<p>Elasticity is the slope after taking a log transform. Throughout this post $\log$ refers to the <em>natural logarithm</em>. That is, the logarithm with base $e$, the <code>ln</code> function in Excel. The logarithm transforms multiplication into addition. Log differences are something akin to percent changes (and they're equivalent for small changes).</p>

<p>Let's say we have:</p>

<p>$$ Q_t = A P_t^b $$</p>

<p>Take the logarithm of both sides:</p>

<p>$$ \log Q_t = \log A + b \log P_t$$</p>

<p>Log differences are conceptually similar to percent changes (for small percent changes). The slope of the line using the log transformed variables is $b$. This is the elasticity.</p>

<p>$$ \frac{d \left( \log Q_t \right) }{\log P_t} = b $$</p>

<h2>Why do people treat elasticities as percent changes?</h2>

<p>$$\log(Y) - \log(X) \approx \frac{Y-X}{X} \quad \quad \text{ for }\frac{y}{x} \approx 1$$</p>

<p>Log differences are approximately the percent change. Why? Take the first-order taylor expansion of the log to approximate it with a line near 1.</p>

<p>\begin{align*} 
\log(Y) &amp;\approx \log(1) + \frac{1}{\log(1)}\left(Y - 1 \right) \\
        &amp;\approx Y - 1 
\end{align*}
Hence $\log\left(\frac{Y}{X} \right) \approx \frac{Y-X}{X}$ for $Y \approx X$. This works well for small changes, eg. $\log(1.03) = .0296$, but it gets quite off for big changes $\log(1.5) = .4055$. This is probably where your stuff is getting off... </p>

<h2>Example where logs are nice...</h2>

<p>Let's say we have $1, 2, 4, 8$. Each is a 100% increase from the previous value. Taking logs we have $0, .693, .1386, 2.079$. The numbers increase linearly by $\log(2)$. Looking at percent increases though, it's less clean: $\frac{2-1}{1} = 100\%$ and $\frac{4-2}{2} = 100\%$ but $\frac{4-1}{1} = 300%$.</p>

<p>$$ \left(1 + R_1 \right) \left(1 + R_2 \right) = 1 + R_1 + R_2 + R_1R_2$$
Dealing with percent changes gets messy when $R_1R_2$ is big enough to matter. Taking logs though, let $r_1 = \log(1+R_1)$ etc...</p>

<p>$$ \log\left((1+R_1)(1+R_2) \right)= r_1 + r_2$$</p>

<h2>What I would do if I were you:</h2>

<p>Take a log and then it will be clean, and simple. Add a few columns to your spreadsheet.</p>

<ul>
<li>$q_t = \log Q_t$.</li>
<li>$a = \log A$</li>
<li>$p_t = \log P_t$</li>
</ul>

<p>Then the equation is:
$$ q_t = a + b p_t$$</p>

<p>$$\frac{dq_t}{dp_t} = b $$</p>

<p>$$ \begin{array}{rrrrr} 
P &amp; Q &amp; \log P &amp; \log Q &amp; \Delta \log P &amp; \Delta \log Q  &amp; \frac{\Delta \log P}{\Delta \log Q} \\
1 &amp; 100 &amp; 0 &amp; 4.605 &amp; \\
2 &amp; 25 &amp; .6931 &amp; 3.2189 &amp; .6931 &amp; -1.3863 &amp; -2.0\\
3 &amp; 11.11 &amp; 1.0986 &amp; 2.4079 &amp; .4055 &amp; -.8109 &amp; -2.0\\
4 &amp; 6.25 &amp; 1.3863 &amp; 1.8326 &amp; .2877 &amp; -.5754 &amp; -2.0\\
5 &amp; 4 &amp; 1.6094 &amp; 1.3863 &amp; .2231 &amp; -.4463 &amp; -2.0
\end{array}
$$</p>

<p>$\Delta$ means "change in", hence $\Delta \log Q_i = \log Q_i - \log Q_{i-1}$.</p>

CC BY-SA 3.0
2016-10-05T22:18:35.683Z
2016-10-08T04:07:15.010Z
2016-10-08T04:07:15.010Z
Matthew Gunn
<p>The problem with your calculations is that you have very big changes in price   </p>

