You should not take the absolute value of the coefficient--although this would let you know the effect of a 1-unit decrease in X. Think of it this way:
Using the original negative coefficient, this equation shows the percentage change in Y for a 1-unit increase in X:
(exp[−0.0564*1]−1)⋅100=−5.48
Your "absolute value" equation actually shows the percentage change in Y for a 1-unit decrease in X:
(exp[-0.0564*-1]−1)⋅100=5.80
You can use a percentage change calculator to see how both of these percentages map onto a 1-unit change in X. Imagine that a 1-unit change in X were associated with a 58-unit change in linear Y:
- Our linear version of Y going from 1,000 to 1,058 is a 5.8% increase.
- Our linear version of Y going from 1,058 to 1,000 is a 5.482% decrease.
The question concerns models of the form
$$\log(y) = \cdots + \beta \log(x) + \cdots$$
(where none of the omitted terms involves anything that changes with $x$). When we change $x$ by $100\delta\%$ we multiply it by $1+\delta$. According to the laws of logarithms, when $1 + \delta \gt 0$,
$$\log(x(1+\delta)) = \log(x) + \log(1 + \delta).$$
Therefore if such a change in $x$ changes $y$ to $y^\prime$,
$$\eqalign{
\log(y^\prime) &= \cdots + \beta \log(x(1+\delta)) + \cdots \\
&= \cdots + \beta\left(\log(x) + \log(1 + \delta)\right) + \cdots \\
&= \cdots + \beta\log(x) + \beta\log(1 + \delta) + \cdots .
}$$
The change in $\log(y)$ is $$\log(y^\prime) - \log(y) = \beta\log(1 + \delta).$$
When $\delta$ is nearly zero (say, $10\%$ or smaller in size), $\log(1 + \delta) \approx \delta$ is a good approximation. When in turn $\beta\delta$ is also close to zero, this is the basis for the approximate interpretation "a $\delta$ percent change in $x$ corresponds to a $\beta\delta$ percent change in $y$."
For larger $\delta$ or $\beta$, however, this approximation fails. The fully general relationship is obtained by noting that adding $\beta\log(1+\delta)$ to $\log{y}$ is tantamount to multiplying $y$ by
$$\exp(\beta\log(1+\delta)) = (1+\delta)^\beta.$$
Therefore, when working with logarithms, think multiplicatively. We may memorialize the result of this analysis with a simple rule:
When $x$ is multiplied by any positive amount $c$, $y$ is multiplied by $c^\beta$.
In other words, log-log relationship are power relationships. Let's look at some examples:
When $\beta=2$, multiplying $x$ by $c$ multiplies $y$ by $c^2$. For instance, tripling $x$ will multiply $y$ by $9$.
When $\beta=1/3$, multiplying $x$ by $c$ multiplies $y$ by $c^{1/3} = \sqrt[3]{c}$. For instance, doubling $x$ will only multiply $y$ by $\sqrt[3]{2}\approx 1.26$.
When $\beta = -1$, multiplying $x$ by $c$ multiplies $y$ by $c^{-1} = 1/c$; that is, $y$ is divided by $c$.
Best Answer
The interpretation is the same on both sides of 0. Ignoring the rest of the equation, with $b > 0$, $-1*b > -2*b$ so an increase from -2 to -1 in $X_1$ also corresponds to an increase in $Y$. The thing is, while it might feel a little strange when writing it, going from -2 to -1 is an increase and going from -1 to -2 is a decrease.
Now, if you have theoretical reasons to expect the relationship not to hold for negative values, i.e. if you think that a decrease, say from 0 to -1, should result in a higher $Y$, then a simple linear regression might not be appropriate (transforming the variable could be solution but a quadratic term, splines or an interaction with an ad hoc binary variable could also capture such relationships).