--- In

mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:

> Therefore there probably aren't any solutions?

>

> --- In mathforfun@yahoogroups.com, Abel Castillo <abel@s...> wrote:

> > At 02:32 p.m. 01/05/2003 -0000, you wrote:

> > >--- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:

> > >> y(x^y + x^(y+1)) = x^(y+2)

> > >> x(y^x + y^(x+1)) = y^(x+2)

> > >>

> > >> Solve for x and y

> > >

> > >So, I think that x=0 and y=0 are the only solutions. I hope I

> > >haven't made a mistake above.

> > >

> >

> > x=0 and y=0 can't satisfy the original equations since 0^0 is

> > indeterminable. That's the only thing wrong I can see.

> >

> > -Abel Castillo

> >

Here's another way of looking at it:

Equation 1 yields:

y*(x^y)(1+x) = x^(y+2)

y = x²/(1+x) (dividing by x^y is okay since it can't be zero)

y = x - 1 + 1/(1+x)

dy/dx = 1 - 1/(1+x)²

d²y/dx² = 2/(1+x)³ > 0

So this graph passes through the origin and is concave up at all

points.

Equation 2 is the inverse of equation 1, so combining this with the

above, they can only intersect at most at two points. One of them

is at (0,0). Since equation 1 is concave up, and has slope = 0 at

the origin, the other point of intersection must be in the first

quadrant. Note that dy/dx is positive in the first quadrant but

less than 1. The inverse has a positive slope always greater than 1

in the first quadrant. So they can't intersect in the first

quadrant. So (0,0) is the only solution. But Abel says that (0,0)

is automatically thrown out, so there are no solutions.

I don't think I've made a mistake because I've graphed it, and

matched my conclusions.