<p>Differentiating both sides of the equation with respect to <span class="math-container">$x$</span>, using the <a href="https://en.wikipedia.org/wiki/Chain_rule" rel="noreferrer">chain rule</a> for the left hand side and noting that, since <span class="math-container">$a$</span> is a parameter, <span class="math-container">$da/dx=0$</span>:
<span class="math-container">$$\frac{1}{y}\frac{dy}{dx}=b\frac{1}{x}$$</span>
Rearranging:
<span class="math-container">$$\frac{dy/y}{dx/x}=\eta=b$$</span></p>

CC BY-SA 4.0
2019-01-13T14:25:33.233Z
2019-01-13T17:57:34.410Z
2019-01-13T17:57:34.410Z
Adam Bailey
<p>Because <span class="math-container">$a$</span> is a parameter, and so
<span class="math-container">$$
\eta = \frac{ d \log y}{d \log x} = \frac{ d \log a + d \ b \log x}{d \log x} = 0 + b.
$$</span></p>

CC BY-SA 4.0
2019-01-13T13:21:40.223Z
2021-04-06T13:22:13.810Z
2021-04-06T13:22:13.810Z
Giskard
<p>Let's consider a relationship between <span class="math-container">$ y $</span> and <span class="math-container">$ x $</span>, <span class="math-container">$ y = a x^b $</span>. Taking log on both sides, we have <span class="math-container">$$ \log y = \log a + b \log x $$</span></p>

<p>Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:</p>

<p><span class="math-container">$$ \eta = b = \frac{ d \log y}{d \log x} $$</span></p>

<p>Now, I understand that <span class="math-container">$ d \log y = \frac 1y dy $</span> and <span class="math-container">$ d \log x = \frac 1x \ dx $</span>. So I understand why we can write <span class="math-container">$ \eta = \frac {d \log y}{d \log x} $</span>. What I don't understand is: why does <span class="math-container">$ b $</span>, which is the power on the variable <span class="math-container">$ x $</span>, equal <span class="math-container">$ \eta $</span>?</p>

<p>Here's a snapshot from the book:</p>

<p><a href="https://i.sstatic.net/2TRHv.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/2TRHv.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2019-01-13T13:24:16.643Z
About half of your [questions](https://economics.stackexchange.com/users/6486/worldgov) have been answered, consider accepting some.
Giskard
CC BY-SA 4.0
2019-01-13T13:44:10.960Z
Thank you for pointing that out. I'll do so right away.
WorldGov
<elasticity>
Elasticity and logarithms
WorldGov
# Elasticity and logarithms

(Asked by: username_0 on 2019-01-13)

Let's consider a relationship between $ y $ and $ x $, $ y = a x^b $. Taking log on both sides, we have $$ \log y = \log a + b \log x $$

Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:

$$ \eta = b = \frac{ d \log y}{d \log x} $$

Now, I understand that $ d \log y = \frac 1y dy $ and $ d \log x = \frac 1x \ dx $. So I understand why we can write $ \eta = \frac {d \log y}{d \log x} $. What I don't understand is: why does $ b $, which is the power on the variable $ x $, equal $ \eta $?

Here's a snapshot from the book:

[![enter image description here](https://i.sstatic.net/2TRHv.png)](https://i.sstatic.net/2TRHv.png)

**Tags:** elasticity

> **Comments:**
> **username_1**: About half of your [questions](https://economics.stackexchange.com/users/6486/username_0) have been answered, consider accepting some.  
> **username_0**: Thank you for pointing that out. I'll do so right away.  

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## 2 answers

### Answer by username_1 on 2019-01-13. Score: 8 | Accepted answer

Because $a$ is a parameter, and so
$$
\eta = \frac{ d \log y}{d \log x} = \frac{ d \log a + d \ b \log x}{d \log x} = 0 + b.
$$


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### Answer by username_2 on 2019-01-13. Score: 7

Differentiating both sides of the equation with respect to $x$, using the [chain rule](https://en.wikipedia.org/wiki/Chain_rule) for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
$$\frac{1}{y}\frac{dy}{dx}=b\frac{1}{x}$$
Rearranging:
$$\frac{dy/y}{dx/x}=\eta=b$$


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