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Given the following model:

$$\ln{\left(W\right)}=\beta_0+\beta_1e+\beta_2e\ast x+\beta_3e\ast\ln{\left(P\right)} +\beta_4\ln{\left(P\right)}\ast x+ error.$$

Setting $x = 4$, and $e =3$, a $1$% increase in $P$ will lead to a __________ % increase in $W$?

I took the derivative with respect to $P$ and got:

$$\frac{\beta_3e}{P}+ \frac{\beta_4x}{P}$$

But don't know how to proceed.

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You must also take the derivative of the right hand side: $$\frac{dW}{dP} \frac{1}{W} = \frac{\beta_3e}{P}+ \frac{\beta_4x}{P} $$ so with $\delta W$ the variation of $W$ you obtain: $$\frac{\delta W} {W} \simeq (\beta_3 e+\beta_4 x) \frac{\delta P}{P}$$


Note that you can obtain directly the result noticing that: $$d \ln(W) =\frac{dW}{W}, d \ln(P) =\frac{dP}{P}$$

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