I tried to ask this question on Math Stackexchange, but got no answer. I'll try here.

I am a bit confused as to how to see elasticity of a function with respect to a variable from logarithm. Lets say we have the following function:

$$y^* =\beta^{\frac{1}{1-a}}(\frac{s}{n+\delta})^{\frac{a}{1-a}} \leftrightarrow$$ $$ln y^* = \frac{1}{1-a} * ln (\beta) + \frac{a}{1-a}*ln(\frac{s}{n+\delta})$$

How does one see from here what the elasticity of y with respect to $ n + \delta $ is?

Kind regards


Elasticity is defined as "In economics, elasticity measures the percentage change of one economic variable in response to a change in another" via Wikipedia. To find the elasticity of y with respect to $n+\delta$, you can use the following formula:

$$ \frac{\partial y}{\partial (n+\delta)} * \frac{n+\delta}{y} = -\frac{a}{1-a} $$

But my books suggests I should be able to spot this from the $lny*$ equation. Anyone got any idea how?


1 Answer 1


Elasticity of any differentiable function $f(x)$ wrt $x$ by definition is:

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

For a particular type of function given by:

$$f(x) = Ax^e$$

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

Moreover, for this sort of function if you take a log of both sides you will find that:

$$\ln f(x) = \ln A + e \ln x$$

where the coefficient before logged variable gives you the elasticity.

In your case, if you take log of the following:

$$y^* =\beta^{\frac{1}{1-a}}(\frac{s}{n+\delta})^{\frac{a}{1-a}}$$

you will get:

$$\ln y^* = \frac{1}{1-a} \ln (\beta) + \frac{a}{1-a}\ln s - \frac{a}{1-a} \ln (n+\delta)$$

where $\frac{1}{1-a}$, $\frac{a}{1-a}$ and $- \frac{a}{1-a}$ are all elasticities of the original function $y$ wrt $\beta$, $s$ and $n+\delta$ respectively.

You can verify that by explicitly calculating:

$$\epsilon_{\beta y} = \frac{\partial y}{\partial \beta}\frac{\beta}{y} = \frac{1}{1-a} $$

$$\epsilon_{s y} = \frac{\partial y}{\partial s}\frac{s}{y} = \frac{a}{1-a} $$

$$\epsilon_{(n-\delta); y} = \frac{\partial y}{\partial (n-\delta)}\frac{(n-\delta)}{y} = -\frac{a}{1-a} $$

  • $\begingroup$ Thank you, I appreciate it. $\endgroup$
    – Zebraboard
    Jan 7, 2022 at 20:35

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