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Math Help - Geometric Series Question

  1. #1
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    Geometric Series Question

    you are geometrically diluting/mixing 0.1 g of powder A with 100g of powder B, how many times do you have to mix the 2 together to finish the process?
    *each time you can only mix an equal portion of powder B to what you currently have mixed.

    Eg.
    1:1
    2:2
    4:4

    3. The attempt at a solution

    So the first time you mix them it will be
    0.1 g + 0.1 g --> 2^0
    then
    0.2 g + 0.2 g --> 2^1
    0.4 g + 0.4 g -->2^2???

    so with the formula is it? 101.1 = 0.1*2^n ???

    I don't know what to do from here?

    help??
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  2. #2
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    Re: Geometric Series Question

    Quote Originally Posted by mikemd View Post
    , how many times do you have to mix the 2 together to finish the process?

    How do you know when you are finished?

    Quote Originally Posted by mikemd View Post
    , how many times do you have to mix the 2 together to finish the process?
    *each time you can only mix an equal portion of powder B to what you currently have mixed.
    So after the first mix you have 100.1g in total, then you can add another 100.1g of powder being giving 200.2g in total which 0.1g will be powder A?
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  3. #3
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    Re: Geometric Series Question

    Kind of an oddly-worded question. I think you almost have it.

    Maybe,
    0.1*(2+4+8+16+...+What) >= 100.1

    0.1*(2^1 + 2^2 + 2^3 + 2^4 +...+ 2^n) >= 100.1

    (2^1 - 2^(n+1))/(1-2) >= 1001
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  4. #4
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    Re: Geometric Series Question

    Quote Originally Posted by pickslides View Post
    How do you know when you are finished?
    Basically you start with 0.1 g and you mix it with 0.1 g of powder B -------> 0.2 g total
    take 0.2 g of powder B and mix it with your 0.2 g -----> 0.4 g total
    take 0.4 g of powder B and mix it with you 0.4 g -----> 0.8 g
    continue untill you used up all of the 100 g of powder B, so I have to figure out how many "mixes" I did before getting to the final mass of 100 g powder B + 0.1 g powder A.

    Thanks
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  5. #5
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    Re: Geometric Series Question

    Quote Originally Posted by TKHunny View Post
    Kind of an oddly-worded question. I think you almost have it.

    Maybe,
    0.1*(2+4+8+16+...+What) >= 100.1

    0.1*(2^1 + 2^2 + 2^3 + 2^4 +...+ 2^n) >= 100.1

    (2^1 - 2^(n+1))/(1-2) >= 1001
    Thanks TKHunny, which formula did you used (or do you have a link?)?

    thanks
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  6. #6
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    Re: Geometric Series Question

    I try not to use formulas, per se, for this sort of thing. You must need to know how to add a finite geometric series. That's why I left it in that ugly form, so you could see it.
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  7. #7
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    Re: Geometric Series Question

    do you get 9.97?
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  8. #8
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    Re: Geometric Series Question

    Well, most truthfully, I didn't get anything. Having established the reliable procedure, there was nothing left for me to do. It's your homework.

    Let's see: \frac{2 - 2^{n+1}}{1-2} = 2\cdot \left(2^{n}-1\right) = 1001 and n = 8.97.

    This suggests you may have missed somehting - perhaps that the original exponent was "n+1" rather than "n". Fix that.

    In any case, that is not all. We must then answer the question.

    "how many times do you have to mix the 2 together to finish the process?"

    Ask yourself, "Can I REALLY mix things 0.97 times?" I'm thinking you need only whole numbers. What do you think?
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  9. #9
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    Re: Geometric Series Question

    I tried a different approach using

    an = ar^(n 1)

    so treating it as
    0.2, 0.4, 0.8, 1.6.....100.1

    plugging in a = 0.2, an = 100.1 and r = 2. I ended with with the 9.9 value. Would this approach work as well?
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