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Thread: 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 $\displaystyle f(x)=3^x$ and $\displaystyle g(x)=100^x$.

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

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

    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; Dec 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 $\displaystyle f(x)=3^x$ and $\displaystyle g(x)=100^x$.

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

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



    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:

    $\displaystyle b_m = 3^{m-1}$

    $\displaystyle a_{100} = 100^{99}$


    So:

    $\displaystyle 3^{m-1} > 100^{99}$

    $\displaystyle log_3{3^{m-1}} > log_3{100^{99}}$

    $\displaystyle m - 1 > log_3{100^{99}}$

    $\displaystyle m > log_3{100^{99}} + 1$

    $\displaystyle m > \lceil 415.98 \rceil$

    $\displaystyle m > 416$
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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:

    $\displaystyle b_m = 3^{m-1}$

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

    $\displaystyle a_2 = f(a_1) = f(3) = 3^{3}$
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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 $\displaystyle a_{100} \neq 100^{99} $, but $\displaystyle 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:
    $\displaystyle a_1 = 3 , a_2 = 27, a_3 = 762,55974,84987 $
    $\displaystyle 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?

    $\displaystyle 5b_n < a_{n+2} < b_{n+1}$

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

    Thanks for your help guys.
    Last edited by Myrc; Dec 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:

    $\displaystyle b_n < 5b_n < a_{n+2} < b_{n+1}$

    The First Bit: (A very difficult proof)

    Since $\displaystyle 5 > 1, 5b_n > b_n $(obviously )

    The Second Bit: (An Induction Proof)

    First, let's check $\displaystyle 5b_1$ and $\displaystyle a_{1+2=3}$:
    $\displaystyle
    5b_1 = 500$ and $\displaystyle a_3 = 3^{27}$

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

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

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

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

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

    The Third Bit: (An Induction Proof)

    Again, we can check a base case.

    $\displaystyle a_3 = 3^{27}$ and $\displaystyle b_2 = 100^{100}$

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

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

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

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

    The Fourth Bit: (The Answer (Finally))

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

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

    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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