If $Q*P$ is constant, why is $\frac{\Delta Q/Q}{\Delta P / P}$ constant?

Question

The title says it all; I don't quite get why $Q*P$ being constant implies that $\frac{\Delta Q/Q}{\Delta P / P}$ is constant. Or does it?

This is about iso-elastic demand curves - why is it that $Q*P=const.$ implies that $\frac{\Delta Q/Q}{\Delta P / P} = -1$?

$Q$ and $R$ are both elements of $\mathbb{R}_+$, respectively.

Answer 1

Actually, in mathematics that should be $ \dfrac{dQ/Q}{dP/P} $ not $\dfrac{\Delta Q/Q}{\Delta P/P}$: $\Delta Q$ and $\Delta P$ would indicate finite changes, which would not give a constant result, while $dQ$ and $dP$ are differentials. Economics may use different notation.

You get it from differentiating: if $PQ$ is constant, $$ 0 = d(PQ) = Q\; dP + P \; dQ $$

so $$ \frac{dQ}{Q} = - \frac{dP}{P} $$

Answer 2

There are (at least) two ways to approach this. In either case, suppose $$ P \cdot Q = k $$ for some nonzero constant $k$.

Method 1:

Recall "logarithmic differentiation". And notice $Q = k/P$.

\begin{align*} \frac{\Delta Q/Q}{\Delta P/P} &= \frac{\Delta(\ln Q)}{\Delta(\ln P)} \\ &= \frac{\Delta(\ln (k/P))}{\Delta(\ln P)} \\ &= \frac{\Delta(\ln (k) - \ln (P))}{\Delta(\ln P)} \\ &= \frac{-\Delta(\ln (P))}{\Delta(\ln P)} \\ &= -1 \text{.} \end{align*}

Method 2:

Here, just notice $Q = k/P$ and $\Delta Q = \Delta\left( \frac{k}{P} \right) = \frac{-k}{P^2}\Delta P$, the latter being an application of implicit differentiation (with, as always, the chain rule).

\begin{align*} \frac{\Delta Q/Q}{\Delta P/P} &= \frac{\frac{\frac{-k}{P^2}\Delta P}{\frac{k}{P}}}{\frac{\Delta P}{P}} \\ &= \frac{\frac{\frac{-k}{P^2}\Delta P}{\frac{k}{P}}\cdot \frac{P/k}{P/k}}{\frac{\Delta P}{P}} \\ &= \frac{\frac{-\Delta P}{P}}{\frac{\Delta P}{P}} \\ = -1 \text{.} \end{align*}