Tuesday, January 28, 2014

11: More Exponents

Can you show if this is true?

$2^6 * 2^6 = 2^{11} + 2^{11}$

Can we generate more like this one ... simple and elegant?




How about one with 3 as a base?

Source:

4 comments:

  1. Nice.
    3^8 * 3^8 = 3^15 + 3^15 + 3^15

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  2. and how about 3^3 * 3^3 = 3^5 + 3^5 + 3^5 ?

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  3. n^x * n^x = n(n^(2x-1)) looks like a generalization of the above, with appropriate restrictions on n & x.

    So for n = 2, and x = 6, we get 2^6 * 2^6 = 2(n^(2*2-1) = n^11 + n^11

    and for n = 3 and x = 8 we get 3^8 * 3^8 = 2(3^(15) = 3^15 + 3^15

    and for n= 3 and x = 3 we get 3^3 * 3^3 = 3(3^5) = 3^5 + 3^5 + 3^15

    This allows us to generate infinite examples of this sort of thing.

    e.g., 5^6 * 5^6 = 5(5^11) = 5^11+5^11+5^11+5^11+5^11

    You can see why this works if you factor the right side as 5^11(1 + 1 + 1 + 1 + 1) =
    5^11 * 5 = 5^`12

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