Using Logs to Solve

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SimonM
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Using Logs to Solve

Postby SimonM » Sat Nov 03, 2007 2:01 pm UTC

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

xx = 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?
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Re: Using Logs to Solve

Postby Mathmagic » Sat Nov 03, 2007 3:33 pm UTC

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

xx = 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?
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Re: Using Logs to Solve

Postby Goplat » Sat Nov 03, 2007 3:34 pm UTC

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)

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Re: Using Logs to Solve

Postby SimonM » Sat Nov 03, 2007 5:07 pm UTC

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?
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Re: Using Logs to Solve

Postby Gwydion » Sat Nov 03, 2007 11:35 pm UTC

You can start by saying for nonzero values of x (00 = 1 =/= 0 can't be a solution), xx = x simplifies to xx-1 = 1. ab = 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.

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Re: Using Logs to Solve

Postby MartianInvader » Sun Nov 04, 2007 9:22 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

-xx=-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).
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Re: Using Logs to Solve

Postby gmalivuk » Sun Nov 04, 2007 10:05 pm UTC

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

-xx=-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 xx is even defined for negative x in general, you're already working in C, in which case you can take the logarithm of a negative number without any trouble (at least, no trouble after picking the branch to use).

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 -(xx) = (-x)x for negative x. We have to first restrict ourselves to { x : (-1)x = -1 }, which restricts us to those rational x whose denominators, when written in lowest terms, are odd. (Can you easily show that none of the ruled-out numbers might be a solution to the original equation?) Though once we've done that, your method works for showing the only options are 1 and -1. (Assume it's not -1, and you divide by the logarithm to get x=1. Then assume it is -1 and check that this works in the original equation.)
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