This has an important application in statistical physics. The entropy S(W) of a system of particles depends on the number of accessible configurations W of the system. Any thermodynamical potential must be additive so the total entropy S12 of two uncorrelated systems 1 and 2 with W1, respectively W2, configurations must be the sum S1+S2. Since the number of configurations of the combined system equal W1*W2 it follows that S(W1*W2)=S(W1)+S(W). From thus it follows that S(W) = k ln W with k a constant, which is the Boltzmann constant.

Fun problem that’s directly related to the motivation for Shannon’s entropy

That the result involves logarithms is clear by inspection. Filling in the details is the hard part.

2:58 You rather mean g: R -> R, i.e., the domain of g should be all of R, not just the positives.

e^x > 1 for x > 0 so the domain of g should be all reals to include (0, 1) in the domain of f.  This later fixes itself, in this case, when we compose e^ln(x) for x>0.

Me looking at initial question: "Isn't that just log?"

It is easy to show that f(x) = A(ln(x)) where A: R -> R is an additive function. That is, A satisfies A(x+y) = A(x) + A(y). It can be shown that there are lots of Lebesgue non-measurable additive functions whose graph {(x, A(x)) for x in R} is dense in R^2. but as soon as you put any mild condition such that the function is measurable in some range or continuous at a single point then A is linear.

I did this:
First we have f(x) = f(x.1) = f(x) + f(1); whence f(1) = 0
Then for x=/=1, we have  0 = f(1) = f(x.1/x)  = f(x) + f(1/x); whence f(1/x) = -f(x) Call this [*]
f: x |-> 0 is an obvious solution. Suppose, then, that f(x) =/= 0 for x=x*. By [*], we may assume f(x*) > 0
By a simple induction argument f(x*^n) = n.f(x*); whence 1 < f(x*^n) = n.f(x*) for some n [Archimedian prop of the Reals]
We also have f(1/x*) < 0 <1. Thus, by continuity of f, f(b) = 1 for some b st 1/x* < b < x*^n
Then m = f(b^m) = f((b^(m/n))^n) = n.f(b^(m/n)); whence f(b^(m/n)) = m/n
We're close now, stick with it.
We want to show f(b^y) = y, for general (not nec rational) y. We will then have f(x) = f(b^log x) = log x [the log here is the log to base b]. I.e., we will have shown that if f is not identically zero, the solutions are all of the form is f(x) = log x (log to some base b). Setting a = 1/In b  and noting that log x = in x / in b = a.in x, and allowing a to = 0. we recover the Prof's solution [OK, those last steps are a bit ragged]
To show f(b^y) = y, we let r(n) be a sequence of rationals converging to y (i.e., a seq n -> r(n) st lim r(n) = y) and use the continuity of f. Thus:
0 = f(1) = f(b^0) = f(b^(lim y - r(n)) = lim f[b^(y - r(n))] =  lim f[b^y.b^(-r(n))] = lim { f(b^y) + f[b^(-r(n))] } =
lim{ f(b^y) + f[1/b^r(n)] } = lim{ f(b^y) - f[(b^r(n)) } = f(b^y) - lim f[(b^r(n)) = f(b^y) - lim r(n) = f(b^y) - y, which is what we need

Great video, and no I don't think it's too hard, but it does take time to break down and understand everything.

And really interesting result. This seems like an extremely useful and interesting tool to have.

I think you need the domain of g to include negative in order to guarantee the range includes numbers in the interval (0,1)

WOW!!  I also loved Spivak Calculus and I still have my copy from high school. Just hearing that name again gave me so much nostalgia! Thank you muchly!

One fun little technique is to note that we can use the defining property to show that f(x^2)=2f(x), and repeating this process we get values for all natural powers. Then we can show, using the fact that f(xyz)=f(x)+f(yz)=f(x)+f(y)+f(z) that the same is true for all 1/n powers, and hence for all rationals. Then, continuity gives us all the reals for free. In essence we then have that f(1)=0, and all other values are determined by any other given value (because e^x is surjective on  positives).

In essence, all possible functions are scalings of one example, e.g the logarithm that we can trivially show works!

Let's go! Was recently craving some functional equations :D

first thing i saw was a group homomorphism and remembered a very special class of morphisms from the multiplicative group of the non-negative reals to the additive group of the reals

Nice video, but you should have mentioned around 3:18 that the equation you got is the famous Cauchy functional equation (and so in the remaining part of the video you're deriving again the fact that the only say continuous solutions to the Cauchy functional equation are the linear functions).

In order to solve this problem, we will first magically bring in the solution at 1.11.

(Or an inverse of the solution, as it happens here)

I like the yellow arrowed comments added in post. very cool.

The continuity get in the role very smoothly!!

Let's not forget that a=0 also works. I think that should be called out separately. It's a bit strange to call the zero function "some sort of logarithm." After all, it's also "some sort of polynomial," etc.

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