Solve for x, y

  • #1
murshid_islam
442
17
is there any way to solve the following set of equations algebraically

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

[tex]x^y = b[/tex]
 

Answers and Replies

  • #2
eaboujaoudeh
150
0
yes there is...now you try it, and tell us how far you get
 
  • #3
whatta
256
0
cough. how many more you have there?
 
  • #4
eaboujaoudeh
150
0
i take back what i said...what on Earth is that equation :)
 
  • #5
whatta
256
0
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?
 
  • #6
murshid_islam
442
17
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.

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:
  • #7
whatta
256
0
hmm. no it's me who takes back what I said. "any" x will not do.
 

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