<p>Let's first make some substitution. Define <span class="math-container">$g = f^\varepsilon s$</span>. This gives:
<span class="math-container">$$
\begin{align*}
L &amp;= \frac{\alpha g[1 - Ak(1 - \alpha + \alpha g)^{1/\varepsilon} g^{-1/\varepsilon}]}{1 - \alpha(1 + g)},\\
&amp;= \frac{\alpha g[1 - AK(1 - \alpha(1 - g))^{1/\varepsilon}g^{-1/\varepsilon}]}{1 - \alpha(1 + g)},\\
&amp;= \frac{\alpha \left[1 - Ak\left(\frac{1- \alpha(1 - g)}{g}\right)^{1/\varepsilon}\right]}{\frac{1 - \alpha(1+g)}{\alpha g}}
\end{align*}
$$</span>
The two terms on <span class="math-container">$\alpha$</span> and <span class="math-container">$g$</span> in the denominator and numerator are almost the same except from a sign reverseal. So maybe this is a typo. I will continue by assuming that it is.</p>
<p>Assume that the signs on <span class="math-container">$g$</span> were flipped, then define
<span class="math-container">$$
h = \frac{1 + \alpha(1- g)}{g}
$$</span></p>
<p>This allows the condensed form:
<span class="math-container">$$
L = \frac{\alpha[1 - Ak h^{1/\varepsilon}]}{h} \tag{1}
$$</span></p>
<p>Then:
<span class="math-container">$$
\ln L = \ln \alpha + \ln(1 - A k h^{1/\varepsilon}) - \ln h
$$</span></p>
<p>And:
<span class="math-container">$$
\frac{\partial \ln L}{\partial t} = \frac{\partial \ln(1 - A k h^{1/\varepsilon})}{\partial t} - \frac{\partial \ln h}{\partial t}
$$</span></p>
<p>The first derivative on the right hand side equals:
<span class="math-container">$$
\begin{align*}
\frac{\partial (1 - A k h^{1/\varepsilon})}{\partial t} &amp;= 
-\frac{A k h^{1/\varepsilon} \dfrac{\dot k}{k} + \frac{1}{\varepsilon} A k h^{1/\varepsilon}\dfrac{\dot h}{h}}{1 - A k h^{1/\varepsilon}},\\
&amp;=- \frac{A h^{1/\varepsilon}k}{1 - A k h^{1/\varepsilon}}\frac{\dot k}{k}- \frac{1}{\varepsilon} \frac{A k h^{1/\varepsilon} }{1 - A k h^{1/\varepsilon}} \frac{\dot h}{h}
\end{align*}
$$</span>
So:
<span class="math-container">$$
\frac{\dot L}{L} = - \frac{A k h^{1/\varepsilon}}{1 - Ak h^{1/\varepsilon}} \frac{\dot k}{k} - \frac{1}{\varepsilon} \frac{A k h^{1/\varepsilon}}{1 - A k h^{1/\varepsilon}} \frac{\dot h}{h} - \frac{\dot h}{h}
$$</span>
Now we know from <span class="math-container">$(1)$</span> that:
<span class="math-container">$$
1 - A k h^{1/\varepsilon} = \frac{L h}{\alpha},
$$</span>
So:
<span class="math-container">$$
\frac{\dot L}{L} = - \dfrac{1-\dfrac{L h}{\alpha}}{\dfrac{Lh}{\alpha}} \frac{\dot k}{k} - \frac{1}{\varepsilon} \frac{1-\dfrac{Lh}{\alpha}}{\dfrac{Lh}{\alpha}} \frac{\dot h}{h}- \frac{\dot h}{h},\\
= - \left(\frac{\alpha}{Lh} - 1\right)\frac{\dot k}{k} - \frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1\right)\frac{\dot h}{h} - \frac{\dot h}{h},\\
= \left(1- \frac{\alpha}{Lh}\right)\frac{\dot k}{k} - \frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{\dot h}{h}
$$</span>
Next, <span class="math-container">$h = \dfrac{1 + \alpha(1 -g)}{g}$</span> so taking logs gives:
<span class="math-container">$$
\ln h = \ln (1  + \alpha(1 - g)) - \ln g
$$</span>
taking the derivative with respect to time gives:
<span class="math-container">$$
\frac{\dot h}{h} = \frac{-\alpha g}{1 + \alpha(1 - g)}\frac{\dot g}{g} - \frac{\dot g}{g},\\
= - \frac{1 + \alpha}{1 + \alpha( 1 - g)}\frac{\dot g}{g},\\
= -\frac{(1+\alpha)}{hg} \frac{\dot g}{g},\\
= -\frac{hg + \alpha g}{hg} \frac{\dot g}{g},\\
= \frac{h + \alpha}{h} \frac{\dot g}{g}
$$</span>
Finally, <span class="math-container">$g = f^\varepsilon s$</span> so:
<span class="math-container">$$
\frac{\dot g}{g} = \varepsilon \frac{\dot f}{f} + \frac{\dot s}{s}
$$</span>
Together this gives:
<span class="math-container">$$
\begin{align*}
\frac{\dot L}{L} &amp;= \left(1 - \frac{\alpha}{Lh}\right) \frac{\dot k}{k} ,\\
&amp;+ \left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{h + \alpha}{h}\frac{\dot f}{f},\\
&amp;+\frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{h + \alpha}{h}\frac{\dot s}{s}
\end{align*}
$$</span></p>

