How to prove that the elasticity of the revenue function is $E_R(p)=\frac{E_R(q)}{E_R(q)-1}$?

Question

The Problem

Let the demand function be $ap+bq=k$.

Prove that this equation (elasticity of revenue) is true: $$E_R(p)=\frac{E_R(q)}{E_R(q)-1}$$

---

Definitions

Demand Function

The Demand Function is defined as the relation between the price $p$ of the good and the demanded quantity $q$ of the good which in our example is: $ap+bq=k$. Note that $D^{-1}(p) = G(q)$

Revenue Function

The Revenue Function is defined as $R = p q$, where R is the total revenue, $p$ is the selling price per unit of sales, and $q$ is the number of units sold

Elasticity of a function

The Elasticity of a function $f(x)$ approximates the change of $f$ given the change of $x$ and is defined as:

$$ E_f(x) = \frac{df}{dx} \frac{x}{f(x)}$$

---

My solution attempt

We need to prove $E_R(p)=\frac{E_R(q)}{E_R(q)-1}$

1. The demand function can be written as: $ap+bq=k \iff \boxed{D(p) = q = \frac{k-ap}{b}} \:\:(1)$ and $\boxed{G(q) = p = \frac{k-bq}{a}}\:\: (2)$
2. Therefore we can write the revenue function as $R(q) = pq = pD(p) \iff \boxed{ R(q) = \frac{kp-ap^2}{b} } \:\:(3)$ and $R(p) = pq = G(q)q \iff \boxed{R(p) = \frac{kq-bq^2}{a}}\:\: (4)$

Hence, $E_R(p) = \frac{R(q)}{dq} \frac{q}{R(q)} \stackrel{(3)}{=} \left(\frac{kp-ap^2}{a}\right)'\cdot \frac{q}{\frac{kp-ap^2}{a}} = \frac{k-2ap}{a}\cdot \frac{q}{\frac{kp-ap^2}{a}} = \frac{\left(k-2ap\right)q}{kp-ap^2}$

$$ \boxed{ E_R(p) = \frac{\left(k-2ap\right)q}{kp-ap^2}}\:\: (5)$$

And, $E_R(q) = \frac{R(p)}{dq} \frac{p}{R(p)} \stackrel{(4)}{=} \left( \frac{kq-bq^2}{a} \right)' \cdot \frac{p}{\frac{kq-bq^2}{a}} = \frac{k-2bq}{a} \cdot \frac{p}{\frac{kq-bq^2}{a}} = \frac{p\left(k-2bq\right)}{kq-bq^2}$

$$ \boxed{E_R(q) = \frac{p\left(k-2bq\right)}{kq-bq^2}} \:\:(6)$$

---

So, at last:

$$E_R(p)=\frac{E_R(q)}{E_R(q)-1} \iff \\ \frac{\left(k-2ap\right)q}{kp-ap^2} = \frac{\frac{p\left(k-2bq\right)}{kq-bq^2}}{\frac{p\left(k-2bq\right)}{kq-bq^2}-1}$$

Which is overly complicated but it must hold, if there were no trivial calculation mistakes.

---

The Question

Given the fact that this was an exam subquestion, I am confident that there is an easier way to prove the elasticity equation (maybe by using elasticity function properties?), but if there is, I can't spot it.

Any ideas?

Answer 1

Note that elasticity of revenue can be written $$ E_R(p)E_R(q)=E_R(p)+E_R(q) $$ and by the definition of elasticity of a function $$ p\frac{R'(p)}{R(p)}\cdot q\frac{R'(q)}{R(q)}=p\frac{R'(p)}{R(p)}+q\frac{R'(q)}{R(q)}. $$ Given that $R(p)=R(q)=pq$ (the only difference is which variable is considered independent), we can simplify all denominators and the factors $p,q$ on the LHS, obtaining $$ R'(p)R'(q)=pR'(p)+qR'(q)\tag1 $$ Now \begin{alignat}{2} R'(p) &= (pD(p))'&&=D(p)+pD'(p)&&=q+pD'(p)\\ R'(q) &= (qG(q))'&&=G(q)+qG'(q)&&=p+qG'(q)\\ \end{alignat} Substituting in $(1)$ $$ [q+pD'(p)][p+qG'(q)]=p[q+pD'(p)]+q[p+qG'(q)]\\ pq+p^2D'(p)+q^2G'(q)+pqD'(p)G'(q)=2pq+p^2D'(p)+q^2G'(q)\\ $$ but, being $D$ and $G$ each other inverse, then $D'(p)G'(q)=1$ and the relation is proved, without any need to use the actual relation between $p$ and $q$.

Answer 2

Note you may equivalently define elasticity (using your notation) as

$$E_f(x)=\frac{d \log f(x)}{d \log x}.$$

Thus,

$$\small E_R(q)=\frac{d \log pq}{d \log q}=\frac{d \log pq}{d \log p}\frac{d \log p}{d \log q}=\frac{d \log pq}{d \log p}\left(\frac{d (\log p+\log q)}{d \log q}-1\right)=\frac{d \log pq}{d \log p}\left(\frac{d \log pq}{d \log q}-1\right)=E_R (p) (E_R(q)-1),$$

and rearranging gives us the desired result.