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Math Help - 2 simple Algebra questions

  1. #1
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    2 simple Algebra questions

    PROVE THAT:
    12^3=(9+ sqrt{5})^3+(9- SQRT{5})^3
    2. e^ln36=36 why?
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  2. #2
    Super Member Matt Westwood's Avatar
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    The first one's giving me a headache, so I'll waffle at the second.

    By definition, the "taking the logarithm of" function and the "raising e to the power of" are direct inverses.

    That is, by definition, x = \ln y \iff y = e^x, if y > 0 (it's undefined otherwise).

    So we have e^x = e^{\ln y} (doing the same thing to both sides of the first of the above).

    Then we note that from the second of the above, y = e^x and so it follows directly that:

    y = e^{\ln y}

    That holds for all y > 0 and so definitely holds for y = 36.
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    thanx!
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  4. #4
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    Quote Originally Posted by blertta View Post
    PROVE THAT:
    12^3=(9+ sqrt{5})^3+(9- SQRT{5})^3
    Did you try just doing the arithmetic on the right?

    (a+b)^3= a^3+ 3a^2b+ 3ab^2+ b^3

    So (9+ \sqrt{5})^3= 9^3+ 3(9^2)(\sqrt{5})+ 3(9)(\sqrt{5})^2+ (\sqrt{5})^3
    = 729+ 243\sqrt{5}+ 135+ 5\sqrt{5}= 964+ 248\sqrt{5}

    (9- \sqrt{5})^3= 9^3- 3(9^2)(\sqrt{5})+ 3(9)(\sqrt{5})^2- (\sqrt{5})^3
    = 729- 243\sqrt{5}+ 135- 5\sqrt{5}= 964- 248\sqrt{5}
    so the terms involving \sqrt{5} cancel and the sum is 1728.

    And, of course, 12^3= 1728

    You could also have seen this by observing that the odd powers of \sqrt{5} will cancel while the even powers add:
    2(9^3+ 3(9)(5))= 2(9)(9^2+ 15)= 2(9)(81+5)[tex]= 2(9)(96)= 2(3^2)(4)(24)= 2(3^2)(4)(4)(2)(3)= (2^6)(3^3)= (4^3)(3^3)= ((4)(3))^2= (12)^3
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