Linear regression (adding constant to variables) I'm running a multiple linear regression. Let's suppose I really need to use the logarithmic transformation. However, all values of one variable are negative. I assume I have to do the following: X1 + constant. After that, I can use the logarithmic transformation and run a multiple regression.
I'd like to mention that I have done that before without the logarithmic transformation, running a simple linear regression and it has affected only an alpha coefficient (makes perfect sense for me).
For example, I have got the following results:

*

*y = 1,08 + 0,56*x1, original x1

*y = -0,03 + 0,56*(x1 + 2), x1 + constant

So I can use both equations for making predictions, getting the same results.
Is it still possible to interpret Beta coefficients? I am used to relying on elasticity and logarithmic transformation, showing how independent variables influence "Y". Do I need to take into account that I have added a "constant"? If I do, how?
 A: In regression equation
$$
y = \alpha + \beta x + \varepsilon
$$
the $\beta$ parameter is about the slope of the regression line, while $\alpha$ is about moving it vertically along $y$-axis. Since a picture is worth a thousand words, you can see this on a picture below showing $x$, and shifted variants of it, on $x$-axis, and $0.56 x$ on $y$-axis. As you can see on the image, in each case the slope is the same, just the lines are shifted. So the slope, and it's interpretation, remain the same.

As you noticed, in your example both regression formulas give same results, because they are the same:
$$\begin{align}
y &= -0.03 + 0.56 (2 + x) \\&= -0.03 + 0.56 \times 2 + 0.56 x \\&= -0.03 + 1.12 + 0.56 x \\&\approx  1.08 + 0.56 x
\end{align}$$ 
where this is approximate only because of rounding error, on unrounded values it's exact, so you get same results. The slope would take care of all such constants, also in multiple regression.
Honestly, I don't follow what you mean by the part where you mention logarithms, but if you transform $x+c$ with some function, like logarithm, the influence on the outcome gets more complicated. However adding a constant should not make any drastic differences in terms of interpretability. If in doubt, try plotting such functions against different values to check what happens (see two examples below).

Since it may be unclear, not every function is additive, so if you add a constant and transform, this does not mean that you could always copy and paste the slope to the equation with different transformation and get the same result. If by "same interpretation" you mean "same slope", this would not be the case for every transformation.
A: I would not do this. The problem is that what you choose to add to make x positive is arbitrary and can have a huge effect on the parameter estimates.
First, let's set up x and y and the model:
set.seed(1234)  #Sets a seed

x <- rnorm(100, -10, 1) #Normal mean = -10, sd = 1
y <- 3*x + rnorm(100)

Now, we'll adjust x to be positive so that logs can be taken. Usually, people choose to make the smallest adjusted x close to 0, but how close? Let's try two variations:
xadj1 <- x-min(x) + 0.01
xadj2 <- x-min(x) + 0.1

Now, we fit models:
m1 <- lm(y~log(xadj1))
summary(m1)  #-32.52 + 3.29*log(xadj1)

m2 <- lm(y~log(xadj2))
summary(m2)  #-33.90 + 4.88*log(xadj2)

And the results are quite different. 
A: My recommendation is to apply a Box-Cox analysis of transformation using the two-parameter option (as the modeling will suggest the proper power transform and additive constant). Here is an easy discussion source and also at here.
More advanced is this paper and also the discussion in Wikipedia, especially the first example noting plots (b) and (c) which illustrates the selection process for the best power transform and additive constant based on the log-likelihood.
