Prove the existence of logarithms

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SUMMARY

The discussion centers on proving the existence of logarithms by establishing that for any base \( b > 1 \) and any positive \( y > 0 \), there exists a unique real number \( x \) such that \( b^x = y \). The proof involves demonstrating that \( b^n - 1 \geq n(b - 1) \) for any positive integer \( n \), leading to the conclusion that \( b^{1/n} < t \) when \( n \) exceeds a certain threshold. The Intermediate Value Theorem is suggested as a key tool in visualizing and confirming the existence of such \( x \).

PREREQUISITES
  • Understanding of exponential functions, specifically \( b^x \) where \( b > 1 \)
  • Familiarity with the Intermediate Value Theorem in calculus
  • Basic knowledge of inequalities and their applications in proofs
  • Concept of limits and continuity in real analysis
NEXT STEPS
  • Study the proof of the Intermediate Value Theorem and its applications in real analysis
  • Explore the properties of exponential functions and their graphs
  • Learn about the uniqueness of solutions in mathematical proofs
  • Investigate the relationship between logarithms and their corresponding exponential functions
USEFUL FOR

Mathematics students, educators, and anyone interested in understanding the foundational concepts of logarithms and their proofs in real analysis.

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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?
 
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my approach:
draw graph of b^x .. and use some theorem (I forgot it's name) .. I think intermediate value theorem
 

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