<p>Suppose we have the function <span class="math-container">$y=ln(x)$</span> Take the first order derivative:
<span class="math-container">$$\frac{d y}{d x}=\frac{1}{x}$$</span> Multiply both sides by <span class="math-container">$dx$</span>:
<span class="math-container">$$dy=\frac{dx}{x} $$</span> In words:a change in the natural logarithm due to an infinitesmal small change in <span class="math-container">$x$</span> is equal to the relative change in <span class="math-container">$x$</span> due to an infinitesmal change in <span class="math-container">$x$</span>, in your example this means that <span class="math-container">$dln(q)=\frac{dq}{q}$</span> This is exactly what you're interested in, namely the relative change in <span class="math-container">$q$</span>. So indeed <span class="math-container">$\frac{dln(q)}{dln(p)} \approx \frac{\Delta q/q}{\Delta p/p} $</span></p>

CC BY-SA 4.0
2019-03-25T11:07:28.483Z
2019-03-25T14:37:14.803Z
2019-03-25T14:37:14.803Z
user18214
<p>The definition of elasticity of demand with respect to price is: <span class="math-container">$\varepsilon_{q,p} = \frac{dq}{dp} \cdot \frac{p}{q}$</span>. So in your demand function we have:</p>

<p><span class="math-container">$$q = kp^{-\epsilon}$$</span>
<span class="math-container">$$\frac{dq}{dp} = -\epsilon kp^{-\epsilon - 1}$$</span>
<span class="math-container">$$\varepsilon_{q,p} = \frac{dq}{dp}\cdot \frac{p}{q}  = -\epsilon kp^{-\epsilon - 1} \cdot \frac{p}{kp^{-\epsilon}} = -\epsilon$$</span> </p>

<p>And we have the answer you're looking for.</p>

CC BY-SA 4.0
2019-04-07T00:33:56.500Z
2019-04-07T00:33:56.500Z
Marcelo Gelati
<p>Given the demand function, <span class="math-container">$ q = kp^{-\epsilon} $</span>, how do I calculate the elasticity? As a result, I do know that the elasticity when the demand function is in this form is <span class="math-container">$ - \epsilon $</span>. But I'd like to know how. I also found a derivation online that proceeded like this:</p>

<p>(1) Take logarithm on both sides
(2) Differentiate on boths ides
(3) You'll get: <span class="math-container">$$ \frac {\text{d} \ln(q)}{\text{d} \ln (p)} = - \epsilon $$</span>
(4) The LHS of the above equation is simply elasticity. </p>

<p>How does <span class="math-container">$$  \frac {\text{d} \ln(q)}{\text{d} \ln (p)} $$</span> represent elasticity?</p>

<elasticity>
Elasticity when the demand function is given
WorldGov
# Elasticity when the demand function is given

(Asked by: username_0 on 2019-03-25)

Given the demand function, $ q = kp^{-\epsilon} $, how do I calculate the elasticity? As a result, I do know that the elasticity when the demand function is in this form is $ - \epsilon $. But I'd like to know how. I also found a derivation online that proceeded like this:

(1) Take logarithm on both sides
(2) Differentiate on boths ides
(3) You'll get: $$ \frac {\text{d} \ln(q)}{\text{d} \ln (p)} = - \epsilon $$
(4) The LHS of the above equation is simply elasticity.

How does $$ \frac {\text{d} \ln(q)}{\text{d} \ln (p)} $$ represent elasticity?

**Tags:** elasticity

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## 2 answers

### Answer by username_2 on 2019-04-07. Score: 2 | Accepted answer

The definition of elasticity of demand with respect to price is: $\varepsilon\_{q,p} = \frac{dq}{dp} \cdot \frac{p}{q}$. So in your demand function we have:

$$q = kp^{-\epsilon}$$
$$\frac{dq}{dp} = -\epsilon kp^{-\epsilon - 1}$$
$$\varepsilon\_{q,p} = \frac{dq}{dp}\cdot \frac{p}{q} = -\epsilon kp^{-\epsilon - 1} \cdot \frac{p}{kp^{-\epsilon}} = -\epsilon$$

And we have the answer you're looking for.


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### Answer by username_1 on 2019-03-25. Score: 1

Suppose we have the function $y=ln(x)$ Take the first order derivative:
$$\frac{d y}{d x}=\frac{1}{x}$$ Multiply both sides by $dx$:
$$dy=\frac{dx}{x} $$ In words:a change in the natural logarithm due to an infinitesmal small change in $x$ is equal to the relative change in $x$ due to an infinitesmal change in $x$, in your example this means that $dln(q)=\frac{dq}{q}$ This is exactly what you're interested in, namely the relative change in $q$. So indeed $\frac{dln(q)}{dln(p)} \approx \frac{\Delta q/q}{\Delta p/p} $


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