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Re: Problem 295

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  • clooneman
    Inspection suggests that x=y is a solution set.
    Message 1 of 9 , May 1, 2003
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      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
    • clooneman
      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
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        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
      • slim_the_dude
        ... 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
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          --- 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.
        • slim_the_dude
          ... 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
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            --- 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
            haven't made a mistake above.
          • Abel Castillo
            ... 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
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              >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


              ---
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            • Abel Castillo
              ... 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
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                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

                ---
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              • clooneman
                Therefore there probably aren t any solutions? ... virus system]
                Message 7 of 9 , May 1, 2003
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                  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]
                • slim_the_dude
                  ... 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
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                    --- 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.
                  • clooneman
                    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
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                      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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