Find real values of a and b satisfying a^b=b^a

[josephcollins]

Hi ppl, could someone offer some help on the following problem:

Obtain real values for a and b which satisfy the equation

a^b=b^a

I took the logs of both sides to give

bloga=alogb

Is there any way of simplifying the equation from here?

Joe

 

[Zurtex]

Perhaps taking log base a would yield more useful results, I'm not sure, but a simple solution would obviously be a=b for $a,b \neq 0$ (I'm not sure there are actually anymore).

 

[Muzza]

There are others, for example, take a = 2. Then b = (approx.) -0.77, 2 and 4 all fulfil a^b = b^a.

 

[Dr.ThinkDeep]

bloga=alogb
Is there any way of simplifying the equation from here?

Divide by a*b and you have log(a)/a = log(b)/b.
Now lokk at that function f(x) = log(x)/x .
Let us restrict ourselves to positive x for the moment.

For x>0, this function is differentiable with derivative (1-log(x))/x^2
It has one extremum at x=e f(e)=1/e and one zero at x=1, the asymptote for x->infinity is the x-axis.
The extremum is a maximum, obviously as 1/e > 0.
So there are three ranges:
1. 0<x<1 with f(x)<0. These values f(x) do NOT occur at any other x.
2. 1<x<e with 0<f(x)<1/e These values do occur at one other value of x in
e<x<infinity.
3. e<x<infinity with 0<f(x)<1/e. These values do occur at one other value
of x in 1<x<e.

I do not have time now for the inverse function of log(x)/x to compute the other x that has the same value of f(x).
You may want to look at the branches of the Lambert W function -

LambertW(x) * exp(LambertW(x)) = x

I would peek into Abramovich and Stegun as a first guess.

 

[Tide]

All $a = b$ for which $a^b$ and $b^a$ are defined are solutions.

 

[Zurtex]

All $a = b$ for which $a^b$ and $b^a$ are defined are solutions.

As I said earlier a=b=0 is not a solution and the reason being as 0⁰ is undetermined.

 

[fourier jr]

what if a& b are integers? I think there was a putnam problem that said that, and the only ones that work are a=1 & b=2

 

[Tide]

Zurtex,

Sorry - I missed that on my first pass.

 

[uart]

I took the logs of both sides to give

bloga=alogb

Is there any way of simplifying the equation from here?

Yes there is. Place a space between the multiplier and the log and a pair of brackets around the log argument.

b log(a) = a log(b)

There, that's much simpler (to read).


Ok this reply was a joke, Dr Thinkdeep's reply gives a really good explanation IMHO. :)

 

[dextercioby]

what if a& b are integers? I think there was a putnam problem that said that, and the only ones that work are a=1 & b=2

a=1 requires b=1,and a=2 requires b=4 and viceversa.For the second one,it's only in the case of integer/whole numbers.

 

[jcsd]

what if a& b are integers? I think there was a putnam problem that said that, and the only ones that work are a=1 & b=2

*non-solution removed*, I guess that it and a=2, b =4 are the only 'non-trivial' solutions

 

[jcsd]

Oops yes your right why did I think that?

 

[jcsd]

Well I don't mind people knowing that I make very stupid mistakes, because I do frequently! But I've removed the offfedning part to avoid confusion. :)

 

[arildno]

To get back to the problem, we see that positive solutions may be written in another form:
Let $$b=a^{k}$$
Hence, solutions must obey the slightly different equation:
$$a^{k}=ka$$
Not that I know whether this is simpler to solve, though..

 

[dextercioby]

To get back to the problem, we see that positive solutions may be written in another form:
Let $$b=a^{k}$$
Hence, solutions must obey the slightly different equation:
$$a^{k}=ka$$
Not that I know whether this is simpler to solve, though..

Okay,so let me give my final version this problem:
Problem:
Find the positive solutions of the equation:
$$a^{b}=b^{a}$$

Attempt to solving it:
The fact that we search the positive (which means also different from 0) solutions means that we can logarithm (in any base actually,but let's pick the natural logarithm) the equation,obtaining:
$$b\ln{a}=a\ln{b}$$
.Since we search for nonzero numbers,we can divide by the product $ab$ to obtain
$$\frac{\ln{a}}{a}=\frac{\ln{b}}{b}$$
.Now pick an arbitrary "a".Compute the number in the LHS of the prior equation and call it "A".The initial problem is reduced to the one of finding all real "b-s" (if they exist) who verify the equation $Ab=\ln{b}$,where "A" is known.The ways to solving the last equation cannot be analytic,because the equation is transcendent.The best known way to solving these equations is graphically (i.e.intersecting the 2 graphics).Since both functions involved (natural logarithm and the linear function passing through (0,0)) are continuous and strictly ascending,the number of solutions (the possible number of "b-s" for "a" fixed) is either 0 or 1.I'm saying the graphics don't intersect more than once regardless od "A".Since "a" is arbitrary,we can say for sure that the number of solutions (pairs "a","b") is infinite (if $0<a<1$ it's obvius).
The solution reads:all pairs (a,b),where b is a solution to the equation "Ab=ln b",where A is given above.


