- #1

- 374

- 1

## Main Question or Discussion Point

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

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

[tex]x^y = b[/tex]

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

[tex]x^y = b[/tex]

- Thread starter murshid_islam
- Start date

- #1

- 374

- 1

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

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

[tex]x^y = b[/tex]

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

[tex]x^y = b[/tex]

- #2

- 150

- 0

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

- #3

- #4

- 150

- 0

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

- #5

- 218

- 0

- #6

- 374

- 1

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.

lol. i have no more. i got the equation in the thread you mention fromcough. how many more you have there?

(x+1)

x

Last edited:

- #7

- 218

- 0

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

- Last Post

- Replies
- 4

- Views
- 1K

- Last Post

- Replies
- 9

- Views
- 30K

- Last Post

- Replies
- 4

- Views
- 719

- Replies
- 1

- Views
- 6K

- Replies
- 11

- Views
- 11K

- Replies
- 5

- Views
- 617

- Last Post

- Replies
- 9

- Views
- 4K

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 20

- Views
- 4K

- Last Post

- Replies
- 3

- Views
- 2K