Proving: a^{log_cb} = b^{log_ca}

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Discussion Overview

The discussion centers around the mathematical expression a^{log_cb} = b^{log_ca} and seeks to establish a proof for this identity involving logarithmic properties. The scope includes mathematical reasoning and proof techniques.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant requests a demonstration of the identity a^{log_cb} = b^{log_ca} for all a, b, and c.
  • Another participant suggests taking the logarithm of both sides to explore the proof, applying the property of logarithms that allows manipulation of exponents.
  • A third participant asserts that the approach is indeed a proof and proposes a more rigorous formulation involving the equality of exponents when bases are the same.
  • A later reply indicates understanding and appreciation for the discussion, acknowledging the contributions of others.

Areas of Agreement / Disagreement

Participants appear to agree on the validity of the proof approach, but there is no explicit consensus on the rigor or completeness of the proof presented.

Contextual Notes

Some assumptions about the properties of logarithms and the conditions under which the identity holds are not fully explored or stated, leaving room for further clarification.

gnome
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Would someone please show me why

[tex]{ a^{log_cb} = b^{log_ca}[/tex]

for all a, b and c.
 
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I don't know if this is really a proof.. but just take the log of both sides:
[tex]{ a^{log_cb} = b^{log_ca}[/tex]
[tex]log_c { a^{log_cb} = log_c b^{log_ca}[/tex]
[tex]({log_cb})({log_ca}) = ({log_ca})({log_cb})[/tex]
using the property that:
[tex]{log_ca^r} = r{log_ca}[/tex]
 
it is a proof. perhaps to make it appear more rigorous, you could write [tex]a^{log_cb}=c^{log_c(a^{log_cb})}[/tex] and similarly for the rhs, and say that the only way for c^xto be equal to c^y is if x=y.
 
Got it!

Thanks, folks.
 

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