Transforming the exponential function of a sum into the sum of functions

Question

Is there a way to transform the function $$\exp(A+B+C),$$ where $\exp(\cdot)$ is the exponential function, into a sum $$f(A)+f(B)+f(C)?$$

Answer 1

If I am understanding correctly you would like to know if there is a suitable function $f$ such that $e^{a+b+c}=f(a)+f(b)+f(c)$ for all values of $a,b,c$.

Notice such a function would satisfy $f(0)+f(0)+f(0)=e^0=1$. So $f(0)=\frac{1}{3}$.

From here we can determine the function uniquely, since we must have $e^x=f(x)+f(0)+f(0)=f(x)+\frac{2}{3}$.

So the function $f$ needs to be $f(x)=e^x-\frac{2}{3}$

We can see easily this doesn't work since if it did we would have $e^3=f(1)+f(1)+f(1)=3(e-\frac{1}{3})=3e-1$ which clearly is not true.

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Shorter solution for generalized version which proves there are no three functions $f,g,h$ so that $f(a)+g(b)+h(c)=e^x: $

Note $f(x)=e^x-g(0)-h(0)$. We can do the same to the rest and get all the functions are $e^x$ minus a constant.This isn't possible since then $f(x)+g(x)+h(x)=3e^x-c$ which is a lot smaller than $e^{3x}$ for large values of $x$.

Answer 2

No. It is multiplicative, not additive. $$ \exp(A + B + C) = \exp(A) \exp(B) \exp(C) $$

Counter example: $$ \exp(x+x+x) = f(x) + f(x) + f(x) \iff f(x) = 1/3 \exp(3x) \\ e = \exp(1 + 0 + 0) =^! 1/3 \exp(3) + 2/3 \exp(0) > 7 $$