# exp(ab) decomposition

How can one write $e^{a(x) \cdot b(x)} = c(x) e^{ b(x) }$ with $c(x)$ not implicitly depending on $b(x)$. I do not believe this is generally possible so alternatively one can use an infinite series or infinite product $$\sum c_k x^k e^{ d_k \cdot b(x)}$$ or $$\prod e^{c_k \cdot x^k} e^{d_k \cdot b(x)}$$

Maybe there is another way using the Lambert $W$ function or some other special function. I'm looking for a quasi power to product formula for exp. e.g., $e^{a \cdot b} = e^{a^b} = Q(a) \cdot e^b$.

Obviously one can write $e^{a b} = e^{(a - 1)b} e^b$ but I'm looking to remove the direct dependence on one factor. Here I'm dealing with functions and the requirements are slightly relaxed. I.e., take the infinite sum and product cases as the main interpretation and not the examples involving constants... which do not have solutions in general.

for example, using the power series exapansion of $a(x)$ we end up with

$$e^{b(x)\sum a_k x^k} = \prod e^{a_k x^k b(x)}$$

but I need to get this to look something like

$$\prod Q(x) e^{a_k b(x)}$$

without $Q(x)$ depending implicitly on $b(x)$ (else I would just go for simpler case $e^{(a(x) - 1) b(x)} e^{b(x)}$.

(the fact that we still allow for dependence between the factors through $a_k$ makes me think that such an expansion is possible but maybe this is wishful thinking...)

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@AndréNicolas Did you not even bother to read the post? I said towards the end THAT IS NOT what I want! – AbstractDissonance Sep 28 '12 at 12:11
@Sorry, missed it. The beginning of the question said you thought it was generally not possible. Since that is false, I stopped reading. – André Nicolas Sep 28 '12 at 13:30
@AndréNicolas heh, sorry too, I guess in my revision I deleted the part where I wanted c(x) to be independent of b(x) in some way. I'll update it to make it clear. – AbstractDissonance Sep 28 '12 at 14:23
You could use something like: $e^{ab} = e^{a\ln(e^b)} = \sum_{k} \frac{a^k}{k!} \prod_{i=1}^k \sum_{m} \frac{(-1)^{m+1}}{m} \prod_{j=1}^m e^b-1$ (use series expansions). But I think what you're really looking for doesn't exist. Since exponential functions grow faster than polynomial functions, it seems unlikely you can succeed in turning exponentiation into multiplication in the way you're looking for. – Bill Cook Sep 28 '12 at 14:49