematically that is not quite true, because a monthly interest rate of r (0 < r < 1, not a percentage) compounds monthly to ##(1+r)^{12}## in one year. So, mathematically, to get 12% = 0.12 in one year we need ##r = 1.12^{1/12}-1 \doteq 0.00948879293##, about 0.9489%. Or, to put it another way, 1% compounded monthly produces ##1.01^{12} \doteq 1.1268,## so a bit more than 12% per year.
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