Question

\[x^{x!} = x!^{x}\]
find the values of x

Using x = x! is refused.

Any help on this ?

++ the legendary mathematical problem

Answer 1

logs maybe?

Answer 2

1, 2 are trivial solutions
for x>=3 show that x^x! > x!^x

Answer 3

yes, but induction might be tricky..

Answer 4

to me, it is good. I tried and got it

Answer 5

The answer is 2 , 1 but I am looking for steps.

Answer 6

me too

Answer 7

You are also looking for the step ? :D

Answer 8

see if this helps
http://math.stackexchange.com/questions/996881/how-to-show-that-nn-gt-nn-for-n-gt-2-in-mathbbn

Answer 9

You mean the "derivative" answer ?

Answer 10

whichever makes sense

Answer 11

The derivative is wrong according to my teacher.

Logs maybe used. On my side , I tried it but achieved nothing. Give it a try.

::: Got some network connection problems.

Answer 12

\(\large n^{n!} = n\times n\times \cdots \times \text{n! times}\)

\(\large n!^n = n^2\text{terms each of which are }\le n\)

Answer 13

since \(\large n! \gt n^2\) when \(n\ge 4\), the value of \(n^{n!}\) will be greater than \(n!^n\)

Answer 14

thats the first reply and it kindof makes sense right ?

Answer 15

Maybe , but how will it help in finding the value of x ?

Answer 16

I can be used in proving that n!^{n} is greater than n^{n!}

Answer 17

a > b

what does that tell you ?

Answer 18

it says that a is NEVER equal to b right ?
proving n^n! > n!^n tells us the same : there are no solutions for n >= 4

Answer 19

You mean if we proved that a > b.
It tells us that n got no possible solutions ?!
or I misunderstood

Answer 20

Exactly ^

Answer 21

if you know that \(\large n^{n!} \gt n!^n\) for all \(\large n \ge 4\),
doesnt that prove there are no solutions for \(\large n\ge 4\) ?

Answer 22

My teacher said " it can be solved by 4 different methods"
He even told me that the answers are ( 0, 1, 2 ) but 0 is refused.

Answer 23

pelletTY network connection -_-

Answer 24

there could be more ways to approach this problem, are you trying to find atleast four different methods to sovle this ?

Answer 25

hmm i have an idea , it might not work though ..

Answer 26

I am trying to find atleast one LAWL. What's your idea ??

Answer 27

ok
x!= nx right ?

thus

x^nx=(x!)^x
(x^n)^x=(x!)^x

x^n=x! gives the only solution

Answer 28

how ever ,
for x>2

x^n = x!

x.x...x = 1.2.3...x

Answer 29

seems like for x! >2
there is at least 2x number of primes ( can't remember how much lol )
but this should solve it :O

Answer 30

I got a few questions, x! = nx : n stands for ?? i mean 4! = 4*3*2*1 do you mean n = 3*2*1 for ex ?

Answer 31

x!=1.2.3.4..x
let 1.2.3.4...x-1 =n
x!=nx ( means x divides x! )

Answer 32

x^n = x! ok , now what will you do on this ?

Answer 33

n = (x-1)!

Answer 34

no note this
x^n means its an x terms only
a number that only have x

Answer 35

Hmm , clarify it a bit

Answer 36

@ganeshie8 help ?xD

Answer 37

im happy with the first solution in mse

Answer 38

ohk :O

Answer 39

To me , it looks like the legendary mathematical problem won't be solved.
All the answers are sticking to the same approach giving no forward answer :/

Answer 40

\[x\approx2.4278\]

Answer 41

(2.4278)! on calculator = Math error, am I mistaken ?
-> x! <- in this case x can only be an integer number.
Correct to me if I am missing something.

Answer 42

\[x!=\Gamma(x+1)\]

Answer 43

http://en.wikipedia.org/wiki/Gamma_function

Answer 44

Indeed, I have no objection to your answer. But I didn't take the Gamma function at least till now.
So I can't use it on exams :/. But I am still confused how can x! be 2.4 ??
When I first took the lesson the prof said "Only integers!".

