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Thread: Help with exponential problem.

  1. #1
    Junior Member
    Dec 2008

    Help with exponential problem.

    I could use some help understanding these problems and setting up the equation using the formulas.
    A single-celled amoeba doubles every 4 days. How long would it take one amoeba to produce a population of about 10,000 amoebae?

    I am a little confused about which formula to use.
    FV= R ((1+i)^n)-1/i
    or A=R(1+r)^n
    or A=R(1+r/k)^kt
    I thought to use the third one, making the equation A=10,000(1+2/4)4t then solve from there.
    Is that formula correct or even the right way to set up that formula?
    thanks a lot.
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  2. #2
    MHF Contributor
    Mar 2007


    Your various equations appear to relate to compounded growth in various contexts. Instead of trying to pick "the right one", use the meaning to find what is helpful.

    In general, compounded growth follows the following formula:

    . . . . .$\displaystyle A\, =\, P\left(1\, +\, \frac{r}{n}\right)^{nt}$

    ...where "A" is the ending amount, "P" is the initial amount ("principal", in a monetary context), "r" is the rate of growth (expressed as a decimal), "n" is the number of compoundings per growth period (usually the number of compoundings per year), and "t" is the number of periods (usually years).

    In your case, you know that, whatever P you start with, you'll have A = 2P in t = 4 days. Let's plug that in:

    . . . . .$\displaystyle 2P\, =\, P\left(1\, +\, \frac{r}{n}\right)^{4n}$

    If we let n = 1, so there is one "compounding" per "period" (because this will simplify our work), then we get:

    . . . . .$\displaystyle 2P\, =\, P\left(1\, +\, r\right)^4$

    Divide through by P to get:

    . . . . .$\displaystyle 2\, =\, (1\, +\, r)^4$

    There are various ways to proceed from here. One would be to take the fourth root of each side, and then subtract the 1 to isolate r:

    . . . . .$\displaystyle \sqrt[4]{2}\, =\, 1\, +\, r$

    . . . . .$\displaystyle \sqrt[4]{2}\, -\, 1\, =\, r$

    Now use the same formula, but plug in the value you have for r. Set P equal to 1, set A = 10,000, and solve:

    . . . . .$\displaystyle 10,000\, =\, 1(1\, +\, \sqrt[4]{2}\, -\, 1)^t$

    . . . . .$\displaystyle 10,000\, =\, (\sqrt[4]{2})^t$

    Use logs to find the value for t.

    . . . . .
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  3. #3
    MHF Contributor

    Apr 2005
    Here's how I would do it: doubling every 4 days means you multiply by 2 every 4 days. In t days, there are t/4 intervals of 4 days so you multiply by 2 t/4 times: $\displaystyle 2^{t/4}$ times the initial number, which was 1.

    The number of amoeba in t days is $\displaystyle P(t)= 2^{t/4}$. To determine how long it will take that one amoeba to reproduce to 10000 you must solve the equation $\displaystyle 2^{t/4}= 10000$. You can do that by taking the logarithm of both sides, $\displaystyle (t/4)log 2= log 10000$, then multiply both sides by 4 and divide by log 2: $\displaystyle t= 4\frac{log 10000}{log 2}$.

    That is exactly stapel's answer.
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