<p>Yes, that is an approximation one does. You have stumbled upon the log-approximation, which is very commonly used in Economics.</p>
<p>It states that <span class="math-container">$log(1+x)≈x$</span>, for very small <span class="math-container">$x$</span>. Hence, typically <span class="math-container">$x$</span> is a fraction. It is also used in macro to approximate growth rates.</p>
<p>There are lots of resources out there that explain it, e.g. <a href="https://planetmath.org/ApproximationOfTheLogFunction#:%7E:text=log(1%2Bx)%E2%89%88x.&amp;text=log-,%E2%81%A1,height%201%20and%20width%20x%20." rel="nofollow noreferrer">here</a>.</p>
<p>The way I remember it is that <span class="math-container">$log(1) =0$</span> so <span class="math-container">$log(1+x)≈x$</span> for small <span class="math-container">$x$</span>. This is mathematically totally wrong, but I find it to be a useful memory device personally.</p>

CC BY-SA 4.0
2021-03-11T05:33:24.573Z
Note if $f(1) = 0$ and $f'(1) = 1$, then $f(1+x) \approx x$. $\log$ satisfies these conditions.
Ramchandra Apte
CC BY-SA 4.0
2021-03-11T10:14:47.867Z
@RamchandraApte How come?
Papayapap
CC BY-SA 4.0
2021-03-11T10:31:11.887Z
@Papayapap, $1=f'(1)=\lim_{x\to 0}\frac{f(1+x)-f(1)}{x}=\lim_{x\to 0}\frac{f(1+x)}{x}$, so $\frac{f(1+x)}{x}\approx 1$ for small $x>0$, hence $f(1+x)\approx x$ for small $x>0$.
VARulle
CC BY-SA 4.0
2021-03-10T21:47:02.247Z
2021-03-10T21:47:02.247Z
BB King
<p>Basic calculus question here! I started reading the textbook Borjas, G. J. (2014). Immigration economics. Harvard University Press and am confused already with the first result.</p>
<p><strong>Background</strong>: He frames the migration decision in terms of wage differences between two countries <span class="math-container">$i \in \{0,1\}$</span> in each country (home is 0)
<span class="math-container">$$ 
log (w_i) = \mu_i + v_i 
$$</span>
where <span class="math-container">$\mu_i$</span> is the average wage in home and potential destination and <span class="math-container">$v_i $</span> is an i.i.d normally distributed random variable with mean 0 and variance <span class="math-container">$\sigma^{2}$</span>. If <span class="math-container">$i=1$</span> is destination country, then migrant will pay a cost <span class="math-container">$C$</span> to move there.</p>
<p><strong>Calculation</strong>: Okay, the simple algebra is then to combine these elements into an index</p>
<p><span class="math-container">$$ 
I = log \left(\frac{w_1}{w_0+C} \right) \approx (\mu_1 - \mu_0 - \pi) + (v_1 - v_0) 
$$</span>
with <span class="math-container">$\pi = C/w_0$</span>.</p>
<p><strong>Steps</strong>: So I did some log manipulations (see below) and think I arrive at this if the <span class="math-container">$log\left(1+\frac{C}{w_0}\right) \approx log\left(\frac{C}{w_0}\right)$</span>. Is this an approximation one does? If yes, do you background at to why? Or is the simplification different?
<span class="math-container">$$ 
log \left(\frac{w_1}{w_0+C} \right) = log (w_1) - log(w_0+C) + log(w_0) - log(w_0) = log (w_1)  - log(w_0) - log\left(1+\frac{C}{w_0}\right) = e^{log(\mu_1 + v_1)} - e^{log(\mu_0 + v_0)} - e^{log\left(1+\frac{C}{w_0}\right)} \approx (\mu_i - \mu_i - \pi) + (v_1 - v_0) 
$$</span></p>

<microeconomics><migration>
Migration wage differential, Borjas (2013) textbook
Papayapap
# Migration wage differential, Borjas (2013) textbook

(Asked by: username_0 on 2021-03-10)

Basic calculus question here! I started reading the textbook Borjas, G. J. (2014). Immigration economics. Harvard University Press and am confused already with the first result.

**Background**: He frames the migration decision in terms of wage differences between two countries $i \in \{0,1\}$ in each country (home is 0)
$$
log (w\_i) = \mu\_i + v\_i
$$
where $\mu\_i$ is the average wage in home and potential destination and $v\_i $ is an i.i.d normally distributed random variable with mean 0 and variance $\sigma^{2}$. If $i=1$ is destination country, then migrant will pay a cost $C$ to move there.

**Calculation**: Okay, the simple algebra is then to combine these elements into an index

$$
I = log \left(\frac{w\_1}{w\_0+C} \right) \approx (\mu\_1 - \mu\_0 - \pi) + (v\_1 - v\_0)
$$
with $\pi = C/w\_0$.

**Steps**: So I did some log manipulations (see below) and think I arrive at this if the $log\left(1+\frac{C}{w\_0}\right) \approx log\left(\frac{C}{w\_0}\right)$. Is this an approximation one does? If yes, do you background at to why? Or is the simplification different?
$$
log \left(\frac{w\_1}{w\_0+C} \right) = log (w\_1) - log(w\_0+C) + log(w\_0) - log(w\_0) = log (w\_1) - log(w\_0) - log\left(1+\frac{C}{w\_0}\right) = e^{log(\mu\_1 + v\_1)} - e^{log(\mu\_0 + v\_0)} - e^{log\left(1+\frac{C}{w\_0}\right)} \approx (\mu\_i - \mu\_i - \pi) + (v\_1 - v\_0)
$$

**Tags:** microeconomics, migration

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## 1 answers

### Answer by username_1 on 2021-03-10. Score: 4 | Accepted answer

Yes, that is an approximation one does. You have stumbled upon the log-approximation, which is very commonly used in Economics.

It states that $log(1+x)≈x$, for very small $x$. Hence, typically $x$ is a fraction. It is also used in macro to approximate growth rates.

There are lots of resources out there that explain it, e.g. [here](https://planetmath.org/ApproximationOfTheLogFunction#:%7E:text=log(1%2Bx)%E2%89%88x.&text=log-,%E2%81%A1,height%201%20and%20width%20x%20.).

The way I remember it is that $log(1) =0$ so $log(1+x)≈x$ for small $x$. This is mathematically totally wrong, but I find it to be a useful memory device personally.

> **Comments:**
> **username_2**: Note if $f(1) = 0$ and $f'(1) = 1$, then $f(1+x) \approx x$. $\log$ satisfies these conditions.  
> **username_0**: @RamchandraApte How come?  
> **username_3**: @Papayapap, $1=f'(1)=\lim_{x\to 0}\frac{f(1+x)-f(1)}{x}=\lim_{x\to 0}\frac{f(1+x)}{x}$, so $\frac{f(1+x)}{x}\approx 1$ for small $x>0$, hence $f(1+x)\approx x$ for small $x>0$.  

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