3

As you say the first step is to take log of both sides after that you are just applying the rules for logarithms and rearrange. For example: $$\ln (XZ)=\ln X + \ln Z$$ $$\ln X/Z= \ln X - \ln Z$$ $$\ln X^a = a \ln X$$ $$\ln 1 = 0$$ Also an important approximations that hold close to zero are applied here as well these are: $\ln(1+x) \approx x $ for $x$ ...


2

$\frac{dlnQ}{dp}=\frac{dlnQ}{dQ} \frac{dQ}{dp}$ thus: $\frac{dQ}{dp}=\frac{dlnQ}{dp} \frac{dQ}{dlnQ}$ Since we know that if $f(x)=lnx \Rightarrow f'(x)=\frac{1}{x}\Rightarrow \frac{1}{f'(x)}=x$ We replace $\frac{dQ}{dlnQ}$ by $Q$. You get: $\frac{dQ}{dp}=\frac{dlnQ}{dp} Q$ It is readily found that $\frac{dlnQ}{dp}= \frac{e}{p}$ So our expression for ...


2

The elasticity of Y with respect to X is often estimated by running a regression like this: $$ \ln Y = \alpha + \beta \ln X + error$$ However, this isn't always applicable because it is perfectly fine to ask what is the percent change in Y to a percent change in X even if X or Y are negative. There are other issues too, like the fact that the elasticity may ...


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