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?


1 Answer 1


$\frac{dlnQ}{dp}=\frac{dlnQ}{dQ} \frac{dQ}{dp}$ 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:

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


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