How to interpret log-log regression coefficients for other than 1 or 10 percent change? I have read many threads here on how to interpret coefficients in a regression where the predictor and the dependent variable are log-transformed. Most give an answer for a one or ten percent change. However, I am not sure whether I am right about interpreting other changes.
I have a beta coefficient of $-0.5057$. Given the standard interpretation (as elasticity), a one percent increase in the predictor variable ($X$) leads to an decrease of $0.5057$ percent in the dependent variable ($Y$).  Alternatively a ten percent increase in $X$ leads to a $5.057$ percent decrease in $Y$.
I found the formulas $1.1^{-0.5057}$ and $\exp(-0.5057(\log(1.1)))$ to calculate the affect of any change in $X$ on $Y$. Are these formulas right because I get for a ten percent increase slightly different results than the $5.057$ decrease.
Furthermore, is it right that the percentage values for the changes in $Y$ do not increase proportionally with increases in $X$? If a one percent increase in $X$ leads to an decrease of $0.5057$ percent in $Y$, it is wrong to assume that a $50$ percent increase in $X$ leads to a $(50 \times  0.5057)$ decrease in $Y$? 
$\exp(-0.5057(\log(1.5)))= 0.8146$. Given the difference between $1$ and this value, it suggests that $Y$ decreases by $18.5$ percent. Am I right?
The formulas mentioned above suggest so. Is that because of the nonlinearity in the model?
 A: 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 $c$.
