Simplifying this expression $(e^u-1)(e^u-e^l)$ Is it possible to write the following
$$(e^u-1)(e^u-e^l)$$
as
$$e^{f(u,l)}-1?$$
 A: Yes, it is possible. Define
$$f(u,l) = \log(1+(e^u-1)(e^u-e^l))$$
We then have
$$e^{f(u,l)}-1 = (e^u-1)(e^u-e^l)$$
EDIT
The above is valid/defined provided $(e^u-1)(e^u-e^l) > -1$.
A: The first expression evaluates to $e^{2u}-e^u-e^l+1$, so if your question is whether there is some algebraic manipulation that brings this into the form of an exponential minus $1$ for general $l,u$ then the answer is no.
In fact the range of the function $x\mapsto e^x-1$ in $\Bbb R$ is $(-1,\infty)$, and the product of two values in that range may be outside the range if one is negative and the other sufficiently large. For instance for $u=-1,l=10$ your product will be much less than $-1$, so it cannot be written as $e^x-1$ for any real number $x$. You might want to get around this by using complex numbers for $f(u,l)$, but then consider the example $u=-\ln2,l=\ln3$, for which you have $e^u-1=-\frac12$ and $e^l-1=2$, so the product $(e^u-1)(e^l-1)$ is $-\frac12\times2=-1$, which is not a value $e^x-1$ even for $x\in\Bbb C$, so this shows you just cannot define $f$ for general arguments $u,l$ at all.
