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

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The discussion revolves around solving the equations (x+1)^y = a and x^y = b algebraically. Participants explore the conditions under which these equations can be satisfied, suggesting that if there exists a y such that b^(1/y) = a^(1/y) - 1, any x could potentially work. They also mention that if a specific relationship involving logarithms holds, then any y would satisfy the equations. However, there is a realization that not just any x will suffice for the equations to hold true. The conversation highlights the complexity and nuances involved in finding solutions to these algebraic equations.
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is there any way to solve the following set of equations algebraically

(x+1)^y = a

x^y = b
 
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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 b^(1/y) = a^(1/y) - 1, any x will satisfy these equations?
 
whatta said:
Hmmm... it appears that if there is such y that b^(1/y) = a^(1/y) - 1, any x will satisfy these equations?
maybe you mean if there is such y that b^{\frac{1}{y}} = a^{\frac{1}{y}} - 1, any x will satisfy these equations?

then it also appears that if there is such x that \frac{\ln a}{\ln (x+1)} = \frac{\ln b}{\ln x}, 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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