Help me solving this maths problem?

Question

R = a^b = b^a
If R is any positive real number. How would you found how many a b pairs are there. And what are they

Like 4^2 = 2^4 = 16
and 2.7454^2.7454 =16

For example i found 2 pairs for 16.
Are there any way to find the pairs for any real number (like 16)

Answer 1

The guys who answered the problem previously had the right idea:

If a^b = b^a, then, taking the log base a of both sides:

loga a^b = loga b^a
b loga a = a loga b
b/a = loga b

(where loga is the log to the base a)

So b and a have the property that, the ratio of one to the other is equal to the power to which you must raise one in order to get the other.

2 and 4 work because 4/2 = log2 4 (and 2/4 = log4 2).

If a = b, it works trivially, so the answer to your question is definitely that there are an infinite number of pairs of the form (a, a) that work.

But are there others besides (2, 4) and (4, 2)? That's the interesting question. Thinking about it....

The only way I could think of to figure out if (2, 4) and (4, 2) are the only other answers is to use a tool to graph

f(x, y) = logx y

And:

f(x, y) = y/x

and see where the functions intersect. I don't have mathematica or maple or matlab though. Going to play with Excel and see how that goes...

OK, I played around with it. I didn't get a good graph, but I did get something that should show you where the solutions lie:

http://www.utdallas.edu/~jimburnell/answer.JPG

this shows the ratio of x log y to y log x. The tan areas are where y log x is much bigger than x log y. The blue areas are where x log y is much bigger than y log x. The green areas are where the ratio is near 1, indicating that they are almost even.

You can see that there's a diagonal line, which represents all the pairs of the form (a, a).

There's also a hyperbolic-ish curve, as you can see. I don't know how to find values on the curve, but I found some points that come really close to working:

19.575 and 1.2, 12.525 and 1.3, 9.9 and 1.375,17.15 and 1.225, 15.3 and 1.25, 1.325 and 11.5, 2.125 and 3.65, 43.55 and 1.1.

If someone can come up with an equation to find these points, I'd love to know what it is.

Answer 2

Do you need any conditions that they be distinct?

Because if a = b, there would be infinitely many. Any real number r where a = r and b = r.

If you wanted to know how many distinct pairings there were, that's another story that I cannot help you with, as it is too advanced for me.

Answer 3

If a=b, then a^b = b^a.
Otherwise, only 2 and 4 work.

Answer 4

it is infinite because, say

(4)(4)=16

(sqrt(4)sqrt(4)
sqrt(4)sqrt(4))=16=2^4

sqrt(2)sqrt(2)sqrt(2)sqrt(2)
sqrt(2)sqrt(2)sqrt(2)sqrt(2)
=16

=sqrt(2)^8=16

likewise we sqrt this and double exponent

[sqrt(sqrt(2))]^16=16, and then you can square root this and make the exponent double

{sqrt[sqrt(sqrt(2))]}^32=16,

see the pattern thre are infinite pairs of numerals, not whole numbers that can be make to equal 16 with repeating multiples

sqrt() means square root()

Answer 5

i think that a and b have to be equal to each other for this to be true. ex. 2^2=2^2
or 3^3=3^3