# Constant elasticity proof for log-linear demand curve

From Perloff 2008e solved 2.2:

Q: Show that the elasticity of demand is a constant e if the demand function is log-linear, ln Q=ln A+e ln p. A: Differentiating with respect to p, we find that (dQ/dp)/Q=e/p.

Where did 'dQ/dp' come from in the numerator?

$$\frac{dlnQ}{dp}=\frac{dlnQ}{dQ} \frac{dQ}{dp}$$ 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 the derivative of $$Q$$ wrt $$p$$ now reads:

$$\frac{dQ}{dp}=\frac{e}{p} Q$$

Divide both sides by $$Q$$ and you get $$\frac{dQ/Q}{dp}=\frac{dQ/dp}{Q}=\frac{e}{p}$$

Elasticity is now found by simply multiplying both sides by $$p$$, you get:

$$\frac{dQ}{dp} \frac{p}{Q} \equiv \eta=e$$