How do I tackle this question? p^q = to q^p THANKS?

Answer 1

Hi,

2^4 = 4^2

I hope that helps!! :-)

Answer 2

are r,q reals, natural or integer numbers?
Solve over which set?
Important: Suppose the set is nonzero natural numbers.

You do as suggested.
q ln p = p ln q

q/(ln q) = p /(ln p)

Let f(x) = x/ (ln x), x>0, x different than 1.
f'(x) = (ln(x) - 1)/(ln(x)^2)
On interval x>e, the function f is increasing and decreasing on (0,1) and (1,e)
So e is a minimum point.
Also on (0,1) the function f is negative.

Therefore f(x_1)=f(x_2), x_1 different than x_2 can be attained only for values
1e
The only natural number in (1,e) is 2

Comments:
1)When x = 2 , we have 2^q = q^2. We see that q= 4 is a solution and the only one since on interval (e, infinity) , f is increasing .
2) We still have to check p=1 since we didn't define f in 1.
If p= 1 then 1= 1^q=q^1=q, so q=1.
3) if p = 0 then 0=0^q=q^0=1, contradiction
4)Obviously, in general the equality works for p=q. 0^0 is not defined so p different than 0.
5) The above are solutions for natural numbers( that is positive integer numbers) but for positive real numbers, we have an infinity of solutions because the limit of f when x approaches 1 from right is infinity and the limit of f when x approaches infinity is infinity.



Conclusion: over natural numbers the solutions are p=2, q=4;
p=4, q=2 and all pairs (p,p), that is p=q, p non-zero
Over positive real numbers, we have an infinity of solutions of pairs (p,q), where p is in (1,e) and q is in (e,infinity)

Answer 3

One solution would be if p and q are the same number.

There are other solutions too, such as 2 and 4. 2^4=4^2

also:
1.5 and 7.409
2.5 and 2.970
3.0 and 2.478
3.5 and 2.190
4.5 and 1.865
5.0 and 1.765

For every value of p>1 except e (=2.718281828459) there are two possible values for q. One of them is p itself.

Answer 4

What are you sovling for? What do you intend to make the subject?

I would start with taking logs

Use natural logs to make things easier

q ln p = p ln q

and continue from there......