<p>This is a standard log-linear model.
<span class="math-container">$log(y)=b_0+b_1log(x)$</span></p>

<p>We know that elasticity is equal to 
<span class="math-container">$e=\frac{dy}{dx}.\frac{x}{y}$</span></p>

<p>We can differentiate the regression equation with respect to <span class="math-container">$x$</span> and then substitute the value of <span class="math-container">$\frac{dy}{dx}$</span> in the elasticity function to get the value.</p>

<p><span class="math-container">$\frac{1}{y}.\frac{dy}{dx}=\frac{b_1}{x}$</span></p>

<p>Therefore <span class="math-container">$\frac{dy}{dx}=b_1\frac{y}{x}$</span></p>

<p>Now substitute it into the elasticity function
<span class="math-container">$e=b_1\frac{y}{x}\frac{x}{y}$</span>
which equals <span class="math-container">$b_1$</span>.</p>

<p>Notice that this is a constant. </p>

CC BY-SA 4.0
2019-02-21T14:11:51.580Z
2019-02-21T14:11:51.580Z
S.Rana
<p>I'm struggling with this question. Please help out with the correct formula. </p>

<p>how to find the elasticity of log(y) =β0+β1log(x).</p>

<econometrics>
how to find the elasticity of log(y) =β0+β1log(x)
Mohuya 
# how to find the elasticity of log(y) =β0+β1log(x)

(Asked by: username_0 on 2019-02-21)

I'm struggling with this question. Please help out with the correct formula.

how to find the elasticity of log(y) =β0+β1log(x).

**Tags:** econometrics

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## 1 answers

### Answer by username_1 on 2019-02-21. Score: 1

This is a standard log-linear model.
$log(y)=b\_0+b\_1log(x)$

We know that elasticity is equal to
$e=\frac{dy}{dx}.\frac{x}{y}$

We can differentiate the regression equation with respect to $x$ and then substitute the value of $\frac{dy}{dx}$ in the elasticity function to get the value.

$\frac{1}{y}.\frac{dy}{dx}=\frac{b\_1}{x}$

Therefore $\frac{dy}{dx}=b\_1\frac{y}{x}$

Now substitute it into the elasticity function
$e=b\_1\frac{y}{x}\frac{x}{y}$
which equals $b\_1$.

Notice that this is a constant.


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