Problem of The Day

[BigBeachBanana]

Solve [imath]x^x=x[/imath]​

 

[Little_Louis]

x = 1?

1^1 = 1

 

[Dr.Peterson]

x = 1?

1^1 = 1

That's one solution; and Desmos implies it:

Spoiler

But there is another ...

 

[BigBeachBanana]

x = 1?

1^1 = 1

You guessed the trivial solution. There are actually 2 of them. The problem can be solved algebraically.

Spoiler: Hint

Take log on both sides

 

[Little_Louis]

Ok so if we have x^ x = x
log x^ x = log x
log x^ x/ log x = x
log x / log x = 1
x = 1

Am I close?

 

[BigBeachBanana]

log x^ x = log x
log x^ x/ log x = x

I'm not sure how you go from line from line 1 to line 2 here, but try this:
[math]\log x^x=\log x\\ x\log |x|=\log x\\ x\log |x|-\log x=0\\ [/math]

 

[Dr.Peterson]

I'm not sure how you go from line from line 1 to line 2 here, but try this:
[math]\log x^x=\log x\\ x\log |x|=\log x\\ x\log |x|-\log x=0\\ [/math]

Is there a reason you introduced an absolute value in one place, but not in another? Sign issues are exactly what makes this tricky to solve, so this is important!

 

[BigBeachBanana]

Is there a reason you introduced an absolute value in one place, but not in another? Sign issues are exactly what makes this tricky to solve, so this is important!

You're right Dr P. It's a habit of mine when using the log power rule to introduce the absolute value to preserve the domain, but isn't necessary for this problem.

 

[Dr.Peterson]

That's one solution; and Desmos implies it:

Spoiler

View attachment 31846

But there is another ...

The trouble is that [imath]x^x[/imath] is not defined for all negative numbers, so it can't be graphed in that case; and also, solving with logs is problematic without paying careful attention to special cases. For example, it is not defined when x is a rational number with an even denominator in lowest terms, and when the denominator is odd, it can be positive or negative depending on the denominator. But these considerations can lead to an algebraic solution.

Here is a graph, where I have manually inserted some discrete points on the graph, and the dotted curves represent where such points all lie:

Spoiler

 

[Dr.Peterson]

Correction:

The trouble is that x^x is not defined for all negative numbers, so it can't be graphed in that case; and also, solving with logs is problematic without paying careful attention to special cases. For example, it is not defined when x is a negative rational number with an even denominator in lowest terms, and when the denominator is odd, it can be positive or negative depending on the numerator. But these considerations can lead to an algebraic solution.

 

[AvgStudent]

My attempt:
[math] x^x=x\\ x\log |x| = \log |x| \\ x\log |x| - \log |x|=0\\ \log |x|\times(x-1)=0\\ \text{Equate the factors to 0:}\\ [1]: x-1=0 \implies x=1\\ [2]: \log |x|= 0 \implies x=\pm 1\\[/math]Therefore, the solutions are [imath]x=1, -1[/imath]. Is this correct?

 

[Steven G]

My attempt:
[math] x^x=x\\ x\log |x| = \log |x| \\ x\log |x| - \log |x|=0\\ \log |x|\times(x-1)=0\\ \text{Equate the factors to 0:}\\ [1]: x-1=0 \implies x=1\\ [2]: \log |x|= 0 \implies x=\pm 1\\[/math]Therefore, the solutions are [imath]x=1, -1[/imath]. Is this correct?

I don't know if it is correct. Can you please check it for me and me know? Thanks.

 

[AvgStudent]

I don't know if it is correct. Can you please check it for me and me know? Thanks.

For [imath]x=1[/imath], it's trivial [math]1^1=1 ✔[/math]For [imath]x=-1[/imath], [math](-1)^{-1}=\frac{1}{-1}=-1✔[/math]

 

[Steven G]

For [imath]x=1[/imath], it's trivial [math]1^1=1 ✔[/math]For [imath]x=-1[/imath], [math](-1)^{-1}=\frac{1}{-1}=-1✔[/math]

Good job! Now are those the only two results? Why?

 

[Dr.Peterson]

My attempt:
[math] x^x=x\\ x\log |x| = \log |x| \\ x\log |x| - \log |x|=0\\ \log |x|\times(x-1)=0\\ \text{Equate the factors to 0:}\\ [1]: x-1=0 \implies x=1\\ [2]: \log |x|= 0 \implies x=\pm 1\\[/math]Therefore, the solutions are [imath]x=1, -1[/imath]. Is this correct?

The one problem I see is that you have started by taking absolute values. What is your justification for doing that? How do you know that doesn't introduce extraneous solutions, as it would in other problems? And are you sure it doesn't cause you to miss any solutions?

 

[AvgStudent]

Good job! Now are those the only two results? Why?

Those are the only ones I found. How can you tell if there's more, or not?

The one problem I see is that you have started by taking absolute values. What is your justification for doing that? How do you know that doesn't introduce extraneous solutions, as it would in other problems? And are you sure it doesn't cause you to miss any solutions?

The reason I introduced the absolute value right off the bat is that, without it, I'm only assuming x>0, which cut off the other branch of the solution, i.e. x<0. I didn't consider that it would introduce the extraneous solution(s). However, from the check, in this case, it didn't. Is there a better way to solve it?

 

[Steven G]

Those are the only ones I found. How can you tell if there's more, or not?

The reason I introduced the absolute value right off the bat is that, without it, I'm only assuming x>0, which cut off the other branch of the solution, i.e. x<0. I didn't consider that it would introduce the extraneous solution(s). However, from the check, in this case, it didn't. Is there a better way to solve it?

What you did was fine. There are only two solutions. Good job.

 

[topsquark]

Here's my thought. For all of this let [imath]x = r e^{i \theta }[/imath].

[imath]x^x = x[/imath]

[imath]\left | x^x \right | = |x|[/imath]

[imath]\left | (r e^{i \theta } ) ^{ r e^{i \theta } } \right | = |r e^{i \theta } |[/imath]

or
[imath]\left | r^r \right | = r[/imath]

(where r is, or course, real and positive) so r = 1.

Now. [imath]x^x = x[/imath] is the same as the system
[imath]\begin{cases} r ln(r) cos( \theta ) - r \theta ~ sin( \theta ) = ln(r) \\ r ln(r) ~ sin( \theta ) + r \theta ~ cos( \theta ) = \theta \end{cases}[/imath]

Since r = 1
[imath]\begin{cases} - \theta ~ sin( \theta ) = 0 \\ \theta ~ cos( \theta ) = \theta \end{cases}[/imath]
which has the solutions [imath]\theta = 0, ~ \pi[/imath], so the only complex solutions are [imath]x = r e^{i \theta } = -1, ~ 1[/imath].

I think that covers it.

-Dan

Addendum: This assumes the only real solutions to [imath]x^x = x[/imath] are [imath]x = \pm 1[/imath], which I think has been pretty well established above.

 

[Steven G]

[imath]\left | (r e^{i \theta } ) ^{ r e^{i \theta } } \right | = |r e^{i \theta } |[/imath]

or
[imath]\left | r^r \right | = r[/imath]

(where r is, or course, real and positive) so r = 1.

How did you get this last equality above?

 

[topsquark]

How did you get this last equality above?

Ew! Now that I ran a couple of examples through W|A I feel a little silly. That is a lot uglier than I had thought.

Back to the drawing board. Thanks for the catch!

-Dan