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

*

*The demand function can be written as: $ap+bq=k \iff \boxed{D(p) = q = \frac{k-ap}{b}} \:\:(1)$ and $\boxed{G(q) = p = \frac{k-bq}{a}}\:\: (2)$

*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?
 A: 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 $q$.
A: 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.
