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?


2 Answers 2


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.


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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