Solve the equation



in (a) the set of naturals.

(b) the set of rationals for .

i know the solution in natural numbers but not in reals,

lets x=(2^p1)(3^p2).........{prime factorization}.......(1)

 

and y=(2^q1)(3^q2)...........     where pi's and qi's are whole numbers.......(2)

x^y=[2^(p1*y)][3^(p2*y)]..........

 

y^x=[2^(q1*x)][3^(q2*x)].........

since 2,3,5......are prime factors,we can happily equate their powers ans simplify as follows:

p1(y)=q1(x)    ;    p2(y)=q2(X)   ;     pi(y)=qi(x)(general)

therefore, 

now,x=ky,and from (1),(2),X=Y^K

obviously,k=1 satisfies with x=y

we observe if k=2,we get x=4,y=2 and x=-4 and y=-2

for x(natural number) =3,we observe that  x=3y and x=y³

which means every prime factor in y repeats itself thrice in x with one extra 3 which is not possible.

therefore k>3,its not possible by applying the above arguement.

thats all i know,hope you find it useful

solving in naturals:

not considering x=y as it is trivial.

It can be easily seen that x & y must have the same prime factors.

let p be one such prime.

let x=p^ak  & y=p^bk'  g.c.d(p,k)=(p,k')=1

since x^y=y^x,we have  p^{a(p^b)}=p^{b(p^a)}.

equating the exponents,we have ap^b=bp^a.

without loss of generality,let us assume a>b.

a/b=p^a-b.

let a/b=2+j;(j>=0)

2+j=p^b+j.

as p>=2,we can safely conclude that 2+j>b+j;forcing b to be 1.

so,we get p^1+j  -(1+j)=1.

drawing the graph of pⁿ  &  n ,

we can easily see that pⁿ-n is an increasing function in +ve integers

so, there is a unique solution p=2 & j=0.

so we get a=2,b=1.

this prorerty of p must be shown by all prime divisors of x& y.

but we have proved that only 2 satisfies the conditions.

hence 2 must be the only prime divisor of x & y.

so we get the unique soln in naturals:(x=4,y=2) & (x=2 ,y=4).

What about rational solutions?

let x=a/b, (gcd (a,b)=1)

y=c/d,gcd(c,d)=1.(a ,c &/or b,d not equal){a,b,c,d are naturals}

(a/b)^c/d=(c/d)^a/b

or,a^bc.d^ad=b^bc.c^ad

due to this equality,we conclude,

prime factorization of a must be same as c

same for b & d.

now,I assert a^bc=c^ad.

let it not be so.let a^bc=k.c^ad.

that means the prime factors of k is a subset of the set of

 primefactors of a & c.

but it forces that b^bc=k.d^ad

which means,prime factors of k is a subset of the set of

primefactors of b & d.but it is not possible.so k=1.

so,a^bc=c^ad.........................(1)

& b^bc=d^ad.........................(2)

from 1,we get a=c^ad/bc & from 2,d=b^bc/ad.

but a & d are naturals. so ad/bc  & bc/ad must be integers too.

(integer to the power non-integer cant be integer.)

that means  i)a=c & b=d,not acceptable.

ii)a=c=1 .this forces b^b=d^d.=>b=d^d/b.if d/b>1,we get b>d,absurd.

so this forces b=d.again not acceptable.

iii)b=d=1.=>x & y are naturals,

we get the unique soln. x=4,y=2 or viceversa.

TO BE EDITED.......IGNORE THE POST

 

'(integer to the power non-integer cant be integer.)'???

WHERE FROM U GET THIS THEORY?????

4 = 16^1/2

2= 16^1/4.

ACCORDING TO U 1/2, 1/4 ARE INTEGERS????

nicely pointed out Hamba,that was indeed careless of me.

i had only primes in my mind when i used that statement.......

I thank you.

I will edit my post as soon i have solved the prob. correctly.

Re:Solving an equation....

jishnu is perfectly ok.

only one addition: b can be any integer other than 0 and 1.

Just my tuppence worth on this discussion.

While considering the question whether there are naturals a and b such  and ,

 it may help to rewrite it as

It is well known that the the sequence  behaves as

i.e. it monotonically increases up to n = 3 and then monotonically decreases. We know that a₂ = a₄ and further a_i>1 for all i>1. Hence a=2 and b = 4 is the only ordered solution set.

Moving to rationals,  monotonically increases up to x = e and is monotonically decreasing for x>e. Which means infinitely many rationals in the interval (1,e) get mapped to rationals in the interval . So that means infinitely many solutions exist among rationals and it looks quite dense to be covered by a closed form.

A very nice observation Bhatt sir.