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Math Help - How close each other 2^x and 10^y can situate if x and y are natural numbers?

  1. #1
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    How close each other 2^x and 10^y can situate if x and y are natural numbers?

    If we have 2^3, itīs distance from 10^1 is only 2.
    If we have 2^10, itīs distance from 10^3 is 24 (as 2^10=1024 and 10^3=1000).

    Is there any other natural number pair x,y which results distance between 2^x and 10^y to be only 2?

    Is there any formula, where to insert desired distance and find x and y as a result, or does it only requires raw computer power to find an answer?

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  2. #2
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    Re: How close each other 2^x and 10^y can situate if x and y are natural numbers?

    I think it will never again be so close.

    If on the other hand you consider 2^x / 10^y or vice versa you can get this close to 1.

    For example 2^{196} \approx 1.004 \times 10^ {59} and this gives 59/196 as a good approximation of \log_{10}2 for whatever that's worth.
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  3. #3
    Super Member ILikeSerena's Avatar
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    Re: How close each other 2^x and 10^y can situate if x and y are natural numbers?

    No, it won't be as close as 2.

    For numbers x, y \ge 1, we have:

    2^x - 10^y=\pm 2

    2^{x-1} - 2^{y-1}5^y=\pm 1

    This can only be true if 2^{x-1} and 2^{y-1}5^y are relatively prime (Euclid).
    That means that y-1=0.
    So the only solution for a distance of 2 is: 2^3 and 10^1.
    Last edited by ILikeSerena; January 26th 2013 at 08:07 AM.
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