price elasticity of output, makes sense?
I use some abbreviations and drop the index t.

$\ln(G)=\beta_0+\beta_1\ln(L)+\beta_2\ln(P)+\epsilon$

Taking both sides as an exponent of e

$e^{\ln(G)}=e^{\beta_0+\beta_1\ln(L)+\beta_2\ln(P)+\epsilon}$

$G=e^{\beta_0+\beta_1\ln(L)+\epsilon}\cdot e^{\beta_2\ln(P)}$

$e^{\beta_0+\beta_1\ln(L)+\epsilon}=C$ is a constant if you differentiate w.r.t $P$.

$G=e^{\beta_0+\beta_1\ln(L)+\epsilon}\cdot e^{\ln(P^{\beta_2})}$

$G=e^{\beta_0+\beta_1\ln(L)+\epsilon}\cdot P^{\beta_2}$

$\frac{\partial G}{\partial P}=e^{\beta_0+\beta_1\ln(L)+\epsilon}\cdot \beta_2\cdot P^{\beta_2 -1}$

The elasticity is $\large{\epsilon}\normalsize{=\frac{\partial G}{\partial P}\cdot \frac{P}{G}}$

$\large{\epsilon}=\normalsize{e^{\beta_0+\beta_1\ln(L)+\epsilon}\cdot \beta_2\cdot P^{\beta_2 -1}\cdot \frac{P}{e^{\beta_0+\beta_1\ln(L)+\epsilon}\cdot P^{\beta_2}}}$

=$C\cdot \beta_2\cdot P^{\beta_2 -1}\cdot\frac{P}{C\cdot P^{\beta_2}}$

You can finish by simplifying the term.

In general it is true that 

$\large{\epsilon}=\normalsize{\frac{\partial \ln(G)}{\partial\ln(P)}}$

Thus the short way is to differentiate $\ln(G)$ w.r.t $\ln(P)$.