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[Witherweaver]

Or you could, you know, think about how you define the function x -> log(x).  Using the non-natural base is silly.

log(x) := (x -> 2^x)^(-1) ?

Seems a bit random to me.  Why didn't you use pi?

actually I thought on using 5...

edit: And seriously, maybe because 2^x is the exponent that naturally appears in your field...
edit2: And what I actually want to hint on is: "How you define log" is not really a reason for one variant, as there are thousands of ways to define log, what's probably more important is "why you define log the way you do", but both "how" and "why" probably depends on context.

So my point is there is a natural way to do things, based on nature independent of us.  It's simply the way of the world.  e isn't a special number to mathematicians, it's a special number to existence.

[Witherweaver]

Or you could, you know, think about how you define the function x -> log(x).  Using the non-natural base is silly.

A function that transforms products into sums?

(i.e, f such as f(a*b) = f(a) + f(b) )

How do you actually show the only such function is a logarithm?

You just defined that as the defining property of the logarithm, how can it not be one?

More importantly is probably the question of existence...

Exsitence is obvious.  2^{x+y} = 2^x 2^y.  So g:x->2^x satisfies g(x+y) = g(x)g(y).  The inverse function then satisfies f(xy) = f(x)+f(y).

Uniqueness is not true.  There is more than one f. 

The question is whether it's a one-dimensional family of functions.  I.e., if you have a function f_0 with f_0(xy) = f_0(x) + f_0(y), is it true that every such f is of the form f = \lambda f_0 for \lambda > 0?

[heron]

Or you could, you know, think about how you define the function x -> log(x).  Using the non-natural base is silly.

A function that transforms products into sums?

(i.e, f such as f(a*b) = f(a) + f(b) )

How do you actually show the only such function is a logarithm?

You just defined that as the defining property of the logarithm, how can it not be one?

More importantly is probably the question of existence...

Exsitence is obvious.  2^{x+y} = 2^x 2^y.  So g:x->2^x satisfies g(x+y) = g(x)g(y).  The inverse function then satisfies f(xy) = f(x)+f(y).

Uniqueness is not true.  There is more than one f. 

The question is whether it's a one-dimensional family of functions.  I.e., if you have a function f_0 with f_0(xy) = f_0(x) + f_0(y), is it true that every such f is of the form f = \lambda f_0 for \lambda > 0?

No, there are other solutions which, similarly to a certain rating system, are too complicated to be put into a formula.
See:

Let g(log(x)) = f(x)
Then we have g(log(xy)) = g(log(x)) + g(log(y))
g(log(x) + log(y)) = g(log(x)) + g(log(y))

The range of the log function is all reals, so we have
g(a + b) = g(a) + g(b) for all reals a and b.
So g satisfies Cauchy's functional equation, which is known to have other solutions than g(x) = cx.
So therefore f(x) has other solutions than f(x) = clog(x).

I guess this requires some assumption that the domain of f is only positive reals. I don't know if there is any solution if the domain of f is all reals.

[SirPeebles]

Or you could, you know, think about how you define the function x -> log(x).  Using the non-natural base is silly.

A function that transforms products into sums?

(i.e, f such as f(a*b) = f(a) + f(b) )

How do you actually show the only such function is a logarithm?

You just defined that as the defining property of the logarithm, how can it not be one?

More importantly is probably the question of existence...

Existence is easy, just map everything to 0.

[DStu]

Or you could, you know, think about how you define the function x -> log(x).  Using the non-natural base is silly.

log(x) := (x -> 2^x)^(-1) ?

Seems a bit random to me.  Why didn't you use pi?

actually I thought on using 5...

edit: And seriously, maybe because 2^x is the exponent that naturally appears in your field...
edit2: And what I actually want to hint on is: "How you define log" is not really a reason for one variant, as there are thousands of ways to define log, what's probably more important is "why you define log the way you do", but both "how" and "why" probably depends on context.

So my point is there is a natural way to do things, based on nature independent of us.  It's simply the way of the world.  e isn't a special number to mathematicians, it's a special number to existence.

All true, but still people define log in diffferent ways for different reasons, and the question is what does "log" mean. And not what should it mean...

[Witherweaver]

Or you could, you know, think about how you define the function x -> log(x).  Using the non-natural base is silly.

A function that transforms products into sums?

(i.e, f such as f(a*b) = f(a) + f(b) )

How do you actually show the only such function is a logarithm?

You just defined that as the defining property of the logarithm, how can it not be one?

More importantly is probably the question of existence...

Exsitence is obvious.  2^{x+y} = 2^x 2^y.  So g:x->2^x satisfies g(x+y) = g(x)g(y).  The inverse function then satisfies f(xy) = f(x)+f(y).

Uniqueness is not true.  There is more than one f. 

The question is whether it's a one-dimensional family of functions.  I.e., if you have a function f_0 with f_0(xy) = f_0(x) + f_0(y), is it true that every such f is of the form f = \lambda f_0 for \lambda > 0?

No, there are other solutions which, similarly to a certain rating system, are too complicated to be put into a formula.
See:

Let g(log(x)) = f(x)
Then we have g(log(xy)) = g(log(x)) + g(log(y))
g(log(x) + log(y)) = g(log(x)) + g(log(y))

The range of the log function is all reals, so we have
g(a + b) = g(a) + g(b) for all reals a and b.
So g satisfies Cauchy's functional equation, which is known to have other solutions than g(x) = cx.
So therefore f(x) has other solutions than f(x) = clog(x).

