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Math Help - Power Problem

  1. #1
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    Power Problem

    Hey guys, I was just wondering if any of you have any thoughts on this problem. I've exhausted all mine

    Let's say we have 2 functions: f and g with f(x)=3^x and g(x)=100^x.

    Then define two sequences:
    1) a_1 = 3 and a_{n+1} = f(a_n) for n \geq 1
    2) b_1 = 100 and b_{n+1} = g(b_n) for n \geq 1

    What is the smallest positive integer m for which b_m > a_{100}?

    I've tried connecting the two series with an inequality but I can't quite get one that I can prove. I've probably overlooked something blindingly obvious

    Thanks in advance if you guys can help
    Last edited by Myrc; December 15th 2008 at 07:27 PM. Reason: typo
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  2. #2
    Super Member Aryth's Avatar
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    Quote Originally Posted by Myrc View Post
    Hey guys, I was just wondering if any of you have any thoughts on this problem. I've exhausted all mine

    Let's say we have 2 functions: f and g with f(x)=3^x and g(x)=100^x.

    Then define two sequences:
    1) a_1 = 3 and a_{n+1} = f(a_n) for n \geq 1
    2) b_1 = 100 and b_{n+1} = g(b_n) for n \geq 1

    What is the smallest positive integer m for which b_m > a_{100}?



    I've tried connecting the two series with an inequality but I can't quite get one that I can prove. I've probably overlooked something blindingly obvious

    Thanks in advance if you guys can help
    Did you try this:

    b_m = 3^{m-1}

    a_{100} = 100^{99}


    So:

    3^{m-1} > 100^{99}

    log_3{3^{m-1}} > log_3{100^{99}}

    m - 1 > log_3{100^{99}}

    m > log_3{100^{99}} + 1

    m > \lceil 415.98 \rceil

    m > 416
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  3. #3
    Lord of certain Rings
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    Quote Originally Posted by Aryth View Post
    Did you try this:

    b_m = 3^{m-1}

    a_{100} = 100^{99}
    Hmm... b_2 = g(b_1) = g(100) = 100^{100}

    a_2 = f(a_1) = f(3) = 3^{3}
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  4. #4
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    Sorry for the late reply, i was away for a few days.

    Isomorphism (see above post) was right btw, it's powers to the power of power, which makes it harder, so  a_{100} \neq 100^{99} , but  a_{100} = 100^{100^{100^{100^{100^{...}}}}} which makes it rather annoying 'cause i can't work with logs that easily. So sorry Aryth, i think you misunderstood the problem, but thanks for the reply anyway

    And it also makes the terms grow waaay too fast:
    a_1 = 3 , a_2 = 27, a_3 = 762,55974,84987
     b_1 = 100, b_2 = 100^{100}
    (I'm not even going to type that out) etc...

    I was wondering because I've got a possible inequality that could work, but I dunnno how to prove it, would induction work? (I'm a bit iffy on induction proofs of inequalities)

    Would this be proveable?

      5b_n < a_{n+2} < b_{n+1}

    Because it seems to work but I can't prove it

    Thanks for your help guys.
    Last edited by Myrc; December 19th 2008 at 01:32 PM. Reason: typo
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  5. #5
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    Hi guys, I was wondering if anyone could possibly give me a few pointers as to how to prove the above inequality by induction 'cause I can't seem to work it out... But if I can prove that then I can solve the problem without too much difficulty.

    Thanks for your help!

    Oh, and Merry Christmas
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  6. #6
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    Talking A Solution (Hopefully)

    Well, I think I have a solution, found in a last minute rush before this problem was due . Please point out any errors that I (probably) will have made

    Here goes:

    If I can prove the following, I can solve it:

    b_n < 5b_n <  a_{n+2} < b_{n+1}

    The First Bit: (A very difficult proof)

    Since 5 > 1, 5b_n > b_n (obviously )

    The Second Bit: (An Induction Proof)

    First, let's check 5b_1 and a_{1+2=3}:
    <br />
5b_1 = 500 and a_3 = 3^{27}

    Clearly 500 < 3^{27} so this base case is true.

    Now, suppose for some k 5b_k < a_{k+2}

    First, note that 4 < log_{3}100 < 5 and that the difference between log_{3}100 and 5\cdot100 in this case is 81 (and  81 > log_35 ), and this difference will only get bigger as the numbers get bigger.

    Therefore, log_{3}(100)b_k + log_{3}5 < 5b_k
    Implying log_{3}(100)b_k + log_{3}5 < a_{k+2}
    What follows is basic application of the log rules:
    log_{3}5 + log_{3}(100)b_k < a_{n+2}
    log_{3}5 + log_{3}(100^{b_k}) < a_{n+2}
    log_{3}(5 \cdot 100^{b_k}) < a_{n+2}
    5 \cdot 100^{b_k} < 3^{a_{n+2}}
    5 \cdot b_{k+1} < a_{n+3}

    So this must be true for k \geq 1 (integral k of course)

    The Third Bit: (An Induction Proof)

    Again, we can check a base case.

    a_3 = 3^{27} and b_2 = 100^{100}

    Clearly, 3^{27} < 100^{100} (If you don't believe me, check the powers).

    Now, suppose for some k, that a_{k+2} < b_{k+1}

    Then since log_{3}100 > 1, we may write:
    a_{k+2} < (log_{3}100)b_k+1
    a_{k+2} < log{3}100^{b_{k+1}}
    3^{a_{k+2}} < 100^{b_{k+1}}
    a_{k+3} < b_{k+2}

    So this part too must be true for integral k \geq 1

    The Fourth Bit: (The Answer (Finally))

    Now, I have (hopefully) convinced you that for k \geq 1:
    b_k < 5b_k < a_{k+2} < b_{k+1}

    Now we need the smallest integer m such that a_{100} < b_m

    By the above inequality, the smallest such integer is 100 - 1 = 99.

    Therefore, I proudly proclaim that the answer is 99!!!

    Oh, and Merry Christmas!!!
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