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Math Help - HELP!!!Word Problem

  1. #1
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    HELP!!!Word Problem

    If your allowance was a penny doubled every day, how many days would it take for you to have over $1000? On which day would you have over $10000? How much money would you have on the 15th day?
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  2. #2
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    y = (1+r)^2

    y = (1+200%)^2

    y = (1+2) ^2

    y = 3^2

    y = 9

    So, on the fifteenth day, you will have $9

    This is an example of exponential growth (y = x(1+r)^t where x = initial starting amount, r = rate of growth expressed as a percent and t = time passed under specific unit). Plugging in your other numbvers ($1000 and $10000) as y will give you the answer to the other problems.


    ****Note that I may be wrong and it may be $6.25. I'm pretty sure that $9 is right, but for some reason my instinct told me to go with 150% the first time I went through with this. Although 150% is NOT doubled. Please check this with your own work or through someone else.****
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  3. #3
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    ?

    The answer is day 18 $1308.16, day 21 $10465.28, on the 15th day $163.52.

    I don't know how to calculate this. Please help!!!
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  4. #4
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    Quote Originally Posted by learnmath View Post
    The answer is day 18 $1308.16, day 21 $10465.28, on the 15th day $163.52.

    I don't know how to calculate this. Please help!!!
    Day 1: $0.01

    Day 2: $2(0.01)

    Day 3: 2(2(0.01)) = \$ 2^2 (0.01)

    Day 4: 2 (2^2 (0.01)) = \$ 2^3 (0.01)

    etc.

    Day n: \$ 2^{n-1} (0.01).


    Q1 Solve 2^{n-1} (0.01) > 1000 \Rightarrow 2^{n-1} > 100,000.

    The are many ways of solving this. Using logarithms is one way:

    Spoiler:
    \log_{10} 2^{n-1} > \log_{10} 100,000 \Rightarrow (n-1) \log_{10} 2 > 5 \Rightarrow n - 1 > \frac{5}{\log_{10} 2} \Rightarrow n - 1 > 16.6


    Q2 Day 15:
    Spoiler:
    \$ 2^{15-1} (0.01)
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