Any other suitable substitution?

Algebra Level 1

If $x^x = \left( \frac12\right)^{\frac12}$ , does this imply that $x = \frac{1}{2}$ ?

Yes No

2 solutions

Shourya Pandey
Mar 6, 2017

We have $( \frac{1}{4})^{\frac{1}{4}} = (\sqrt {\frac{1}{4}})^{\frac{1}{2}} = (\frac{1}{2})^{\frac{1}{2}}$ ,

So $x$ could be $\frac{1}{4}$ too.

I took the help of a graphing calculator.. is there any other rigourous method to check the same @Pi Han Goh @Shourya Pandey

ABHIJIT DIXIT - 4 years, 3 months ago

One way of doing it is to see that " $0^0$ " $= 1$ and $f(x)= x^x$ has a minima at $x= \frac {1}{e}$ . Since $f(0)> f(\frac{1}{2}) > f(\frac{1}{e})$ , so by Intermediate Value Theorem, there is a $0<c< \frac{1}{e}$ for which $f(c)= f(\frac{1}{2})$ .

Shourya Pandey - 4 years, 3 months ago

Ok, Thanks . I've heard of Intermediate Value Theorem. But I can relate to what you said. Thanks again

ABHIJIT DIXIT - 4 years, 3 months ago

$x = (\frac{1}{2^{n}})$ where $n \in {Z}$

Substituting this in the equation and simplifying gives $2^{n} = 2^{2^{n - 1}} \Rightarrow n = 2^{n - 1} \Rightarrow \boxed{n = 1 , 2}$

For $n = 3 , 2^{n - 1} > n$

So taking this as the base step and inducting on $n$ :

For $n = m , 2^{m - 1} > m$

Also , $2^{m - 1} \geq 1$

Adding the two inequalities , we get $2^{m} > m + 1$

Hence , it s true for $n = m + 1$ and hence our induction stands complete.

$\Rightarrow$ There are no solutions for $n \geq 3$

Ankit Kumar Jain - 4 years, 2 months ago

Huh? What are you trying to show here?

Pi Han Goh - 4 years, 2 months ago

I just wanted to find all the solutions and I have shown that in my comment.

I have shown that $\frac{1}{2} , \frac{1}{4}$ are the solutions.

I hope that this works.

Thanks!

Ankit Kumar Jain - 4 years, 2 months ago

@Pi Han Goh Sir , I have edited it a bit ..Is it fine now??

Ankit Kumar Jain - 4 years, 2 months ago

I still don't understand what you're doing. Sure $n = 2^{n-1}$ gives solutions of $n=1,2$ , but how do you know there isn't any non-integer solution? Or how do you know that there isn't any non-real solution?

Plus, you claimed that the solution for $x$ is in the form of $x = \dfrac1{2^n}$ , but why must $x$ be a rational power of 2 to begin with? How do you know that there isn't a solution for say, $x = \dfrac1{3^n}$ ?

On the other hand, if you want to show that $x= \dfrac1{2^n}$ is a solution for integer $n$ via induction, then shouldn't ALL the solution be $x= \dfrac1{2^1} , \dfrac1{2^2}, \dfrac1{2^3}, \dfrac1{2^4} , \ldots$ ?

(Note that your working for the proof by induction is also wrong, do you know where you went wrong?)

Pi Han Goh - 4 years, 2 months ago

@Pi Han Goh – I presumed that $x = \frac{1}{2^n}$ that was the error , if somehow we can prove that the solutions would be of this form , then I think that the rest of the solution will be fine.

As for the non - integer solutions of $n = 2^{n - 1}$ , I already presumed $n$ to be an integer.

Through induction I wanted to sow that for $n \geq 3$ there are no solutions for $n = 2^{n - 1}$

Where is my induction part wrong ?? Sir , can you please point out the errors?

Ankit Kumar Jain - 4 years, 2 months ago

@Ankit Kumar Jain –

if somehow we can prove that the solutions would be of this form , then I think that the rest of the solution will be fine.

That's the thing: You didn't prove that all of the solutions of $x$ must be an integer power of 2.

Through induction I wanted to sow that for $n \geq 3$ there are no solutions for $n = 2^{n - 1}$

Ah I see. It's not clear what you're trying to achieve here. If you want to prove that there's no integer solution of $n = 2^{n-1}$ for $n\geq 3$ , then it's better to start with the proposition of "Let $P(n) = n - 2^{n-1}$ . I will prove that $P(n) < 0$ for all integers $n \geq 3$ via induction."

On the other hand, (once again) your solution only shows that there is no integer solution of $n = 2^{n-1}$ for $n\geq 2$ . But how do you know that there isn't a non-integer solution, say $n = 3.333333333$ ?

This is a good attempt, but unfortunately, your proof isn't rigorous because you have only shown that if $x$ is a rational power of 2, then the only solutions of are $1/2$ and $1/4$ .

Pi Han Goh - 4 years, 2 months ago

@Pi Han Goh – I got your point . Thanks!!

Is there any way by which we can find all the solutions?

Ankit Kumar Jain - 4 years, 2 months ago

@Ankit Kumar Jain – First show that x^x is not a strictly increasing function in the interval (0,1), then show that it's strictly decreasing in (0, 1/e ] and strictly increasing in [1/e , 1). Now we know that x=1/2 > 1/e is a solution, so there exists precisely one other solution in (0, 1/e ] , trial and error shows that x=1/4 is the (only) other solution.

Pi Han Goh - 4 years, 2 months ago

@Pi Han Goh – Sir , what do you mean by this 'On the other hand, if you want to show that $x= \dfrac1{2^n}$ is a solution for integer $n$ via induction, then shouldn't ALL the solution be $x= \dfrac1{2^1} , \dfrac1{2^2}, \dfrac1{2^3}, \dfrac1{2^4} , \ldots$ ?'??

Ankit Kumar Jain - 4 years, 2 months ago

The question is worded improperly. The asker meant to ask if it implied that x=1/2 is the only solution, which it doesn't, but what he actually asked was if it implies that x=1/2 is a solution, which it does.

Therefore the correct answer here is yes.

Matthew Helm - 4 years, 2 months ago

The question is phrased properly. Consider the following:

If person A is a boy, does this imply that person A is Matthew?

Even though Matthew is a boy, that doesn't necessarily imply that person A can only be Matthew. Just as the answer to this question is no, the answer to the problem is also no.

Calvin Lin Staff - 4 years, 2 months ago

@Shourya Pandey. How did you figure that the function has a minimum at 1/e? Thanks.

Mohammad Hasan - 4 years, 2 months ago

Sorry for the late reply. Differentiate the function $y = x^{x}$ and equate it to zero. Differentiation of $x^{x}$ :

$y = x^{x}$

$ln y = xlnx$

$\frac {d}{dx}(ln y) = \frac{d}{dx} (xlnx)$

$\frac{y'}{y} = 1+lnx$

$y' = y(1+lnx) = x^{x}(1+lnx)$ .

Shourya Pandey - 4 years, 1 month ago

Justin Fitzpatrick
Mar 24, 2017

Using calculus, one has that $x^x$ approaches 1 as $x$ approaches 0 from the right, decreases to a minimum at $x=1/e$ , and increases thereafter. Therefore, for ALL $y \in (\left(\frac{1}{e}\right)^{1/e}, 1)$ , the equation $x^x=y$ has two solutions!

Any other suitable substitution?

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