simple problem

[razorfane]

umm, yeah, since it is a 2nd degree ecuation there are two solutions. Two get both solutions i usually isolate the x and use the cuadratic formula, which doesn't work here, or make the equation equal zero and have a multiplication on the other side, then make each factor equal to zero, which doesn't work either. So that second answer i wasn't able to find algebraically.

[CPL.Luke]

isn't 2^1/2 equal to the square root of two not (1/2)^-1/2

[Ducky Havok]

Well, when you're rewriting it (1/2)^(-1/2) is the same as 1/((1/2)^(1/2)) which is 2^(1/2) so they are the same exact thing. (1/4)^(-1/4) works the same way.

[Algebracus]

I have some remarks to this problem:

 

1) This is not a quadratic equation. A quadratic equation has the form ax^2 + bx + c = 0, but the given equation has not.

2) Razorfane states that the equation has two solution. That is not correct. The equation has in fact three solutions: x = 0, x = 1/4 and a third approximated to 1,44.

3) I register that Razorfane has brought the equation into the form

f(x) = f(1/4), where f(x) = x^(x - 1/2). This is a fine move, but it is not enough to state that x = 1/4 is the only solution (which it in fact isn't). First of all, Raxorfane divided by x in his first step, and therefore cancelled the solution x = 0. Secondly, f(x) = f(1/4) gives x = 1/4 only if and only if f(x) is an injective function, which it isn't.

4) To check how many solutions the equation has, we can use calculus on f(x). First, we state that the function is continuos. Then we find in which intervalls it is increasing and decreasing. Then we find which values it takes in the given intervalls. Then we (by the intermediate value theorem) find for how many intervalls (in which the function is monotone) we have some x such that f(x) = f(1/4) = 2^(1/2).

5) We easily see that there are no solution x < 0, since 2x = x^(2x) = (x^x)^2 >= 0. Therefore, we can (without loosing any solution except 0) write x = 2^v and x^(2x) = 2^(v*2^(v + 1)) = 2^(v + 1). Since the function f(x) = 2^x is one-to-one, we therefore have v * 2^(v + 1) = v + 1. It could be of some interest to work with this problem instead of the original one.

[matt grime]

That depends on what you claim 0 to the 0 is. A lot of people would say that it is undefined, though it makes most sense to declare it to be 1.

[razorfane]

1) This is not a quadratic equation. A quadratic equation has the form ax^2 + bx + c = 0, but the given equation has not.

Well, I actually said it's a second degree ecuation, but now I'm not really sure, since [math]x^2[/math] would make an ecuation second degree, but this is [math]x^{2x}[/math]. About the third answer I'm not really sure either since [math]0^0[/math] would be undefined. I've never seen it equal to 1, I think there's no proof that it equals one so texts take it as undefined.

[Algebracus]

I have to correct myself at some point here: If we are going to define 0^0, then it is most convenient to define 0^0 as 1, up to continuity, since

 

lim x^x = 1, as can be seen by l'Hospitals formula:

x->0+

 

x^x = e^(x ln x), and x ln x = (ln x)/(1/x), of the form (-infinity)/(infinity). Then

 

lim x ln x = lim (1/x)/(-1/(x^2)) = lim (-x) = 0, and e^0 = 1.

x->0+

 

So x = 0 cannot be said to be a solution.

 

When it comes to the number of solutions, check for instance x^(2x) = 0.

[Asimov Pupil]

my algebra comes out to 1/2 let me keep trying

[uncool]

Algebracus, 0^0 cannot be defined except for the limit which is being used. In this case, I believe that the limit successfully gets 0^0 = 0.

-Uncool-

[Algebracus]

I have already proved that the upper limit is 1, so your belief is wrong.