- #1

MathematicalPhysicist

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one (and perhaps the only) example is:2^4=4^2

what more examples are there?

what more examples are there?

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- Thread starter MathematicalPhysicist
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- #1

MathematicalPhysicist

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one (and perhaps the only) example is:2^4=4^2

what more examples are there?

what more examples are there?

- #2

On Radioactive Waves

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Originally posted by loop quantum gravity

one (and perhaps the only) example is:2^4=4^2

what more examples are there?

2! does not = 4!!!!!

1^1=1^1

- #3

MathematicalPhysicist

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what?

- #4

StephenPrivitera

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I think he means to say that 4!=24[x=] 2

Your example 2^4=4^2 doesn't satisfy the condition m!=n (m=4, n=2, 4![x=]2)

If you solve the problem for m, you get m=(m!)^{1/(m-1)!}. I don't know how to find the solutions, but 1 works. Since only positive integers can be a solution, next try 2. You get sqrt2=2, so that won't work for m. I'm pretty sure nothing after that works either. For instance, m cannot be 4, because 4[x=]6thrt(24).

Your example 2^4=4^2 doesn't satisfy the condition m!=n (m=4, n=2, 4![x=]2)

If you solve the problem for m, you get m=(m!)

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- #5

MathematicalPhysicist

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ah, this is from c++ language which means does not equal to (m doesn't equal to n).Originally posted by StephenPrivitera

I think he means to say that 4!=24[x=] 2

Your example 2^4=4^2 doesn't satisfy the condition m!=n (m=4, n=2, 4![x=]2)

my mistake.

- #6

bogdan

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Use VB...!= is <>...hehehe...

- #7

bogdan

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n^(k*n)=(n*k)^n

(n^k)^n=(n*k)^n

n^k=n*k...

n^(k-1)=k....

if n=1...k=1...wrong...m=n...

if n=2...2^(k-1)=k...if k>2 then it's wrong...so k=2...m=4...

if n>2...n^(k-1)>2^(k-1)>k...(for k>1...)

So...n=2...m=4...it's the only solution...

- #8

STAii

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Your conclusion yeilds that there is no solution other than m=2, n=4.

Well, i have a second solution, which means that ur solution is wrong bogdan (or at least, not complete).

Here is the other solution :

m=-4, n=-2

You said :

Prove it.Well...it is quite simple to prove that m must be equal to k*n...where k is an integer...

I don't see why m=n*k, m and n can be any two numbers, like ... 1.5 and 2, those two are not related by an integer k.

- #9

bogdan

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Ooops...sorry...I thought m and n were integers...sorry again...stupid me...

- #10

On Radioactive Waves

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Originally posted by bogdan

n^(k*n)=(n*k)^n

(n^k)^n=(n*k)^n

n^k=n*k...

n^(k-1)=k....

if n=1...k=1...wrong...m=n...

if n=2...2^(k-1)=k...if k>2 then it's wrong...so k=2...m=4...

if n>2...n^(k-1)>2^(k-1)>k...(for k>1...)

So...n=2...m=4...it's the only solution...

Yes, I was to tired to find the [x= symol. 2![x=]4 and 4![x=]2, so how does this work? When you plug the numbers back in it dosn't work.Silly me, I was also thinkg of inregers. Back to learning!

And how is x! defined when it isn't postive or integer? [gamma](x)=x! ?

can someone explain this? and what the hell is superfactorials used for?

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- #11

STAii

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Well, even if they were integers, it is not always possible to express one integer as k*n.Ooops...sorry...I thought m and n were integers...sorry again...stupid me

Take n=3, m=2, you can NOT express m as k*n (nor n as k*m), can you ?

So even if m and n where integers, you idea still doesn't hold !

I personally do not support writting it in mathematical expressions (unless talking about C language of course).

- #12

There are actually and infinite number of solutions. Let me show how:

The given are as follows:

(1) a^{b} = b^{a}

(2) a != b

To this I would add the following because I don't know what raising a negative base to an irrational power means:

(3) a, b >= 0

From the original equation it follows that neither a nor b can equal zero:

assume a = 0

0^{b} = b^{0}

0 = 1 (contradiction, proved similar for b=0)

So we can add the following (which will help us divide by a and b):

(4) a,b != 0

Now for the actual proof:

assume a^{b} = b^{a}, then

ln( a^{b} ) = ln( b^{a} ) (since we know both sides are positive)

b ln(a) = a ln(b)

ln(a)/a = ln(b)/b

Note that we can reverse these implications:

assume ln(a)/a = ln(b)/b, then

b ln(a) = a ln(b)

ln(a^{b}) = ln(b^{a})

a^{b} = b^{a} ( ln(x) is one-to-one )

So if we can find an a and b for which ln(a)/a = ln(b)/b and a != b, then we have found a solution to the original problem. If we take a look at the graph of f(x) = ln(x)/x we can note that this continous function rises, hits a maximum, and then goes back down. This implies there exist an infinite number of values a != b for which f(a) = f(b) or ln(a)/a = ln(b)/b. As we already determined, this is sufficient to imply there exist an infinite number of solutions to the original problem.

In short, (a,b) = (2,4) is not the only solution as far as real numbers are considered.

edit: how can i get an image on these boards? it's easy to make one and it's easy to use the img tag, but i don't have a place to put an image on the net. i wanted to put up a visual of the graph.

