<p>Elasticity of any differentiable function <span class="math-container">$f(x)$</span> wrt <span class="math-container">$x$</span> by definition is:</p>
<p><span class="math-container">$$\epsilon = \frac{df(x)}{dx}\frac{x}{f(x)}$$</span></p>
<p>For a particular type of function given by:</p>
<p><span class="math-container">$$f(x) = Ax^e$$</span></p>
<p><span class="math-container">$$\epsilon = \frac{df(x)}{dx}\frac{x}{f(x)} = e $$</span></p>
<p>Moreover, for this sort of function if you take a log of both sides you will find that:</p>
<p><span class="math-container">$$\ln f(x) = \ln A + e \ln x$$</span></p>
<p>where the coefficient before logged variable gives you the elasticity.</p>
<p>In your case, if you take log of the following:</p>
<p><span class="math-container">$$y^* =\beta^{\frac{1}{1-a}}(\frac{s}{n+\delta})^{\frac{a}{1-a}}$$</span></p>
<p>you will get:</p>
<p><span class="math-container">$$\ln y^* = \frac{1}{1-a}  \ln (\beta) + \frac{a}{1-a}\ln s - \frac{a}{1-a} \ln (n+\delta)$$</span></p>
<p>where <span class="math-container">$\frac{1}{1-a}$</span>, <span class="math-container">$\frac{a}{1-a}$</span> and <span class="math-container">$- \frac{a}{1-a}$</span> are all elasticities of the original function <span class="math-container">$y$</span> wrt <span class="math-container">$\beta$</span>, <span class="math-container">$s$</span> and <span class="math-container">$n+\delta$</span> respectively.</p>
<p>You can verify that by explicitly calculating:</p>
<p><span class="math-container">$$\epsilon_{\beta y} = \frac{\partial y}{\partial \beta}\frac{\beta}{y} = \frac{1}{1-a} $$</span></p>
<p><span class="math-container">$$\epsilon_{s y} = \frac{\partial y}{\partial s}\frac{s}{y} = \frac{a}{1-a} $$</span></p>
<p><span class="math-container">$$\epsilon_{(n-\delta); y} = \frac{\partial y}{\partial (n-\delta)}\frac{(n-\delta)}{y} = -\frac{a}{1-a} $$</span></p>

CC BY-SA 4.0
2022-01-07T20:30:33.603Z
2022-01-10T18:03:46.137Z
2022-01-10T18:03:46.137Z
1muflon1
<p>I tried to ask this question on Math Stackexchange, but got no answer. I'll try here.</p>
<p>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:</p>
<p><span class="math-container">$$y^* =\beta^{\frac{1}{1-a}}(\frac{s}{n+\delta})^{\frac{a}{1-a}} \leftrightarrow$$</span>
<span class="math-container">$$ln y^* = \frac{1}{1-a} * ln (\beta) + \frac{a}{1-a}*ln(\frac{s}{n+\delta})$$</span></p>
<p>How does one see from here what the elasticity of y with respect to <span class="math-container">$ n + \delta $</span> is?</p>
<p>Kind regards</p>
<p>Edit:</p>
<p>Elasticity is defined as &quot;In economics, elasticity measures the percentage change of one economic variable in response to a change in another&quot; via Wikipedia. To find the elasticity of y with respect to <span class="math-container">$n+\delta$</span>, you can use the following formula:</p>
<p><span class="math-container">$$
\frac{\partial y}{\partial (n+\delta)} * \frac{n+\delta}{y} = -\frac{a}{1-a}
$$</span></p>
<p>But my books suggests I should be able to spot this from the <span class="math-container">$lny*$</span> equation. Anyone got any idea how?</p>

<elasticity>
Finding the elasticity of a function with respect to a variable from logarithm
Zebraboard
# Finding the elasticity of a function with respect to a variable from logarithm

(Asked by: username_0 on 2022-01-07)

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

Edit:

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?

**Tags:** elasticity


---

## 1 answers

### Answer by username_1 on 2022-01-07. Score: 5 | Accepted answer

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:

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

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

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


---

