Originally posted by neo_maya

 

HI,

 

I had this small question about this problem. Here it is -

 

I was just wondering - I always thought that .....amm... it takes at least two equations to solve a two variable problem. I mean, shouldn't there be another equation like y=x or y = 5x ?

 

I thought that we can draw a graph with one equation and from there we can get the point where the curve intersects the x - axis. (Though I am not quite sure how we can draw one in this case)

 

I mean I tried this - ln(x^y) = ln(y^x) => ylnx = xlny but couldn't go further to draw a graph.

 

But I think graph is not the thing u want, right ?.

 

So, my question is - can we solve one equation with two variables? If so, how ? If the procedure is complicated u don't have to explain it - just give the name of the preocedure or better if u can post a link.

 

THANKS

 

[ PS : How did u write y^x = x^y this way - y^x=x^y ]

 

I don't think the purpose is to find a numerical solution, but rather to re-arrange the equation such that you get a function y = f(x).

Excuse me for using blue logx for log with base x---I just can't find the sub button!

 

Take logx on both side, it gives

ylogx x=xlogx y

y=xlogx y

 

let y=kx

kx=xlogx kx

k=logx k+1

k-1=logx k

 

when k=1

y=x

which is a solution

 

when k does not equal to 1

1=(1/(k-1))logx k

1=logx (k^(1/(k-1)))

 

Therefore

x=k^(1/(k-1))

y=k^(k/(k-1))

 

It works for x=2, y=4 (k=2), but it doesn't work for x=-2, y=-4.

t works when k=4, but it fails when k=3.

 

I am confused. Could anyone explain why?

I get:

 

y>(arrow) xproductlog[Log[x]/x/Log[x].

 

I get from transposing f[x]. as there is not a numerical answer.

Lynn said in post #28 :

let y=kx

 

How can you make that statement?

 

I also asked my algebra prof. this question. He was stumped and then said something about Newton

How can you make that statement?

 

Well, if you assume that x and y are both numbers, the y devided by x must also be a number, right? Well we don't know what that number is (since we don't know both x and y) so we can just assign that number a variable, let's use k. NOw we find that we have;

 

y/x=k

 

we can then rearange this equation (by multiplying both sides of the equation by x) so that we have;

 

y=kx

 

 

and now we see why we can make this statement. Really it just stems from the fact that if y is number, then we can pick two numbers that, when multiplied to gether, will give y. THen we call these numbers x and k. That is all.

 

Cool

Does mine not work???

I don't think it's possible to rearrange this equation. That's all I can say really, but I'm not entirely convinced by that substitution either tbh.

It's not possible according to my calc2 professor, who is the queen of algebraic rearrangements (99% of the time spent doing problems in class is her going through ways to algebraically manipulate problems, that actual calculus only takes a couple seconds)... that's where I got this problem. If it is possible to come up with a generalized solution, it's certainly not just an simple substitution. (Further evidence for that statement is that none of the CASs out there can solve it for the general case)

Thought so. It seems like a bit of a beast of a problem to be honest, and perhaps it needs to be broken down into seperate cases that can be dealt with individually. Even then, there's no guarantee that it'll actually have any generalized solution at all.

VendingMenace said in post #31 :

 

Well, if you assume that x and y are both numbers, the y devided by x must also be a number, right? Well we don't know what that number is (since we don't know both x and y) so we can just assign that number a variable, let's use k. NOw we find that we have;

 

y/x=k

 

we can then rearange this equation (by multiplying both sides of the equation by x) so that we have;

 

y=kx

 

 

and now we see why we can make this statement. Really it just stems from the fact that if y is number, then we can pick two numbers that, when multiplied to gether, will give y. THen we call these numbers x and k. That is all.

 

Cool

 

But then you're assuming that y & x have a linear relationship. I don't know if you read over on the other forums they posted here, but 2⁴ = 4² also works. As well as 1¹ = 1¹ .

1 / 1 = 1. => k = 1

but 4 / 2 :neq: 1

[quote name='But then you're assuming that y & x have a linear relationship.

 

Yup. But it doesn't matter if i assume that. I make no assumption about what k is' date=' so it iwll not affect the answer any.

 

If you don't quite see this, then lets take a look at the two answers you posted....

 

2⁴ = 4² also works.

 

OK for this solution can set x=4 and y=2, right?

 

then for x/y=k

we have

2/4=k

 

so k=0.5

 

next answer...

 

As well as 1¹ = 1¹ .

1 / 1 = 1. => k = 1

 

WEll, you already did this one. And you are right, for this X and Y, k=1.

 

 

but 4 / 2 1

 

well, there is no reason why k has to equal 1. It is a variable. Just like x and y. There may be other solutions to for x and y, we don't just assume there is one. Likewise, depending on what x and y are, k could by (prolly will be) something diferent.

 

I hope that helps.

Oh, I thought k was constant. hm...

 

I'll check out the mechanics of your work later today...

This equation can be manipulated into the form:

 

ln(x)/x = ln(y)/y.

 

This means that x and y are two values that give the same value of the function f(z) = ln(z)/z. If you plot w = f(z) versus z in thezw -plane, you will find a horizontal asymptote is w = 0, and avertical asymptote is z = 0. A horizontal line w = c intersects thiscurve in exactly two points if 0 < c < 1/e, one point (e,1/e) if c = 1/e or c <= 0, and no points if c > 1/e. The z-coordinates of the two points give you the solutions (x,y) where x and y are not equal. If x and y are not equal, then y/x = a will be different from 1. That means that x = y/a, and then:

 

 

 

ln(y/a)/(y/a) = ln(y)/y

a*[ln(y)-ln(a)] = ln(y)

(a-1)*ln(y) = a*ln(a)

ln(y) = [a/(a-1)]*ln(a)

 

y = a^[a/(a-1)]

 

x = a^[1/(a-1)]

 

Every solution with x and y unequal does correspond to one of these values of a, namely a = y/x.

There are still problems

 

Whatever k (or 'a' in wolfson's post) is, it gives positive x and y.

But x=-2. y=-4 works.

So k=-4/-2=2

But substituting it into the parametric equations will give

x=2, y=4

 

And, y=kx would not work if x and y are complex numbers...

It only gives the positive values because logarithms only deal with positive numbers (in the set of real numbers anyway).

 

It's like asking why |x| always equals a positive.

Did you read my paragraph about its functions?

how about x=1 and y=2? Won't that work

aommaster said in post # :

how about x=1 and y=2? Won't that work

 

2 doesn't equal 1 duder.

..and we're looking for a general solution, since specific solutions exist and are not too difficult to find.

Y^x = x^y

 

The derivative of f(x) is f'(x) = (1 - ln(x)) / x^2, so that f'(x) = 0 gives x = e. From that we conclude that f(x) is increasing for 0<x<e and decreasing for x>e.

 

For any integer n>4 the horizontal line y=f(n) is "lower" than the line y=f(2) and thus intersects the graph of f(x) as second time for some 1<x<2, so that there is no integer m with f(m)=f(n).

 

When we started with x/ln(x) = y/ln(y), we restricted ourselves toPositive x and y (by definition of ln). But the original question does not exclude negative solutions. It is easy to show that there are no solutions with one positive and one negative integer. Also, when x^y = y^x, then (-x)^(-y) = +/- (-y)^(-x)………

sorry guys! I was tired when i wrote that! Of couse 1 to the power of two is a hundred!

Wasnt the very first question to solve for Y in terms of X...In other words get y by itself, Y=???.

 

why are all of you looking for actual numerical answers?

simple, the yth root of y^x = x

Oops, I meant the xth root of x^y

there should only be a y on the left, and only x functions on the right