The given are as follows:

(1) a

(2) a != b

To this I would add the following because I don't know what raising a negative base to an irrational power means:

(3) a, b >= 0

From the original equation it follows that neither a nor b can equal zero:

assume a = 0

0

0 = 1 (contradiction, proved similar for b=0)

So we can add the following (which will help us divide by a and b):

(4) a,b != 0

Now for the actual proof:

assume a

ln( a

b ln(a) = a ln(b)

ln(a)/a = ln(b)/b

Note that we can reverse these implications:

assume ln(a)/a = ln(b)/b, then

b ln(a) = a ln(b)

ln(a

a

So if we can find an a and b for which ln(a)/a = ln(b)/b and a != b, then we have found a solution to the original problem. If we take a look at the graph of f(x) = ln(x)/x we can note that this continous function rises, hits a maximum, and then goes back down. This implies there exist an infinite number of values a != b for which f(a) = f(b) or ln(a)/a = ln(b)/b. As we already determined, this is sufficient to imply there exist an infinite number of solutions to the original problem.

In short, (a,b) = (2,4) is not the only solution as far as real numbers are considered.

edit: how can i get an image on these boards? it's easy to make one and it's easy to use the img tag, but i don't have a place to put an image on the net. i wanted to put up a visual of the graph.

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- #13

MathematicalPhysicist

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so what other solutions are there?

- #14

Originally posted by loop quantum gravity

so what other solutions are there?

For those of you who don't care to read below, I put these approximate solutions first (generated by a program I built using a method described below). Because the program I built stores the numbers in a relatively small memory space the answers aren't as good as they could be.

(1.5, 7.40876)

(1.13, 32.0162)

(2.5, 2.97029)

(12.1, 1.30982)

(8.4343, 1.43866)

I neglected to mention any specific numbers earlier because I didn't know how to exactly solve ln(a)/a = ln(b)/b for either a or b. Remember, all we have to do to find a solution to a

assume a > 0 is given (we want to find b):

ln(a)/a = ln(b)/b

b ln(a)/a = ln(b)

b ln(a)/a - ln(b) = 0

Now from here we could simply go on the solve the equation numerically for b by applying Newton's Method. Unfortunantly, not all given values of a will produce a solution. In addition, Newton's Method requires an initial guess to get the ball rolling and if we guess inappropriately we might just end up with b = a (which we already know would solve the equation but we ruled out by a != b because it was too trivial).

In order to get a better idea of what's going on we will define a function f(x) which will equal the left-hand side of the equation that is zeroed above and analyze when this function has roots, and if it does, where they exist because these roots are the solutions we are looking for.

b ln(a)/a - ln(b) = 0; a,b > 0 (from above)

f(x) = x ln(a)/a - ln x, x > 0

To get an idea of what f(x) looks like, we plug in some values and analyze the derivative:

f'(x) = ln(a)/a - 1/x, x > 0

f'(x) = ( x ln a - a )/( ax ) (possible extrema at x = a/ln a, if a > 1, otherwise a/ln a would be negative and outside the domain of f(x))

f'(x < a/ln a) < 0

f'(x > a/ln a) > 0 (last two confirm x = a/ln a is an absolute minumum if a > 1)

Limit[x->0, f(x)] = +inf

Limit[x->+inf, f(x)] = +inf if a > 1

Limit[x->+inf, f(x)] = -inf if a < 1

These limits inform us that if an absolute minumum exists (a > 1) and is also negative, then f(x) will cut the x-axis twice and therefore have two roots. One of the roots would just be x=a as we mentioned earlier. The other would be the one we're looking for. On the other hand, the absolute minumum being zero would imply only one root at the trivial x=a (the abs. min can't be positive because the trivial root always exists). If a < 1 and there is no absolute minumun then the limits inform us only one root exists (which would be the trivial x=a). So let's assume a > 1 and find when the abs. min is negative.

Let us see what the absolute minumun is:

f(a/ln a) = (a/ln a)(ln(a)/a) - ln(a/ln a)

f(a/ln a) = 1 - ln(a/ln a)

When is this absolute minumun less than zero:

1 - ln(a/ln a) < 0

1 < ln(a/ln a)

e < a/ln a (e

The last is true whenever a > 1 and a != e.

Okay, so far we have established if we are given a > 1 and a != e, then f(x) will have two roots and one of these will be the solution for b that we are in search for. To actually find this root we will use Newton's method, but which root should we go after? The one to the left or the right of the minumum? One is a "dud" and the other is the actual solution. We know the dud is at x = a. This will be to the left of the minumum (at x = a/ln a) if ln a < 1 (or 1 < a < e) and to the right of the minumum if ln a > 1 (or 1 < e < a). So if a < e, our initial guess to Newton's method should be to the right of the minumum and vice-versa for a > e.

We have always talked about 'a' being given, but it since the original equation is "symmetrical", we could just as well have been talking as if 'b' were given. To be fair to both sides, well say the given is x. So in summary, given any number x > 1, x != e, we can find another number b for which x

if x < e, start with b

if x > e, start with b

b

- #15

MathematicalPhysicist

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do you have examples of whole numbers?

- #16

Originally posted by loop quantum gravity

do you have examples of whole numbers?

No, other than 2 and 4, they don't exist because one of the natural numbers would have to be between 1 and e (2.718). The only natural number that exists to satisfy this condition is 2.

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