Solve Algebraically: (x+1)^y = a and x^y = b

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Discussion Overview

The discussion revolves around the algebraic solution of the equations (x+1)^y = a and x^y = b. Participants explore potential relationships between the variables and conditions under which solutions may exist.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about the possibility of solving the equations algebraically.
  • Another participant suggests that there may be a solution and encourages others to attempt it.
  • A participant expresses confusion about the complexity of the equations.
  • Some participants propose that if there exists a y such that b^(1/y) = a^(1/y) - 1, then any x will satisfy the equations.
  • Further, it is suggested that if there exists an x such that ln(a)/ln(x+1) = ln(b)/ln(x), then any y will satisfy the equations.
  • One participant retracts their earlier statement about the generality of x satisfying the equations, indicating that not "any" x will work.

Areas of Agreement / Disagreement

Participants express differing views on the conditions under which solutions exist, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There are limitations regarding the assumptions made about the variables and the specific conditions under which the proposed relationships hold true. The mathematical steps involved in deriving these conditions remain unresolved.

murshid_islam
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is there any way to solve the following set of equations algebraically

[tex](x+1)^y = a[/tex]

[tex]x^y = b[/tex]
 
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yes there is...now you try it, and tell us how far you get
 
cough. how many more you have there?
 
i take back what i said...what on Earth is that equation :)
 
Hmmm... it appears that if there is such y that [tex]b^(1/y) = a^(1/y) - 1[/tex], any x will satisfy these equations?
 
whatta said:
Hmmm... it appears that if there is such y that [tex]b^(1/y) = a^(1/y) - 1[/tex], any x will satisfy these equations?
maybe you mean if there is such y that [tex]b^{\frac{1}{y}} = a^{\frac{1}{y}} - 1[/tex], any x will satisfy these equations?

then it also appears that if there is such x that [tex]\frac{\ln a}{\ln (x+1)} = \frac{\ln b}{\ln x}[/tex], any y will satisfy these equations.

whatta said:
cough. how many more you have there?
lol. i have no more. i got the equation in the thread you mention from
(x+1)y = 216
xy = 125
 
Last edited:
hmm. no it's me who takes back what I said. "any" x will not do.
 

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