x

^{x}= x

Taking logs

xlog(x)=log(x)

x=1

However, clearly (-1)

^{(-1)}=-1

I know that this is because logs don't apply to certain problems, but is there an easier way to resolve this issue?

**Moderators:** gmalivuk, Moderators General, Prelates

I was recently thinking about logs, and I decided to solve

x^{x} = x

Taking logs

xlog(x)=log(x)

x=1

However, clearly (-1)^{(-1)}=-1

I know that this is because logs don't apply to certain problems, but is there an easier way to resolve this issue?

x

Taking logs

xlog(x)=log(x)

x=1

However, clearly (-1)

I know that this is because logs don't apply to certain problems, but is there an easier way to resolve this issue?

mosc wrote:How did you LEARN, exactly, to suck?

- Mathmagic
- It's not as cool as that Criss Angel stuff.
**Posts:**2926**Joined:**Thu Nov 30, 2006 12:48 am UTC**Location:**In ur fora posting in teh threads

SimonM wrote:I was recently thinking about logs, and I decided to solve

x^{x}= x

Taking logs

xlog(x)=log(x)

x=1

However, clearly (-1)^{(-1)}=-1

I know that this is because logs don't apply to certain problems, but is there an easier way to resolve this issue?

It's not that they "don't apply to certain problems". In this case, it's because when you take the log of x, you need to say that x > 0, because you can't take the log of a negative number.

What "issue" are you trying to resolve? A different way of solving a problem like that?

Axman: That, and have you played DX 10 games? It's like having your corneas swabbed with clits made out of morphine.

Pathway: cocks cocks cocks

Pathway: cocks cocks cocks

The log equation is still correct, if you use a weird multi-valued complex kind of log.

log(-1) = πi, -πi, 3πi, -3πi, ...

so log(-1) = -log(-1)

log(-1) = πi, -πi, 3πi, -3πi, ...

so log(-1) = -log(-1)

mathmagic wrote:What "issue" are you trying to resolve? A different way of solving a problem like that?

Exactly, I can determine another solution by looking at the problem, but are there any otherways of looking at it so it will work?

mosc wrote:How did you LEARN, exactly, to suck?

You can start by saying for nonzero values of x (0^{0} = 1 =/= 0 can't be a solution), x^{x} = x simplifies to x^{x-1} = 1. a^{b} = 1 when a is 1 and b is noninfinite, or when a is nonzero and b is 0, or when a is -1 and b is even. Thus, the set of solutions is {1, -1}.

Edit: I guess I forgot about a = e and b = 2n*pi, but that's not feasible here because of the way the problem is set up.

Edit: I guess I forgot about a = e and b = 2n*pi, but that's not feasible here because of the way the problem is set up.

- MartianInvader
**Posts:**805**Joined:**Sat Oct 27, 2007 5:51 pm UTC

SimonM wrote:xlog(x)=log(x)

x=1

Technically you can't even quite do this, since when you divide by log(x) you're assuming x isn't 1 (otherwise you're dividing by zero). So this actually shows a contradiction and proves that x can't be any positive number OTHER than 1, which you can then plug in and find that it is a solution.

If you wanted to check negative numbers, you could try assuming x < 0, then multiplying both sides by -1. Then you know both sides are positive, so you'll get

-x

xlog(-x)=log(-x)

x = 1

Dividing by log(-x) here assumed that x wasn't -1. Since you assumed x < 0, arriving at x = 1 gives a contradiction for any x < 0 OTHER than -1, which you can then plug in and find out it works. This lets you conclude that 1 and -1 are the only real solutions (after checking 0 doesn't work either).

Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

- gmalivuk
- GNU Terry Pratchett
**Posts:**26607**Joined:**Wed Feb 28, 2007 6:02 pm UTC**Location:**Here and There-
**Contact:**

MartianInvader wrote:SimonM wrote:xlog(x)=log(x)

x=1

Technically you can't even quite do this, since when you divide by log(x) you're assuming x isn't 1 (otherwise you're dividing by zero). So this actually shows a contradiction and proves that x can't be any positive number OTHER than 1, which you can then plug in and find that it is a solution.

If you wanted to check negative numbers, you could try assuming x < 0, then multiplying both sides by -1. Then you know both sides are positive, so you'll get

-x^{x}=-x

xlog(-x)=log(-x)

x = 1

Dividing by log(-x) here assumed that x wasn't -1. Since you assumed x < 0, arriving at x = 1 gives a contradiction for any x < 0 OTHER than -1, which you can then plug in and find out it works. This lets you conclude that 1 and -1 are the only real solutions (after checking 0 doesn't work either).

Not exactly. First of all, if you're assuming x

Assuming x to be positive, x log(x) = log(x) is fairly simply dealt with:

Case 1: x is not 1, so you can divide by log(x) and get x=1, a contradiction.

Case 2: x is 1, which works both in the original equation as well as the logarithm thereof (We can't just naively conclude from Case 1 that x=1, because it could have been that our first assumption, that there is any positive solution in the first place, was the one leading to the contradiction)

Taking the negative to start with doesn't immediately give you x log(-x) = log (-x), though, since it's not in general true that -(x

Users browsing this forum: No registered users and 12 guests