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# ok then do this one... [maths] watch

1. This is one of the hardest questions I have come across in my life, im sure there is a trick to do it but dont know wot.

If m and n are connected by:

m^n = n^m

find all pairs of m and n which satisfy this relationship
2. and m and n are different?
3. yes m and n are unequal
4. (Original post by integral_neo)
This is one of the hardest questions I have come across in my life, im sure there is a trick to do it but dont know wot.

If m and n are connected by:

m^n = n^m

find all pairs of m and n which satisfy this relationship
logm (n) = logn (m)

by changing the base:

logm (n) = logm (n) logn (m)

=> logn (m) = 1

=> m == n
5. the question asks for finding all the pairs.. ur answer cannot be right that

m = n

6. (Original post by integral_neo)
the question asks for finding all the pairs.. ur answer cannot be right that

m = n

sorry, i did my logs wrong, you have to have the same base on each side
7. I did it without logs:

m^n = n^m

(m^n)/(m^n) = (n^m)/(m^n)

(n^m)/(m^n) = 1

n^(m-n) = 1 = n^0

m-n = 0

m=n

so m and n can be anything as long as they are equal.
wow I didn't think I'd get that one!
8. m and n do not have to be equal, e.g. m=4, n=2.
9. ok... maybe what i proved was that if they are equal it must work

but now that I think about it that goes without saying.
10. so if u face this question in an exam, any kinda of exam, wot r u supposed to write if it cant be solved?
11. (Original post by integral_neo)
so if u face this question in an exam, any kinda of exam, wot r u supposed to write if it cant be solved?
it would probably come up in a STEP paper, and so it wouldnt be important that you couldnt solve it as long as you attempted to solve it and showed that it couldnt be solved trivially.
12. how can you prove that it can't be solved?
13. I have a feeling it's very hard to show that it can't be solved using exact methods.
14. this is wot 'theone' posted be4
Most equations of this form, i.e. of x^y = y^x are unsolvable directly.

However you can deduce various facts about such equations.

Now consider the function f(t) = ln(t) / t. If you sketch the graph and find it's maximum you see that if x < e, then there are two values of t for which f(t) = x. If x = e, there is one value, if x > e there are no values.

But x^y = y^x is equivalent to y/lny = x/lnx. so lny/y = lnx /x . Now we know lnx/ x < 1/e, so for each value of x there are two values of y which give lny/y = lnx/x, except when x = e, when there is one solution.

This also tells us about the range of the solutions, when there are two, one will always be in the range [0,e] and one in [e,+inf]

i would like to say i follow but it is tricky...how can u do this?
and incidently, wot does 'transcedental' mean/
15. Which bits don't you follow?

Transcendental means that the number is not the solution to any polynomial with integer coefficients.
16. I can't wait to learn about Log and In... I don't have a clue what they mean but they sound pretty important.

edit: and "e" as well. what's e? is it a constant?
17. (Original post by mik1a)
I can't wait to learn about Log and In... I don't have a clue what they mean but they sound pretty important.

edit: and "e" as well. what's e? is it a constant?
It's Ln (or ln) not in

e can be defined in various ways, all equivalent. The most common ones are the constant such that e^x has gradient e^x or the limit of ((n+1)/n)^n as n -> infinity.
18. (Original post by mik1a)
I can't wait to learn about Log and In... I don't have a clue what they mean but they sound pretty important.

edit: and "e" as well. what's e? is it a constant?
e is a constant. its a bit like pi, in that it is an irrational trancendental fundamental constnt of nature.
19. (Original post by mik1a)
I can't wait to learn about Log and In... I don't have a clue what they mean but they sound pretty important.

edit: and "e" as well. what's e? is it a constant?
do u not do physics?

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