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

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#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
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16 years ago
#2
and m and n are different?
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#3
yes m and n are unequal
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16 years ago
#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
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#5
the question asks for finding all the pairs.. ur answer cannot be right that

m = n

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16 years ago
#6
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16 years ago
#7
(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
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16 years ago
#8
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!
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16 years ago
#9
m and n do not have to be equal, e.g. m=4, n=2.
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16 years ago
#10
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.
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#11
so if u face this question in an exam, any kinda of exam, wot r u supposed to write if it cant be solved?
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16 years ago
#12
(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.
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16 years ago
#13
how can you prove that it can't be solved?
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16 years ago
#14
I have a feeling it's very hard to show that it can't be solved using exact methods.
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16 years ago
#15
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/
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16 years ago
#16
Which bits don't you follow?

Transcendental means that the number is not the solution to any polynomial with integer coefficients.
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16 years ago
#17
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?
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16 years ago
#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?
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.
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16 years ago
#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?
e is a constant. its a bit like pi, in that it is an irrational trancendental fundamental constnt of nature.
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16 years ago
#20
(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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