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Math Help - Baby Rudin - Existence of logarithm

  1. #1
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    Baby Rudin - Existence of logarithm

    I'm working out of Baby Rudin, chapter 1 problem 7. The problem is "Fix b >1, y > 0, and prove that there is a unique real x such that  b^x = y by completing the following outline..."

    I'm at part f) "Let A be the set of all w such that  b^w < y , and show that  x = sup(A) satisfies  b^x = y .

    What I'm wondering about - if m is a positive integer, since b > 1, is it a leap to assume that there's an integer k such that  b^k > m ?
    Last edited by Math Major; August 30th 2010 at 02:17 PM.
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    Quote Originally Posted by Math Major View Post
    I'm working out of Baby Rudin, chapter 1 problem 7. The problem is "Fix b >1, y > 0, and prove that there is a unique real x such that  b^x = y by completing the following outline..."

    I'm at part f) "Let A be the set of all w such that  b^w < y , and show that  x = sup(A) satisfies  b^x = y .

    What I'm wondering about - if m is a positive integer, since b > 1, is it a leap to assume that there's an integer k such that  b^k > m ?
    You can easily enough justify that "leap" by writing b = 1 + h, where h > 0. Then b^k = (1+h)^k > 1 + kh (since the right-hand side is just the first two terms of the binomial expansion), and that can clearly be made arbitrarily large.
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    Oh wow, thank you. Never would've thought of the binomial theorem. I kept trying in my head to figure out a way to write k without referencing a logarithm >.<. Thanks!
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