## Re: Problem 295

Expand Messages
• Inspection suggests that x=y is a solution set.
Message 1 of 9 , May 1, 2003
Inspection suggests that x=y is a solution set.

--- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
> x(y^x + y^(x+1)) = y^(x+2)
> y(x^y + x^(y+1)) = x^(y+2)
>
> Solve for x and y
• Sorry, I m wrong, I think. Aye, I am. An x y substitution just gives you the same equation twice, or something......
Message 2 of 9 , May 1, 2003
Sorry, I'm wrong, I think. Aye, I am. An x<--->y substitution just
gives you the same equation twice, or something......

--- 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
• ... It does? Let s try x=y=1 as a subset of that solution set. Both equations become: 1(1^1 + 1^2) = 1^3 Simplifying to: 2 = 1 It appears that in general, x=y
Message 3 of 9 , May 1, 2003
--- In mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:
> Inspection suggests that x=y is a solution set.
>
> --- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
> > x(y^x + y^(x+1)) = y^(x+2)
> > y(x^y + x^(y+1)) = x^(y+2)
> >
> > Solve for x and y

It does? Let's try x=y=1 as a subset of that solution set.

Both equations become:
1(1^1 + 1^2) = 1^3

Simplifying to:
2 = 1

It appears that in general, x=y is not a solution set.
• ... The first equation yields y*(x^y)(1+x) = x^(y+2) y(1+x) = x² y = x²/(1+x) The second equation is just like the first with x y so that yields x =
Message 4 of 9 , May 1, 2003
--- 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

The first equation yields
y*(x^y)(1+x) = x^(y+2)
y(1+x) = x²
y = x²/(1+x)

The second equation is just like the first with x<->y so that yields

x = y²/(1+y)

Substitute the first into the second and you get

x = [x²/(1+x)]² / [1 + (x²)/(1+x)]

x + x³/(1+x) = x^4 / (1+x)²
1 + x²/(1+x) = x³ / (1+x)²
(1+x)² + x²(1+x) = x³
x² + 2x + 1 + x² + x³ = x³
2x² + 2x + 1 = 0
x² + x + 1/4 = -1/4
[x+(1/2)]² = -1/4

x has no solution.

However, not that some division occurred along the way, so we must
consider x=0.

x = 0 yields

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

becomes

y*0 = 0 and
0 = y²
In other words, y=0

So, I think that x=0 and y=0 are the only solutions. I hope I
• ... Substituting y = x yields x(x^x + x^(x+1)) = x^(x+2) -- x^x(x+x^2) = x^x * x^2 so x^x = 0 (indeterminable or impossible?) or x + x^2 = x^2, so x = 0
Message 5 of 9 , May 1, 2003
>Inspection suggests that x=y is a solution set.
>
>Sorry, I'm wrong, I think. Aye, I am. An x<--->y substitution just
>gives you the same equation twice, or something......
>
>--- 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
>

Substituting y = x yields
x(x^x + x^(x+1)) = x^(x+2) --> x^x(x+x^2) = x^x * x^2
so x^x = 0 (indeterminable or impossible?) or x + x^2 = x^2, so x = 0
(indeterminable value upon replacing in original equation).

Substituting y = kx yields
k*x(x^(kx) + x^(kx+1)) = x^(kx+2) --> k(x^(kx+1))(1+x) = x*x^(kx+1)
so x^(kx) = 0 (indeterminable or impossible?) or x = 0 (indeterminable
value upon replacing in original equation) or k(1+x) = x so x = k/(1-k),
and we'd have to solve for y in the original equation and verify in the
second equation...

I feel like I'm drowning in a glass of water, though.

-Abel Castillo

---
[This E-mail was scanned for viruses by the Santa Cruz BBS anti-virus system]
• ... 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 ... [This E-mail was
Message 6 of 9 , May 1, 2003
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
>

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

---
[This E-mail was scanned for viruses by the Santa Cruz BBS anti-virus system]
• Therefore there probably aren t any solutions? ... virus system]
Message 7 of 9 , May 1, 2003
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
>
> ---
> [This E-mail was scanned for viruses by the Santa Cruz BBS anti-
virus system]
• ... 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 +
Message 8 of 9 , May 1, 2003
--- 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
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.
• You mean complex numbers have logs? How does that work? As in, what is, say, 2 to the power of i?
Message 9 of 9 , May 6, 2003
You mean complex numbers have logs? How does that work? As in, what
is, say, 2 to the power of i?

--- In mathforfun@yahoogroups.com, bqllpd <no_reply@y...> wrote:
> This is exactly what I got.
> An interresting note: log(x)=y and log(y)=x (according to my
> calculator)
>
> --- In mathforfun@yahoogroups.com, garple51 <no_reply@y...> wrote:
> > --- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
> > 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
> > >
> > > The first equation yields
> > > y*(x^y)(1+x) = x^(y+2)
> > > y(1+x) = x²
> > > y = x²/(1+x)
> > >
> > > The second equation is just like the first with x<->y so that
> yields
> > >
> > > x = y²/(1+y)
> > >
> > > Substitute the first into the second and you get
> > >
> > > x = [x²/(1+x)]² / [1 + (x²)/(1+x)]
> > >
> > > x + x³/(1+x) = x^4 / (1+x)²
> > > 1 + x²/(1+x) = x³ / (1+x)²
> > > (1+x)² + x²(1+x) = x³
> > > x² + 2x + 1 + x² + x³ = x³
> > > 2x² + 2x + 1 = 0
> > > x² + x + 1/4 = -1/4
> > > [x+(1/2)]² = -1/4
> > >
> > > x has no solution.
> > >
> >
> > I saw no restriction to x and/or y real.
> >
> > [x+1/2]^2=-1/4
> >
> > appears to have solutions:
> >
> > x=-(1+i)/2, and x=-(1-i)/2.
> >
> > RonL
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