Solve for x, y

1. Feb 26, 2007

murshid_islam

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

$$(x+1)^y = a$$

$$x^y = b$$

2. Feb 26, 2007

eaboujaoudeh

yes there is...now you try it, and tell us how far you get

3. Feb 26, 2007

whatta

cough. how many more you have there?

4. Feb 26, 2007

eaboujaoudeh

i take back what i said...what on earth is that equation :)

5. Feb 27, 2007

whatta

Hmmm... it appears that if there is such y that $$b^(1/y) = a^(1/y) - 1$$, any x will satisfy these equations?

6. Feb 27, 2007

murshid_islam

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.

lol. i have no more. i got the equation in the thread you mention from
(x+1)y = 216
xy = 125

Last edited: Feb 27, 2007
7. Feb 27, 2007

whatta

hmm. no it's me who takes back what I said. "any" x will not do.