[Ximo Tamarit]

How can I solve this equation?

a raised to 4.10667 = e raised to a.

e is number e. I know a is 9.0427 (only 4 decimals).

Is there a method?

Thanks

[ebaines]

I don't think there is a closed form solution for this. If you rearrange terms you get:


which may be a little easier to work. There are actually two solutions - in addition to the solution you mentioned there's also one that is approximately 1.4095

[galactus]

Without getting into advanced methods, you could use ol' Newton's method with an initial guess of 1.5

It will converge rather quickly to 1.41. Use an initial guess of 9 and you'll get the naswer you already know.

[Ximo Tamarit]

I don't think there is a closed form solution for this. If you rearrange terms you get:


which may be a little easier to work. There are actually two solutions - in addition to the solution you mentioned there's also one that is approximately 1.4095

How did you get there were two solutions? The correct solution must be higher than e (>e). The initial equation was a=4.10667·ln(a). I wonder if exists a method without graphics or guessing.

Any case, thanks.

[Ximo Tamarit]

Without getting into advanced methods, you could use ol' Newton's method with an initial guess of 1.5

It will converge rather quickly to 1.41. Use an initial guess of 9 and you'll get the naswer you already know.

How did you get there were two solutions? I'd like to know if there is a method without graphics or guessing. Which are those advanced methods you mentioned?
The only solution must be >e, the initial equation was a=4.10667·ln(a)
Thanks a lot

[galactus]

A 'more advanced method' I thought may work is the Lambert W function.

Look into it. There should be lots on the web. Wikipedia, for example.

[ebaines]

How did you get there were two solutions? I'd like to know if there is a method without graphics or guessing. Which are those advanced methods you mentioned?
The only solution must be >e, the initial equation was a=4.10667·ln(a)
Thanks a lot

I realized that there must be two solutions by thinking of how the graph of y=4.10667·ln(x) compares to the graph of y=x. There are three possibilities:

1. The lines don't cross and there are are no solutions. But you already showed that you had a solution, so this isn't the case.
2. There is just 1 solution, which would be the case if the value of the multiplier 4.10667 is precisely the right value to make the two lines just barely touch each other. This seems very unlikely, although admittedly possible.
3. There are two solutions. The most likely case. So I simply tried some values and found that a solution between 1 and 2 would work. Then it's a case of zeroing in on the value.

Why do you say that "the only solution must be >e"?

[Ximo Tamarit]

Why do you say that "the only solution must be >e"?

We could say that "e" is the middle. It is, e/ln(e)=1. Any number n>e divided by ln(n) will give us another number>1. Any number 1<n<e divided by its own ln(n) will give also another number>1. With integers, only 2 and 4 give the same result and both are in different parts from e, it is, e is more or less a mirror, the middle. But if I want to study integers I have to take the part that corresponds to numbers >e.

[galactus]

, not 1.

[Ximo Tamarit]

, not 1.

Sorry, efectively, e/ln(e)=e, what i wanted to say is that it is (with positive numbers) the minimum limit (sorry if my english is not good enough), the smallest possible value. In other words, it seems that n/ln(n) for positive numbers has e as a minimum. If you try with numbers between 1 and e you can get any result you want and, in addition you can get the same result looking for numbers >e; so, we can say there will be always 2 different solutions (with positive numbers), one <e and another >e. The question is, if i know that we have two different solutions for only one equation "a=n·ln(a) and we have the value of n, that can be the value we want... Is there a method to obtain the solutions without graphics or guessing? I'm going to look the Lambert W function, thanks a lot.

[galactus]

There probably is a method using something more advanced, but the best way is to graph it and then possibly use the Intermediate Value Thereom to home in on it.

[galactus]

Hello again:

Equations like this are not solvable by elementary means, like some integrals, etc.

So, as mathematicians can tend to do, we hammer it into some form.

It's been a while since I messed with the LambertW, but here goes. I suggest you investigate yourself further.



Log of both sides gives:



Now, let :



Divide through by :







There's the LambertW solution, I hope:D

Does that help a little?

[Ximo Tamarit]

Hello again:

There's the LambertW solution, I hope:D

Does that help a little?.

Then, we are at the same point that at the beginning, I mean, your answer is quite good for avoiding the problem but, at the end, we are in the same problem but in a different way.
If we have two solutions ("a" and "b" unique solutions) for an equation, we can write the equation as a polinomic expression:

x^2 - x·(a+b) + (a·b)=0. In this case we would have two different equations with the same solutions. For instance:

m=2.88539008·ln(m) has two solutions = 2 and 4
x^2- x(6) + 8=0 has two solutions = 2 and 4

One of these solutions is always >1 and <e, and the other is always >e, but I cannot go further.
Anyway, thanks a lot.

[ebaines]

m=2.88539008·ln(m) has two solutions = 2 and 4
x^2- x(6) + 8=0 has two solutions = 2 and 4

One of these solutions is always >1 and <e, and the other is always >e, but I cannot go further.
Anyway, thanks a lot.

Not too difficult to show that your supposition about one of the solutions is always >e and one is always less is correct. Looking at the function


The first derivative is:

This shows that there is a max at the value x = m. Therefore the two solutions must have one value <m and the other >m.

Now consider the allowable range for m. The minimum allowable value for m is e; otherwise x/m is always greater than ln(x), and there are no solutions. If m = e, there is only one solution for f(x) = ln(x) - x/e, and that is x=e. For m>e there are always two solutions. From the argument above one of these must be >m, hence one of your solutions must always be > e. The other solution is <m, but must it always be <e? Turns out yes, it must, because as you change the value of m, looking at greater and greater values, the lower solution approaches 1. In other words, if is the lower solution, then



Hence the lower solution has an upper value of e and a lower value of 1.

I think this resolves the allowable range of values for the two solutions.