I guess this requires some assumption that the domain of f is only positive reals. I don't know if there is any solution if the domain of f is all reals.

That's interesting, thanks.

[SirPeebles]

There's a really nice result stating that

2 cos(x) = b^ix + b^-ix

The value of b depends on the units you use to measure angles.

If you want b = 10, then a full circle is 2*pi*ln(10)

If you want b = 2, then a full circle is 2*pi*ln(2)

If you want a full circle to be 360, then b = e^360/(2*pi)

It is that special base b = e which gives you that a full circle is tau, i.e. radians.

[Davio]

I like log scales: http://xkcd.com/482/

[pacovf]

Or you could, you know, think about how you define the function x -> log(x).  Using the non-natural base is silly.

A function that transforms products into sums?

(i.e, f such as f(a*b) = f(a) + f(b) )

How do you actually show the only such function is a logarithm?

You just defined that as the defining property of the logarithm, how can it not be one?

More importantly is probably the question of existence...

Exsitence is obvious.  2^{x+y} = 2^x 2^y.  So g:x->2^x satisfies g(x+y) = g(x)g(y).  The inverse function then satisfies f(xy) = f(x)+f(y).

Uniqueness is not true.  There is more than one f. 

The question is whether it's a one-dimensional family of functions.  I.e., if you have a function f_0 with f_0(xy) = f_0(x) + f_0(y), is it true that every such f is of the form f = \lambda f_0 for \lambda > 0?

No, there are other solutions which, similarly to a certain rating system, are too complicated to be put into a formula.
See:

Let g(log(x)) = f(x)
Then we have g(log(xy)) = g(log(x)) + g(log(y))
g(log(x) + log(y)) = g(log(x)) + g(log(y))

The range of the log function is all reals, so we have
g(a + b) = g(a) + g(b) for all reals a and b.
So g satisfies Cauchy's functional equation, which is known to have other solutions than g(x) = cx.
So therefore f(x) has other solutions than f(x) = clog(x).

I guess this requires some assumption that the domain of f is only positive reals. I don't know if there is any solution if the domain of f is all reals.

That's interesting, thanks.

That argument doesn't work for continuous functions, no?

[qmech]

That argument doesn't work for continuous functions, no?

That's right, g(a+b)=g(a)+g(b) forces g(x) to be linear on the rationals, so continuity forces it to be linear on R.

[SirPeebles]

That argument doesn't work for continuous functions, no?

That's right, g(a+b)=g(a)+g(b) forces g(x) to be linear on the rationals, so continuity forces it to be linear on R.

Ok, but it still doesn't pin down g(i).

[pacovf]

That argument doesn't work for continuous functions, no?

That's right, g(a+b)=g(a)+g(b) forces g(x) to be linear on the rationals, so continuity forces it to be linear on R.

Ok, but it still doesn't pin down g(i).

Is there a reasonable extra condition that would force g to be proportional to log?

[SirPeebles]

That argument doesn't work for continuous functions, no?

That's right, g(a+b)=g(a)+g(b) forces g(x) to be linear on the rationals, so continuity forces it to be linear on R.

Ok, but it still doesn't pin down g(i).

Is there a reasonable extra condition that would force g to be proportional to log?

Analyticity.

[pacovf]

Analyticity.

So I was thinking about this. I would make a pretty lousy mathematician, and analysis is not precisely my strongest suit, but please humour my attempt.

We have f(a*b) = f(a) + f(b), with f analytical. We want to prove that f is a logarithm.

The idea is to prove that if f and g are two solutions to the problem that satisfy f(a) = g(a) for some a =/= 1, then f = g. Then, since we can always find k such as k*log(a) = f(a), we've proved what we were set to prove.

So let's take such f and g. f-g still satisfies the conditions of the problem, and f-g(a) = 0. Furthermore, for any n in Z, we have f-g(aⁿ) = n*(f-g)(a) = 0, by induction since f(a*aⁿ) = f(a) + f(aⁿ), and f(1/a) = -f(a), since f(1) = 0.

Here's the part were my lousiness becomes obvious: we've got an analytical function (f-g) null at 1 such that, for any interval around 1, we can find x =/=1 in the interval such that (f-g)(x) = 0. This is because aⁿ goes to 1 when n tends to plus infinity (if a<1)  or minus infinity (if a>1). I would think that this forces the coefficients in the Taylor expansion of f-g to be all equal to zero, but I do not know how to prove it rigorously. If it is indeed true, then QED.

As such, we would have a definition of log that does not involve e in any obvious way!

(of course, you could define its inverse function, notice that the derivative of that function is proportional to itself, and find the specific solution to this problem such that this constant of proportionality happens to be 1, but I would consider this to be far removed enough to say that e does not necessarily fall over the definition of logarithm)

[ehunt]

Yes, log is the only analytic function from positive reals to reals taking times to plus (up to multiplying by a constant i.e. changing the base). In fact it is the only continuous such function. There are infinitely many non continuous functions with this property, but they are non-constructible (i.e. you need the axiom of choice.)

[DStu]

(i.e. you need the axiom of choice.)

or some other axiom, like the axiom of the existence of infinitely many non continuous functions that satisfy f(a*b) = f(a) + f(b).

also cf. xkcd://703

[Zappie]

for me: log means stratigraphic column