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Math Help - Evaluation of two natural numbers

  1. #1
    Member Pranas's Avatar
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    Evaluation of two natural numbers

    \displaystyle \[A = {7^{15}} + {15^7}\]
    \displaystyle \[B = {6^{16}} + {16^6}\]

    Hence
    a) \displaystyle \[A < B\]
    b) \displaystyle \[A > B\]
    c) \displaystyle \[A = B\]


    I remember evaluating something like that at school, but when was asked today unfortunately could not produce a reasonable answer
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Pranas View Post
    \displaystyle \[A = {7^{15}} + {15^7}\]
    \displaystyle \[B = {6^{16}} + {16^6}\]

    Hence
    a) \displaystyle \[A < B\]
    b) \displaystyle \[A > B\]
    c) \displaystyle \[A = B\]


    I remember evaluating something like that at school, but when was asked today unfortunately could not produce a reasonable answer
    When I was at school 55 years ago we would have done this by looking at the logarithms (to base 10) of the numbers:

    \log(7^{15}) = 15\log7\approx 12.7,\quad \log(15^7) = 7\log15\approx 8.2,
    \log(6^{16})=16\log6\approx12.5,\quad\log(16^6) = 6\log16\approx7.2.

    It's clear from the logarithms that 7^{15} is considerably bigger than 6^{16}, and the other two numbers are several orders of magnitude smaller than either of these. So A must be quite a lot bigger than B.
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