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Prove the existence of logarithms

  1. May 21, 2008 #1
    Fix [itex] b >1, \ y >0 [/itex], and prove that there is a unique real [itex] x [/itex] such that [itex] b^{x} = y [/itex].

    Here is the outline:

    (a) For any positive integer [itex] n [/itex], [itex] b^{n}-1 \geq n(b-1) [/itex]. Why do we do this?

    (b) So [itex] b-1 > n(b^{1/n}-1) [/itex].

    (c) If [itex] t>1 [/itex] and [itex] n > (b-1)/(t-1) [/itex] then [itex] b^{1/n} < t [/itex].

    etc..


    Is this the correct process?
     
  2. jcsd
  3. May 21, 2008 #2
    my approach:
    draw graph of b^x .. and use some theorm (I forgot it's name) .. I think intermediate value theorm
     
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