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Math Help - Geometric Progression

  1. #1
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    Geometric Progression

    Here is the question. I've spent about ten hours trying to get this to work, but I just can't crack it. I've changed the numbers to try and keep it simple. I can't really seem to find any similar questions on the net.

    A jewelry firm sells 1000 pieces of jewelry in its first month. If sales increase at a rate of 5% a month, how long will it take to sell 15 000 products in total?

    Here's what I know. It's geometric progression, as the topic title suggests, and the number is larger than one, so the formula is:

    Sn = a((r^n)-1)/(r-1)

    Which will look like:

    15000n = 1000((1.05^n)-1)/(1.05-1)

    I've worked out how long it will take to be sell 15000 a month (by accident) and I even know what n should be (about 11.5 months) but I can't find a mathematical method to prove it. I think the next step should be:

    log(15000)n = 1000 + log((1.05^n)-1) - log(1.05-1)

    But even if that is right, I'm not sure how to solve it because of the ^n on the right hand side of the equation.

    Any help on this matter would be much appreciated.

    Ang
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  2. #2
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    Quote Originally Posted by AngusBurger View Post
    Here is the question. I've spent about ten hours trying to get this to work, but I just can't crack it. I've changed the numbers to try and keep it simple. I can't really seem to find any similar questions on the net.

    A jewelry firm sells 1000 pieces of jewelry in its first month. If sales increase at a rate of 5% a month, how long will it take to sell 15 000 products in total?

    Here's what I know. It's geometric progression, as the topic title suggests, and the number is larger than one, so the formula is:

    Sn = a((r^n)-1)/(r-1)

    Which will look like:

    15000n = 1000((1.05^n)-1)/(1.05-1)

    There's the culprit! The n multiplied by 15000 shouldn't be there.

    I've worked out how long it will take to be sell 15000 a month (by accident) and I even know what n should be (about 11.5 months) but I can't find a mathematical method to prove it. I think the next step should be:

    log(15000)n = 1000 + log((1.05^n)-1) - log(1.05-1)

    But even if that is right, I'm not sure how to solve it because of the ^n on the right hand side of the equation.

    Any help on this matter would be much appreciated.

    Ang
    If you redo the calculations with S=15,000 instead of 15,000n
    everything will work out.

    S_n is the sum, not S(n)
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  3. #3
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    Because \dfrac{15000}{1000}=15 you can solve this: 15\le (1.05)^n.

    Using the floor function  \left\lfloor {\dfrac{{\ln (15)}}{{\ln (1.05)}}} \right\rfloor +1.
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  4. #4
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    Sorry, but I don't see it.

    Archie Meade, I appreciate that I was barking up the wrong tree with respect to 15000(n), but that was not what was causing me the headache.

    Am I right in thinking I therefore have: log15000n = 1000 + nlog(1.05) - 1 - log(1.05 - 1) ?

    Plato, that log natural equation is (similar to) what I did in the first place, and it leaves me with 56.5, which is the number of months it takes for sales to reach 15 000 a month, not sales to reach 15 000 in total.
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  5. #5
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    15000=1000(\frac{1.05^n-1}{1.05-1})

    15=(\frac{1.05^n-1}{1.05-1})

    15\times 0.05+1=1.05^n

    now take log on both sides
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  6. #6
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    Quote Originally Posted by AngusBurger View Post
    Sorry, but I don't see it.

    Archie Meade, I appreciate that I was barking up the wrong tree with respect to 15000(n), but that was not what was causing me the headache.

    Am I right in thinking I therefore have: log15000n = 1000 + nlog(1.05) - 1 - log(1.05 - 1) ?
    No, because your logs are haphazard.

    \displaystyle\ 15000=1000\left[\frac{1.05^n-1}{1.05-1}\right]

    If you want to take logs now, use the log law log(ab)=loga+logb
    then use log(c/d)=logc-logd

    \displaystyle\ log15000=log1000+log\left[\frac{1.05^n-1}{1.05-1}\right]

    log15000=log1000+log\left[1.05^n-1\right]-log\left[1.05-1\right]


    It's far simpler to simplify at the beginning before taking logs

    viz

    \displaystyle\ 15000=1000\left[\frac{1.05^n-1}{1.05-1}\right]\Rightarrow\frac{15000}{1000}=\frac{1.05^n-1}{0.05}

    \Rightarrow\ 15(0.05)=1.05^n-1\Rightarrow\ 0.75+1=1.05^n

    1.05^n=1.75

    Now take logs. It saves unnecessary hassle.

    ln1.05^n=ln1.75

    \displaystyle\ nln1.05=ln1.75\Rightarrow\ n=\frac{ln1.75}{ln1.05}
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  7. #7
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    I see it now, thanks. The examples I've been working through were pretty simple, and then I was hit by this question and was stunned.

    I agree the logs were haphazard (I couldn't understand why the 1000 would be exempt). The equation wasn't my idea, rather someone else's who apparently knew better, but they insisted that was how I should go about it, much to my hesitation.

    Problem solved. Many thanks.
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