<p><span class="math-container">$\frac{dlnQ}{dp}=\frac{dlnQ}{dQ} \frac{dQ}{dp}$</span> thus:</p>

<p><span class="math-container">$\frac{dQ}{dp}=\frac{dlnQ}{dp} \frac{dQ}{dlnQ}$</span></p>

<p>Since we know that if <span class="math-container">$f(x)=lnx \Rightarrow f'(x)=\frac{1}{x}\Rightarrow \frac{1}{f'(x)}=x$</span> </p>

<p>We replace <span class="math-container">$\frac{dQ}{dlnQ}$</span>
by <span class="math-container">$Q$</span>. You get:</p>

<p><span class="math-container">$\frac{dQ}{dp}=\frac{dlnQ}{dp} Q$</span></p>

<p>It is readily found that 
<span class="math-container">$\frac{dlnQ}{dp}= \frac{e}{p}$</span></p>

<p>So our expression for the derivative of <span class="math-container">$Q$</span> wrt <span class="math-container">$p$</span> now reads:</p>

<p><span class="math-container">$\frac{dQ}{dp}=\frac{e}{p} Q$</span></p>

<p>Divide both sides by <span class="math-container">$Q$</span> and you get <span class="math-container">$\frac{dQ/Q}{dp}=\frac{dQ/dp}{Q}=\frac{e}{p}$</span></p>

<p>Elasticity is now found by simply multiplying both sides by <span class="math-container">$p$</span>, you get:</p>

<p><span class="math-container">$\frac{dQ}{dp} \frac{p}{Q} \equiv \eta=e$</span></p>

CC BY-SA 4.0
2019-06-25T21:28:08.650Z
2019-06-25T21:28:08.650Z
user18214
<p>From Perloff 2008e solved 2.2:</p>

<p>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.</p>

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

<supply-and-demand><elasticity><log-linearization>
Constant elasticity proof for log-linear demand curve
huppuguga
# Constant elasticity proof for log-linear demand curve

(Asked by: username_0 on 2019-06-25)

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?

**Tags:** supply-and-demand, elasticity, log-linearization

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## 1 answers

### Answer by username_1 on 2019-06-25. Score: 2

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