Answer 45

Can someone solve it using Logs please ?

Answer 46

Here we go...

\[\LARGE x!^x=x^{x!}\]\[\LARGE x!=x^{(x-1)!}\]

So this means x! is also just a single number raised to a power. But factorials are consecutive numbers multiplied together, so obviously the only possible choices are for when you just begin.

This immediately allows 1 and 2 in because everything can be multiplied by 1, so it doesn't take away 2's "pure" number to a power status. After that, there's no way it can be satisfied.

Answer 47

To clarify a bit, the left side of the equation says: \[\LARGE x*(x-1)*(x-2)*...\] while the right side says \[\LARGE x*x*x*...\]

So the only case it works looks like this (ok, I mean it's sort of half plugged in already, but you can see what the pattern is I'm considering here maybe more clearly by looking.

\[\LARGE 1=1^{0!}\]
and

\[\LARGE 2*(2-1)=2^{1!}\]

It's quite obvious that this and higher orders won't work I think by looking at the exponents.
\[\LARGE 3*(3-1)*(3-2)=3^{2!}\]

Answer 48

\[x^{x!}=x^\frac{x!}{2}x^\frac{x!}{2}=(x^x)^\frac{(x-1)!}{2}(x^\frac{(x-1)!}{2})^x \\ x!^x=(x \cdot (x-1))!^x=x^{x}(x-1)!^x \\ \text{ it is obvious } \frac{(x-1)!}{2} \ge 1 \text{ for all integer } x \ge 3 \\ \text{ so we need to prove } (x-1)! \le x^\frac{(x-1)!}{2} \text{ for } x \ge 3 \\ \text{ this isn't so bad to prove by induction }\]

Answer 49

Kainui , right, x is a single number raised to a power which does nothing (1! = 1, 0! = 1).

Answer 50

but all this is theoretical, I wanna prove it mathematically.

Answer 51

freckles, fine, could you try to do so ?. I failed to prove anything at least on this one :(

Answer 52

\[(3-1)!=2 \le 3=3^\frac{(3-1)!}{2} \\ \text{ Suppose} (x-1)! \le x^\frac{(x-1)!}{2} \text{ for some integer } x \ge 3 \\ \text{ We want to show } x! \le (x+1)^{\frac{x!}{2}} \\ x(x-1)! \le x \cdot x^\frac{(x-1)!}{2} \text{ multiply x on both side the induction hypothesis } \\ x! \le x^{\frac{(x-1)!+2}{2}} \\ (x-1)! \ge 2 \text{ for all integer } x \ge 3 \text{ so we have} \\ x! \le x^{\frac{(x-1)!+(x-1)!}{2}} =x^\frac{2(x-1)!}{2} \\ x! \le x^\frac{x(x-1)!}{2} \text{ since } x \ge 3 \\ x! \le x^\frac{x!}{2} \\ \text{ finally } x+1>x \text{ for all integer } x \ge 3 \\ x! \le (x+1)!^{\frac{x!}{2}}\]
Now think everything I did seems fine.

Answer 53

[Math Processing Error]
I can't view your text.
What do you mean by "seems fine" ?

Answer 54

"but all this is theoretical, I wanna prove it mathematically."

I don't understand how my argument and reasoning is somehow not good enough? Where does your doubt come in? What more do you need to see formalized to make you feel better about it? It will only ever continue to diverge as I showed, but we can wrap it up with a lot of fancy mathematical analysis symbols but it would come down to saying exactly the same thing in only a more convoluted form.

Answer 55

Sis bro whatever , I didn't refuse your argument , but do you think it will be wise of me to write something like that in an exam ? Of course not.
Even through , All the symbols will prove what you have proved theoretically.
It's obvious if x^x! = x!^x then x = x! and x , x! >= 2 but how to prove that mathematically ? that's the matter.

Answer 56

At least , thanks for tying to help.