If think it's not right,please,speak up!

Daniel.

 

[Gokul43201]

.Since both functions involved (natural logarithm and the linear function passing through (0,0)) are continuous and strictly ascending,the number of solutions (the possible number of "b-s" for "a" fixed) is either 0 or 1.I'm saying the graphics don't intersect more than once regardless od "A"

If think it's not right,please,speak up!

Daniel.

I think this is not right (it may be true in this case, but not for the reasons specified). How does continuity and monotonicity require no more than one intersection ?

y=x and y=x^2 are continuous and monotonically increasing for x>=0, yet intersect at x=0 and x=1.

 

[dextercioby]

I think this is not right (it may be true in this case, but not for the reasons specified). How does continuity and monotonicity require no more than one intersection ?

y=x and y=x^2 are continuous and monotonically increasing for x>=0, yet intersect at x=0 and x=1.

I don't mean to be mean,but apparently u didn't not see the mistake i made and u managed to find an exception to a rule that doesn't exist.
You should have argued that those 2 graphics are more likely to intersect twice or not at all,and only in one case Ax is tangent to the natural logaritm graphic (that would correspond to a single solution of the equation).I even now cannot imagine how i could make such an error... Nonetheless the reasons for continuity and strict monotonicity on all of the domains [0,+infinity) and (0,+infinity) are the reasons why the equation can't have more that 2 solutions.Actually strict monotonicity is the key.
So i think it should be all clear now.

 

[shmoe]

Nonetheless the reasons for continuity and strict monotonicity on all of the domains [0,+infinity) and (0,+infinity) are the reasons why the equation can't have more that 2 solutions.Actually strict monotonicity is the key.

Maybe I'm misunderstanding you, but are you saying that if you have two functions, say f(x) and g(x), that are continuous on [0, +infinity) and strictly increasing on (0,infinity) then there are at most two values of x on [0,+infinity) where f(x)=g(x) ? This statement is false in general-let f(x)=2x and g(x)=2x+sin(x). Both satisfy the requirements yet intersect infinitely often. So what are you trying to say again?

 

[dextercioby]

Maybe I'm misunderstanding you, but are you saying that if you have two functions, say f(x) and g(x), that are continuous on [0, +infinity) and strictly increasing on (0,infinity) then there are at most two values of x on [0,+infinity) where f(x)=g(x) ? This statement is false in general-let f(x)=2x and g(x)=2x+sin(x). Both satisfy the requirements yet intersect infinitely often. So what are you trying to say again?

1.You make the same mistake like Gokul.Give an exception to a nonexisting rule.
2.I used continuity to show that the two graphics cannot intersect more than twice.If A (the slope of x) is sufficiently small,then for arguments in the vecinity of 1 (yet not necessarilly,anyway,depending of the magnitude of A,of course) the 2 graphs sould intersect and for a reasonably long distance on the "x" axis,the values of "Ax" would be smaller than the values of "ln x".But as lim(x->+infinity) of lnx/x is zero,there must be another point of intersection.Why only one?Continuity ans strict monotony.For bigger and bigger arguments,the "ln x" flattens (its slope tends asymptotically to 0),while "Ax"increases at the same rate/speed,bacause its slope is constant.
If A is big enough not to get an intersection for reasonably small values of "x",then as,"x"increases,the chances of an intersaction would be nil.

I didn't claim mathematical rigurosity for my proof,because I'm still a physicist after all (though theorist) and it's natural for me to claim mathematical rigurosity where physics applies.

 

[shmoe]

1.You make the same mistake like Gokul.Give an exception to a nonexisting rule.

You stated that monotonicity and continuity were the reason why the graphs intersect no more than twice. My counter example was to show there is more to it than just these two properties. Heck, even in your latest explanation you mention that 'For bigger and bigger arguments,the "ln x" flattens', which is a statement about the derivative of log(x) being monotonic, not the function itself.

It's not difficult to show Ax and log(x) intersect at most twice. In between any two points of intersection, there is a point where the derivative of log(x) equals A by the mean value theorem. Since the derivative of log(x) is monotonic, we see 3 points of intersection are impossible.

 

[Gokul43201]

1.You make the same mistake like Gokul.Give an exception to a nonexisting rule.

You seem to forget that you were the one that came up with the rule. We are only providing counter-examples to show that this rule, is not, in general, true.

 

[arildno]

Let's resrict our attention to find positive solutions to this problem:
1) let $$b=a^{k}$$
2) In order for this to be a solution, we must have:
$$a^{k}=ka$$
3) Or "a" must fulfill:
$$a=k^{\frac{1}{k-1}},0<k\neq{1}$$
4) Hence, we have a solution set (apart from the trivial set a=b)
$$a=k^{\frac{1}{k-1}},b=k^{\frac{k}{k-1}},0<k\neq{1}$$
5) We therefore have an infinite number of non-trivial solutions, for example by letting k=3, we gain:
$$a=3^{\frac{1}{2}},b=3^{\frac{3}{2}}$$
6)Finally, we note that $$a(\frac{1}{k})=b(k)$$