What are the conditions for solving (a+b)^(-c)?

  • Context: High School 
  • Thread starter Thread starter Pietair
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Discussion Overview

The discussion revolves around the mathematical expression (a+b)^(-c) and the conditions under which it can be solved or manipulated. Participants explore specific cases, particularly when c = 1, and examine related equations and their equivalences.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant asks how to work out (a+b)^(-c).
  • Another participant specifies their case with c = 1, simplifying the expression to (a+b)^-1.
  • A question is raised about the meaning of x^(-1), leading to a clarification that it represents 1/x.
  • A participant presents an equation r = 1/(a + bcos(c)) and attempts to relate it to another form r = (1/a) * (1/(1+bcos(c))).
  • Another participant challenges the equivalence of the two forms, stating they are not equal based on the provided statements.
  • A later reply suggests that if there are no conditions for the equivalence, a counterexample should be provided to demonstrate the impossibility of proving it.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the equivalence of the two forms of the equation. There is disagreement regarding the conditions necessary for the expressions to be equal, with some suggesting the need for counterexamples.

Contextual Notes

Participants express uncertainty about the conditions under which the expressions can be manipulated or equated, indicating a need for further clarification on the definitions and assumptions involved.

Pietair
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Good day,

How do I work out (a+b)^(-c)?

Thanks.
 
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In my case, c = 1, so I have got: (a+b)^-1
 


Well, what does x ^{-1} represent?
 


(1/x).

I have got the equation:
r = 1/(a + bcos(c))

This should be equal to:
r = (1/a) * (1/(1+bcos(c)))

I just can't figure out why.
 


To be straight that I understand: you have

<br /> r = \frac 1 {a + b \cos{c}}<br />

and need to show that this equals

<br /> r = \left(\frac 1 a\right) \left( \frac 1 {1 + b \cos{c}}\right)<br />

If your statements are the ones I've written here, they aren't equal.
 


Exactly.
 


Are there some type of conditions? If not, just write down a counterexample, therefore showing it's impossible to prove it (ie. a not = 1, b, c in reals)
 

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