United Kingdom - A Nice Exponential Equation | Math olympiad Question

A Nice Exponential Equation solving

[ytmiguelar]

I think you have problems with some concepts.
1) In the change of variable y = k x, k is not a constant but a parameter.
2) You do a lot of unnecessary steps. For example, in the fifth line it is totally unnecessary to extract the square root.
3) In the tenth line you do not analyze what happens if k = 1. In such a case, both sides of the equation cannot be raised to 1/(k-1).
In the case k = 1, the parameterization y = k x takes the form y = x, which is the identity function that solves the original equation for all x>0 (first quadrant) and for the values x<0 for which there exists the expression (x^x) (third quadrant).
4) It remains to analyze the final expression x = k^(1/(k-1)), y = k^(k/(k-1)), which is a parameterization of a solution function. The parameter k must be a positive real number in which case the curve is in the first quadrant, or a negative even integer in which case they are discrete points in the fourth quadrant.
You can write in geogebra the equation x^y = y^x and you will get the graph of the two solution curves (y = x, and the curve with parameter k) which are in the first quadrant and "intersect" at the point limit of (x, y) = ( k^(1/(k-1)), k^(k/(k-1)) ) as k → 1, which is (e, e).

[irahartoch]

There are an infinite number of solutions. For every positive value of X, x=y is an obvious solution. Also x=2, y=4 is a solution (as is x-4 and y-2). And so forth....

[klauscosmin]

I don't understand why was necessary to square root when it would have been easier to rise direct to power (1/x) instead of (2/x)

[stuarts]

There is no reason to take a square root. The key idea is writing y=kx. After this, just a few steps lead to the result x=k^(1/(k-1): Using y=kx in x^y=y^x leads to x^(kx)=(kx)^x = k^x x^x. Divide both sides by x^x. The result is x^(kx-x)=k^x or x^((k-1)x)=k^x. Now raise both sides to the 1/x to get x^(k-1) = k. If k is not equal to 1 then raise both sides to the 1/(k-1) to get x=k^(1/(1-k)). Video then shows that y =kx = k^(k/(1-k)). When k=1, y=x is a solution for any x. One should also stipulate in the statement of the problem that x and y are positive -- otherwise roots of negative numbers might arise.

[crmn_tv]

x = 1, y = 1 can also be a solution (if there is no other constraints such as x != y)

[Godeau]

This is a bit partial as the answer is the solution to two equations, where the linearity of x and y is hidden. But the equation as it is is not bounded by linearity. If you try a different functional form between x and y, you get new answers. And no functional form is required either. Good question though.

[timm]

As long as x=y, there are infinite number of roots. 👍👍👍

[DaHaiZhu]

Seems there are infinite solutions where x = y, since that condition was not excluded.

[mariorodriguezruiz]

Why assuming y=kx? That only gives a family of solutions

[luciusluca]

The solution is based on a prior guess of a linear relationship y=k x, with
k as a constant parameter (parametrization). A more systematic approach, without guessing, is to introduce polar coordinates. Then the same result follows upon finally identifying Tan[theta] with kappa.

[thasicommunitiyheatlhcarec]

Dad carries his son on his shoulder and when the dad becomes older, the son carries his dad on his shoulder: PROVED

[SomeoneCommenting]

This is kind of a weird question for a test since it has to undergo a lot of analysis for all the trivial vs non trivial solutions. There's not "a solution", single one, so the answer to the exercise would have been much more than just solving for a single value to show proof that "it worked".

[MsRad]

Hi Madame,

I prefer this solution,
1- between X and Y there is a tangent lets say
Y/X = T

Now, Y = TX
The new equation
X ^ TX = XT ^ X

If we take the LN of the equation we get this:
TX Ln x = X Ln TX

Now we reverse again from Ln to the initial formula and we solve it for X
X^ T = TX OR
X X^(T-1) = TX
X ([X^(T-1) - T] = 0

Finaly we get
X = 0 ( not a good solution)
Or
X = e^ [LnT/(T-1)] with
T alwasy +
and T ]0, 1[ ]1, +inf[
T NOT EQUAL 0 OR 1and not a negative number

[theupson]

f(u) = logu/u on u>1 has two continuous monotone branches {(1<u<e), (e<u)} with the same range (0, 1/e). therefore every real number x in the interval (1, e) there is a value y in the other interval such that logx/x = logy/y ==> x^y = y^x

[redroach]

I think a better solution is x=1/e^lambert w(-ln(y)/y)because it doesnt involce a random k variable. If you would like to know how to get this feel free to reply to this comment asking.

[dmitryr]

IMHO the solution is not complete.
It was considered only a case for y = kx, need to consider the other cases, probably need to proof that there are no solution other than y = kx

The solution can be also a little bit simplified by skipping to take a square root, the following step would be the same, just without 1/2 in power...

[longcours]

1. Why do you take square roots initially ?
2. What allows you to take as an assumption that y=Kx ? Why should the ensemble of solutions be limited to this special case ?

[bernardseffah]

Logically, if x=y, then any number satisfies the equation

[orchestra]

Please, correct me if I'm wrong somewhere... We have one equation with x and y. What we can do is to find the function in its explicit form y(x) that satisfies this equation. There's no way how we can find any numerical values, because for that we would need 2 equations!! Here's my solution:
1. Logarithmize of both sides: y * ln(x) = x * ln(y) (here actually any log of any positive base "a" would work)
2. ln(x) / x = ln(y) / y
3. Exponentiate both sides: exp (ln(x)/x) = exp(ln(y) / y). (here I used "e" as a base, but like in Step1 same arbitrary positive "a" as the base would also work)
4. exp(ln(x))^(1/x) = exp(ln(y))^(1/y)
5. x^(1/x) = y^(1/y).
6. Compare LHS and RHS from Step5 and see that the only option for this to hold is y=x.
So, y=x is the only solution. So, whatever values you put as "x", if you put same for "y", the equation will hold. In your case, for y = kx, this means that k can be only 1.
In special case, using a bit of limits and calculus, it can be shown that same holds also for x = 0 and y =0.

UPD: I just realized the following.. If we have expression like f(x) = f(y), this leads to x=y only if the function f is such that a certain value of f(x) corresponds to only one x, like for example if it is monotonic for the whole interval of x of interest. Then, the inverse function f^-1(x) is well-defined. In our case a function f(x) = x^(1/x) is not such a function. As f is below 1, there is only one x that corresponds f(x), however in between [1, e^(1/e)] there are two x values that are mapped to the same function value. You can see, for example, that 1^1 = 1, so that f(1) =1, but at the same time when x is infinetely large (x->+infinity), f also tends to 1 These two branches coalesce at the point (e, 1^(1/e)). For any x<1, we have one and only solution y=x, for 1<x<e, we will have y1=x, and the second solution is defined through W0 and W1 branches of Lambert W function of a parameter t: x(t) = W0(t)/t, y2(t) = W1(t)/t, for -1/e<t<0, and for x>e we will analougously have y = x and the second is solution defined by means of x(t) = W1(t)/t, y2(t) = W2(t)/t, for -1/e<t<0 (same thing, but different combination of W0 and W1 for x and y). Interestingly, these untrivial "secondary" solutions connect with each others very smoothly at x=e.

[dbdb]

Where are you trying to find the solutions? In C, R, Q, Z or N?
For example all the points on the line y = x in the first quadrant are also solutions.