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Prove the existence of logarithms

  • Thread starter tronter
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  • #1
186
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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?
 

Answers and Replies

  • #2
378
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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