Interpreting log-log regression with log(1+x) as independent variable Is interpreting log-log regression with log(1+x) as independent variable the same as having log(x) as independent variable? 1% increase in x results in beta% change in y?
What is x has both negative and positive values with the absolute value less than 1 so that 1+x are all positive?

Definitely not, except when $x$ is much larger than $1.$  This is one reason why the automatic reflex to "just add $1$ to values that might be zero before taking the log" is difficult to justify.
Let's see what is really going on.  Suppose you are modeling a response $y$ in the form
$$\log y = \cdots + \beta \log(1 + x) + \cdots$$
where the missing stuff doesn't vary with $x.$  Then increasing $x$ by $100\delta\%$ changes $y$ to
$$\log y^\prime = \cdots + \beta \log(1 + x(1 + \delta)) + \cdots = \log y + \beta(\log(1 + x + x\delta) - \log(1 + x)),$$
showing that

$\log y$ changes by $\beta(\log(1 + x + x\delta) - \log(1 + x)).$

That's as nasty to understand as it looks, even when (as is usual) $\delta$ is taken to be very small. For small values of $x\delta$ (that is, $x$ isn't huge) we can approximate it by
$$\beta(\log(1 + x + x\delta) - \log(1 + x)) \approx  \beta x\delta;\quad |x\delta| \ll 1+|x|.$$
This is approximately a $100 \beta x \delta \%$ change in $y$ itself.  This covers the case of small negative values of $x,$ too -- but of course they cannot be $-1$ or less, for then $\log(1+x)$ would be undefined.
For large $x \gg 1$ we can approximate this change by neglecting $1$ in comparison to $x,$ giving
$$\beta(\log(1 + x + x\delta) - \log(1 + x)) \approx  \beta \delta;\quad x \gg 1.$$
This is close to the usual log-log relation, reflecting approximately a $100 \beta \delta\%$ change in $y.$
Here, to illustrate and help the intuition, are log-log plots of two such relationships (with $\beta=1/2$):

The straight-line (dotted red) plot is $\log y = \beta \log(x).$  Because it is a line, we can interpret a change in $\log x$ as being related to a fixed multiple of that change in $\log y.$  But the black graph of $\log y = \beta\log(1+x)$ departs pretty strongly from this linear shape when $x$ is small to medium in size.  Its changing curvature means that the relationship between any change in $\log x$ and the value of $\log y$ changes with $x:$ it's small for smaller $x$ but grows as $x$ gets larger.
Thus, the answer is it's complicated: when $x$ ranges from smallish to largish values, the change in $\log y$ ranges from a multiple $x\delta$ to a multiple $\delta$ of $\beta.$  The value depends on $x$ itself, at least until $x$ is sufficiently large. There is no fixed, simple relationship.
