1) If x^a=x^b, it doesn't follow that a=b;

2) (1/16)^x-log(1/16,x) is not monotonously increasing on (0,1).

These problems have been bugging me for quite a long time now. I would appreciate it if anybody could post the answers here.

I think the answer for problem 1 is e roots. But I doubt this to be correct. I used that x^n = n has n complex roots. But I think this only applies when n is a natural number and not an irrational number like e.

I haven't got very far with problem 2. Actually I haven't got anywhere with problem 2.

My respect to maxim for posting these challenging problems. I came up with ideas and did some research that helped me a lot in understanding complex numbers better. But eventually I could not solve the problems.

Thank you
Ariyan

Problem 1.

There is an infinite number of solutions in complex space.

If the solution S is a complex number of modulus R and angle alpha (on the argand diagram), then:

S ^ e has a modulus of R ^ e and angle alpha * e. Setting these equal to e (i.e. modulus e, angle 0) we get solutions S:

mod(S) = e^(1/e)
arg(S) = 2k.pi./e (where k is an integer)

For k = 0 the solution is real.

All the solutions lie on a circle in a complex plane, centred on zero, with radius e^(1/e).

I think this is the solution to the first problem (but who knows, I may be wrong...).

Problem 2:

Replace 1/16 with 'a'.

LHS is a^x RHS is ln(x)/ln(a)

Assume 0<a<1 (which is valid for the actual problem specified).

Now we can limit the range of the problem a bit.

1. From the RHS, x must be greater than 0 (because log of zero or negative is invalid).

2. LHS varies from 1 (at x = 0) to 0 (as x gets big).

3. ln(a) is negative (because a < 1). When x > 1, ln(x) is positive, so RHS is negative, so the two sides cannot be equal. When x < a, RHS is greater than 1, so the two sides cannot be equal.

Therefore, we can state tha any solution(s) to the problem occur in the range a < x < 1.

at x = a, RHS = 1 (and is therefore greater than LHS).

at x = 1, RHS = 0 (and is therefore less than LHS).

Both sides are continuous and smooth in the range a < x < 1.

Therefore, there is an odd number of solutions.

You'll notice that I haven't actually found a solution.

I believe that for large values of a (~0.9) there is 1 solution, but for small values of a (~0.01) there are 3 solutions. I believe that there are no cases for which there are more than 3 solutions.

For a = 1/16 I think there are 3 solutions, 0.25 and 0.5 are two of them. There is another.

Not a very good solution of the problem, I'm afraid, but I hope this helps a bit.

KL..

It's interesting that even some pretty sophisticated computer algebra systems give this answer. I'll just say that Ariyan's idea of having e roots is not entirely unreasonable (even though having 'half a root' is certainly a highly untrivial idea): the equation z^a=b (with variable z) cannot have more than abs(a)^2/Re(a)+1 roots if Re(a)>0.

Your solution to problem 2 is correct. In the particular case of a=1/16 we can plot both functions, notice that the curves are symmetrical with respect to the line y=x, and therefore there's a root lying on that line. But since (1/16)^(1/2)=log(1/16,1/2) and (1/16)^(1/4)=log(1/16,1/4), this already gives us three roots (the last two aren't on the line y=x).

What is going on here can be seen in the attached picture; the key point is the slope of the curves at the point of their intersection with y=x. If the exponent decreases faster than the logarithm at this point, then there must be another root at some greater value of x, because eventually logarithm has to go 'down' to -infinity.

It turns out that the root of a^x=x can be found explicitly in terms of certain special functions. Define W(x) as the root of W(x)*e^W(x)=x (think of how ln(x) can be introduced as the solution to e^ln(x)=x). W(x) is called the Lambert W function. For example, 1*e^1=e, so W(e)=1. But strictly speaking, since exponent is periodic in the complex plane, W(x) will be multi-valued just as the logarithm.

Now it's easy to verify that the solution of a^x=x (for 0<a<1) is

x0 = -W(-ln(a))/ln(a)

So this is our point on the line y=x where we have a^x0=log(a,x0)=x0. Now we look at the derivatives at this point:

diff(a^x,x) = a^x*ln(a) = -W(-ln(a)),
diff(log(a,x),x) = 1/(x*ln(a)) = -1/W(-ln(a))

They're, of course, reciprocals, and so they're equal to each other when -W(-ln(a))=-1. This equation can be solved: a0=e^(-e). If a is greater or equal to this value, there will be only one root, otherwise there will be three roots. And 1/16 is very close to e^(-e), but on the side where we have three roots. Note that this is still not a proof, it's more like a heuristic analysis.

For problem 1, it might help to see that the question is not well defined...

The map exp is not one-to-one in the complex plane, so to define (part of) its inverse -a complex version of logarithm-, one must choose a branch. (Geometrically, exp maps the imaginary axis to the unit circle, so log maps the unit circle to... *some* interval of length 2pi on the imaginary axis, which contains the origin.)

Now x^something is e^(something*log x), so here we have 2 or 3 roots depending on our choice. In an ensemble of aleatory choices of a branch, the mean number of roots is about 2.718... (number which happens to be called e, but this has nothing to do with the capitan's age!)

Start with the solution that KL_GB had for problem 1, and note that it correctly gives all the possible moduli and arguments of solutions. However, of course the modulus-argument plane maps multiply to the real-imaginary plane, so that a strip y <= arg < y+2.pi maps to the entire complex plane. Therefore, KL_GB is partly correct in that there are an infinite number of solutions in the mod-arg plane; but this is not the complex plane, and for any choice of coordinate strip you get only a selection of 2 or 3 of these solutions. The rest of them become "phantom solutions" once you have chosen a strip.
>
>Now x^something is e^(something*log x), so here we have 2 or
>3 roots depending on our choice. In an ensemble of aleatory
>choices of a branch, the mean number of roots is about
>2.718... (number which happens to be called e, but this has
>nothing to do with the capitan's age!)

Similarly in terms of strips of the mod-arg plane, there are evenly spaced roots spaced a distance 2pi/e apart along the arg direction, so the expected number of roots in a strip of width 2pi is e. This is just a rephrasing of what you said, in a slightly more geometric language.

Thank you for you interesting post.

--Stuart Anderson