<p>If you made price increases of $1\%$ in price, you would find the quantity fell by about $2\%$ as shown in the table below leading to an elasticity of about $-2$  </p>

<p>If you made the price changes even smaller, the calculated elasticity would get even closer to $2$, and in a sense you are looking for the limit of elasticity as the price change approaches $0$, which will indeed be the constant $b=-2$</p>

<pre><code> a       b      P       Q=a*P^b pct change Q    pct change P   elasticity

100     -2      1       100.00          
100     -2      1.01     98.03      -1.97%          1%          -1.97
100     -2      2        25.00          
100     -2      2.02     24.51      -1.97%          1%          -1.97
100     -2      3        11.11          
100     -2      3.03     10.89      -1.97%          1%          -1.97
100     -2      4         6.25          
100     -2      4.04      6.13      -1.97%          1%          -1.97
100     -2      5         4.00          
100     -2      5.05      3.92      -1.97%          1%          -1.97
</code></pre>

CC BY-SA 3.0
2016-10-05T23:57:07.410Z
2016-10-05T23:57:07.410Z
Henry
<p>"It is well known" that the elasticity of $Q$ relative to $P$ is the constant $b$ in the equation $Q(P)=aP^b$. We can verify it using the convenient definition of elasticity $= \frac{dQ}{dP} \frac{P}{Q}$, we can verify it in Wikipedia.</p>

<p>Based on this, when I create a spreadsheet with parameters $b=-2$, $a=100$ and P ranging from 1-5 by 1 I expected to recover an elasticity of -2, where elasticity =  % change in Q/ % change in P.  This is not the case. Instead the elasticity  varies substantially as the % change in P varies.</p>

<p>I'm forced to conclude that 'constant elasticity' does not mean a fixed value regardless of the magnitude of the price change, but only invariant with regards to the value of price OR that I have a spreadsheet error.</p>

<p>I've pasted an image of the spreadsheet results below.  Obviously, without seeing the formulas, no one can verify that I don't have an error, but I'm hopeful that someone will give this a spin for themselves and get very different numbers and then I'll know my failure to verify the constant elasticity in this function is a spreadsheet error.</p>

<p><a href="https://i.sstatic.net/lE83p.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lE83p.png" alt="spreadsheet image"></a></p>

CC BY-SA 3.0
2016-10-05T22:07:56.723Z
Possible duplicate of [Intuition: Difference in price elasticities of demand due to different bases](http://economics.stackexchange.com/questions/4594/intuition-difference-in-price-elasticities-of-demand-due-to-different-bases)
Giskard
<microeconomics><elasticity>
Verifying the constant elasticity of the demand model Q= aP^b
Ted Lorenzen
# Verifying the constant elasticity of the demand model Q= aP^b

(Asked by: username_0 on 2016-10-05)

"It is well known" that the elasticity of $Q$ relative to $P$ is the constant $b$ in the equation $Q(P)=aP^b$. We can verify it using the convenient definition of elasticity $= \frac{dQ}{dP} \frac{P}{Q}$, we can verify it in Wikipedia.

Based on this, when I create a spreadsheet with parameters $b=-2$, $a=100$ and P ranging from 1-5 by 1 I expected to recover an elasticity of -2, where elasticity = % change in Q/ % change in P. This is not the case. Instead the elasticity varies substantially as the % change in P varies.

I'm forced to conclude that 'constant elasticity' does not mean a fixed value regardless of the magnitude of the price change, but only invariant with regards to the value of price OR that I have a spreadsheet error.

I've pasted an image of the spreadsheet results below. Obviously, without seeing the formulas, no one can verify that I don't have an error, but I'm hopeful that someone will give this a spin for themselves and get very different numbers and then I'll know my failure to verify the constant elasticity in this function is a spreadsheet error.

