x^{x!} = X!^{x}
Find the values of x.

Question

x^{x!} = X!^{x} 
Find the values of x.

anonymous · Answer

one simple solution is x=1

trojanpoem · Answer

Where is the steps ? and as it was said by my teacher it got two possible solutions not just one.

trojanpoem · Answer

0, 1 , 2 and the 0 will be refused. Looking for the steps.

kirbykirby · Answer

For $x=1$, you get 
LHS: $1^{1!}=1$
RHS: $(1!)^1=1$

For $x=2$:
LHS: $2^{2!}=4$
RHS: $(2!)^2=4$

For x=3:
LHS: $3^{3!}=3^6=729$
LHS: $(3!)^3=6^3=216$

If you try 4 and 5, it seems like the difference between the LHS and RHS gets bigger and bigger. My hunch is that $x^{x!} > (x!)^x$ for $x \ge 3$, so I suppose you could prove that statement by induction. If this is the case, there will be no more solutions after x=3. 

I'm assuming this is looking at positive integers only though.

kirbykirby · Answer

And the fact that your teacher said there are only 2 solutions makes me think that the induction should work!

kirbykirby · Answer

I think you can do this actually...
The base base was proven above (x=3).
 Assume it's true for x=k,, so the induction hypothesis becomes: $k^{k!}>(k!)^k$. 

Then,  show it's true for x=k+1:, so: 
$ (k+1)^{(k+1)!}>(k+1)!^{k+1}$

Going on the LHS:
$$\large \begin{align}(k+1)^{(k+1)!}&=(k+1)^{(k+1)k!}\&=(k+1)^{k!k+k!}\&=(k+1)^{k!k}(k+1)^{k!}\&>(k+1)^{k!k}\color{red}{k^{k!}}\&>(k+1)^{k!k}\color{red}{k!^k}, \text{ by the induction hypothesis}\&=\left[(k+1)^{k!}\right]^kk!^k\&=\left[(k+1)^{k!}k!\right]^k\&>\left[(k+1)^2 k! \right]^k\&=\left[(k+1)(k+1)k! \right]^k\&=(k+1)^k(k+1)^kk!^k\&=(k+1)^k(k+1)!^k\&>(k+1)!(k+1)!^k\&=(k+1)!^{k+1}\end{align}$$

The before last line I used the fact that $k^k > k! \implies (k+1)^k >(k+1)!$ at least when k is larger than 2

kirbykirby · Answer

Gosh I hope this is correct. Maybe someone else can confirm this?

kirbykirby · Answer

when k larger than 1*

trojanpoem · Answer

Factorials can't be a 	$$\frac{ x }{ b }$$ so the only possible answers are 2 and 1.

swissgirl · Answer

I cant believe this .... so there was no other answer besides 1 and 2 .... -.-
I was racking my brain to figure out the other 2 instances when this equation is true.

kirbykirby · Answer

@swissgirl I hope my induction is correct. But according to Wolfram, http://www.wolframalpha.com/input/?i=x%5E%28x%21%29%3D%28x%21%29%5Ex , 1 and 2 are the only integer solutions. There is one that is 2.427... but this a non-integer. I am guessing they used the gamma function instead of a factorial to find that additional solution.

swissgirl · Answer

Not sure if your induction is correct but I like the route you took. I got a different proof but landed in the same place :)

trojanpoem · Answer

Assume it's true for x=k

Is not correct as $$2^{4}  - 4^{2}   $$
4 is not equal to 2. If it was true you can easily get the answer by making x! = x.

Our teacher you can solve it using logs. Any suggestions ?

turingtest · Answer

just playing around here, taking the log of both sides gives$$x!\ln x=x\sum_{i=1}^x\ln i\\ln x[(x-1)!-1]=\sum_{i=1}^{x-1}\ln i$$dunno if anyone can make use of that lol

kirbykirby · Answer

@TrojanPoem My induction was to show that $x^{x!}>(x!)^x$. And I started with the base case of x=3, which was shown to be true:
LHS: $3^{3!}=3^6=729$
LHS: $(3!)^3=6^3=216$, 

and  $729 > 216$

The case you are showing seems like you were doing it for x=2, but $2!=2\times 1=2$, and not 4. 

I have shown above that the formula is true for x=2 (and also x=1). But for x=3 and above, I used induction to show that the left side term is always larger than the right side term, meaning that they are never equal to each other. 

For using logs, maybe you can do as follows:
$$\begin{align} x^{x!}&=(x!)^x \
\log\left(x^{x!} \right)&=\log\left( (x!)^x\right) \
x!\log(x)&=x\log(x!)\
\prod_{i=1}^xi\cdot\log(x)&=x\log\left( \prod_{i=1}^xi\right)\
\prod_{i=1}^xi \cdot \log(x)&=x\sum_{i=1}^x \log(i)\end{align}$$

Now hopefully you can check that for x=1 and x=2, the equality holds. But look what happens for x=3:
$$ \begin{align}\prod_{i=1}^3i \cdot \log(3)&=3\sum_{i=1}^3 \log(i)\
\prod_{i=1}^2i\cdot3\cdot \log(3)&=3\left( \sum_{i=1}^2\log(i)+\log(3)\right) \
3\log(3)+3\log(3)&=3\log(2)+3\log(3), ~~~~~~~\text{ since} \log(1)=0 \end{align}$$
Clearly LHS > RHS

Now with x=4:
$$\begin{align}\prod_{i=1}^4i \cdot \log(4)&=4\sum_{i=1}^4 \log(i)\
\prod_{i=1}^3i\cdot 4\cdot \log(4)&=4\left(\sum_{i=1}^3\log(i)+\log(4)  \right)\
4\log(4)+4\log(4)+16\log(4)&= 4\log(4)+4\log(3)+4\log(2)\end{align}$$
using the fact that $\prod_{i=1}^3i=3!=6$, and this multiplied by 4 means you'll have a total of 24 "log(4)" 's 
Again, clearly LHS > RHS

For x=5, you should get something like:
$$5\log(5)+5\log(5)+5\log(5)+105\log(5)=5\log(5)+5\log(4)+5\log(3)+5\log(2) $$

You'll always be able to decompose the left and right-hand sides like this for higher values of x,  meaning that the LHS is always larger than the RHS. 

It seems like this boils down to the same type of induction I originally did above, but here a logarithm was used and I think this makes it more digestible to look at because terms are easily compared :)