# 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.