[![spreadsheet image](https://i.sstatic.net/lE83p.png)](https://i.sstatic.net/lE83p.png)

**Tags:** microeconomics, elasticity
> **Comments:**
> **username_1**: Possible duplicate of [Intuition: Difference in price elasticities of demand due to different bases](http://economics.stackexchange.com/questions/4594/intuition-difference-in-price-elasticities-of-demand-due-to-different-bases)  

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## 2 answers

### Answer by username_3 on 2016-10-05. Score: 2 | Accepted answer

The problem with your calculations is that you have very big changes in price   
If you made price increases of $1\%$ in price, you would find the quantity fell by about $2\%$ as shown in the table below leading to an elasticity of about $-2$  
If you made the price changes even smaller, the calculated elasticity would get even closer to $2$, and in a sense you are looking for the limit of elasticity as the price change approaches $0$, which will indeed be the constant $b=-2$
 a       b      P       Q=a*P^b pct change Q    pct change P   elasticity

100     -2      1       100.00          
100     -2      1.01     98.03      -1.97%          1%          -1.97
100     -2      2        25.00          
100     -2      2.02     24.51      -1.97%          1%          -1.97
100     -2      3        11.11          
100     -2      3.03     10.89      -1.97%          1%          -1.97
100     -2      4         6.25          
100     -2      4.04      6.13      -1.97%          1%          -1.97
100     -2      5         4.00          
100     -2      5.05      3.92      -1.97%          1%          -1.97


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### Answer by username_2 on 2016-10-05. Score: 2

Elasticity is the slope after taking a log transform. Throughout this post $\log$ refers to the *natural logarithm*. That is, the logarithm with base $e$, the `ln` function in Excel. The logarithm transforms multiplication into addition. Log differences are something akin to percent changes (and they're equivalent for small changes).

Let's say we have:

$$ Q\_t = A P\_t^b $$

Take the logarithm of both sides:

$$ \log Q\_t = \log A + b \log P\_t$$

Log differences are conceptually similar to percent changes (for small percent changes). The slope of the line using the log transformed variables is $b$. This is the elasticity.

$$ \frac{d \left( \log Q\_t \right) }{\log P\_t} = b $$

## Why do people treat elasticities as percent changes?

$$\log(Y) - \log(X) \approx \frac{Y-X}{X} \quad \quad \text{ for }\frac{y}{x} \approx 1$$

Log differences are approximately the percent change. Why? Take the first-order taylor expansion of the log to approximate it with a line near 1.

\begin{align\*}
\log(Y) &\approx \log(1) + \frac{1}{\log(1)}\left(Y - 1 \right) \\
&\approx Y - 1
\end{align\*}
Hence $\log\left(\frac{Y}{X} \right) \approx \frac{Y-X}{X}$ for $Y \approx X$. This works well for small changes, eg. $\log(1.03) = .0296$, but it gets quite off for big changes $\log(1.5) = .4055$. This is probably where your stuff is getting off...

## Example where logs are nice...

Let's say we have $1, 2, 4, 8$. Each is a 100% increase from the previous value. Taking logs we have $0, .693, .1386, 2.079$. The numbers increase linearly by $\log(2)$. Looking at percent increases though, it's less clean: $\frac{2-1}{1} = 100\%$ and $\frac{4-2}{2} = 100\%$ but $\frac{4-1}{1} = 300%$.

$$ \left(1 + R\_1 \right) \left(1 + R\_2 \right) = 1 + R\_1 + R\_2 + R\_1R\_2$$
Dealing with percent changes gets messy when $R\_1R\_2$ is big enough to matter. Taking logs though, let $r\_1 = \log(1+R\_1)$ etc...

$$ \log\left((1+R\_1)(1+R\_2) \right)= r\_1 + r\_2$$

## What I would do if I were you:

Take a log and then it will be clean, and simple. Add a few columns to your spreadsheet.

* $q\_t = \log Q\_t$.
* $a = \log A$
* $p\_t = \log P\_t$

Then the equation is:
$$ q\_t = a + b p\_t$$

$$\frac{dq\_t}{dp\_t} = b $$

$$ \begin{array}{rrrrr}
P & Q & \log P & \log Q & \Delta \log P & \Delta \log Q & \frac{\Delta \log P}{\Delta \log Q} \\
1 & 100 & 0 & 4.605 & \\
2 & 25 & .6931 & 3.2189 & .6931 & -1.3863 & -2.0\\
3 & 11.11 & 1.0986 & 2.4079 & .4055 & -.8109 & -2.0\\
4 & 6.25 & 1.3863 & 1.8326 & .2877 & -.5754 & -2.0\\
5 & 4 & 1.6094 & 1.3863 & .2231 & -.4463 & -2.0
\end{array}
$$

$\Delta$ means "change in", hence $\Delta \log Q\_i = \log Q\_i - \log Q\_{i-1}$.


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