CC BY-SA 4.0
2021-11-25T14:08:30.783Z
2021-11-25T14:08:30.783Z
tdm
<p>What are you trying to achieve?
if your aim is simply log linearizing the equation, the best you can get is</p>
<p><span class="math-container">$$ 
ln(L) = ln(a) + {\epsilon}ln(f)+ ln(s)+ ln[1-Ak(1-\alpha+\alpha f^{\epsilon}s)^{\frac{1}{\epsilon}} f^{-1} s^\frac {-1}{\epsilon}] - ln[ 1-\alpha(1+f^{\epsilon}s)]
$$</span></p>

CC BY-SA 4.0
2021-11-22T16:53:22.907Z
2021-11-22T16:53:22.907Z
Macosso
<p>As a solution to an economic model, I end up with the following equation:
<span class="math-container">$$L=\frac{\alpha f^{\epsilon}s\left [ 1-Ak(1-\alpha+\alpha f^{\epsilon}s)^{\frac{1}{\epsilon}} f^{-1} s^\frac {-1}{\epsilon}\right ]}{\left [ 1-\alpha(1+f^{\epsilon}s) \right ]}$$</span></p>
<p>Is there any way to simplify it further and make it more readable, e.g. by differentiation or log-linearisation? How would it become in the latter case using the approximation that <span class="math-container">$ln(1+x)\approx x$</span>?</p>
<p>I thank in advance anyone who takes a minute to help me.</p>

<macroeconomics><microeconomics><mathematical-economics>
Help with a (simple) log-linearisation
Alessandro
# Help with a (simple) log-linearisation

(Asked by: username_0 on 2021-11-22)

As a solution to an economic model, I end up with the following equation:
$$L=\frac{\alpha f^{\epsilon}s\left [ 1-Ak(1-\alpha+\alpha f^{\epsilon}s)^{\frac{1}{\epsilon}} f^{-1} s^\frac {-1}{\epsilon}\right ]}{\left [ 1-\alpha(1+f^{\epsilon}s) \right ]}$$

Is there any way to simplify it further and make it more readable, e.g. by differentiation or log-linearisation? How would it become in the latter case using the approximation that $ln(1+x)\approx x$?

I thank in advance anyone who takes a minute to help me.

**Tags:** macroeconomics, microeconomics, mathematical-economics


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## 2 answers

### Answer by username_1 on 2021-11-25. Score: 2

Let's first make some substitution. Define $g = f^\varepsilon s$. This gives:
$$
\begin{align\*}
L &= \frac{\alpha g[1 - Ak(1 - \alpha + \alpha g)^{1/\varepsilon} g^{-1/\varepsilon}]}{1 - \alpha(1 + g)},\\
&= \frac{\alpha g[1 - AK(1 - \alpha(1 - g))^{1/\varepsilon}g^{-1/\varepsilon}]}{1 - \alpha(1 + g)},\\
&= \frac{\alpha \left[1 - Ak\left(\frac{1- \alpha(1 - g)}{g}\right)^{1/\varepsilon}\right]}{\frac{1 - \alpha(1+g)}{\alpha g}}
\end{align\*}
$$
The two terms on $\alpha$ and $g$ in the denominator and numerator are almost the same except from a sign reverseal. So maybe this is a typo. I will continue by assuming that it is.

Assume that the signs on $g$ were flipped, then define
$$
h = \frac{1 + \alpha(1- g)}{g}
$$

This allows the condensed form:
$$
L = \frac{\alpha[1 - Ak h^{1/\varepsilon}]}{h} \tag{1}
$$

Then:
$$
\ln L = \ln \alpha + \ln(1 - A k h^{1/\varepsilon}) - \ln h
$$

And:
$$
\frac{\partial \ln L}{\partial t} = \frac{\partial \ln(1 - A k h^{1/\varepsilon})}{\partial t} - \frac{\partial \ln h}{\partial t}
$$

The first derivative on the right hand side equals:
$$
\begin{align\*}
\frac{\partial (1 - A k h^{1/\varepsilon})}{\partial t} &=
-\frac{A k h^{1/\varepsilon} \dfrac{\dot k}{k} + \frac{1}{\varepsilon} A k h^{1/\varepsilon}\dfrac{\dot h}{h}}{1 - A k h^{1/\varepsilon}},\\
&=- \frac{A h^{1/\varepsilon}k}{1 - A k h^{1/\varepsilon}}\frac{\dot k}{k}- \frac{1}{\varepsilon} \frac{A k h^{1/\varepsilon} }{1 - A k h^{1/\varepsilon}} \frac{\dot h}{h}
\end{align\*}
$$
So:
$$
\frac{\dot L}{L} = - \frac{A k h^{1/\varepsilon}}{1 - Ak h^{1/\varepsilon}} \frac{\dot k}{k} - \frac{1}{\varepsilon} \frac{A k h^{1/\varepsilon}}{1 - A k h^{1/\varepsilon}} \frac{\dot h}{h} - \frac{\dot h}{h}
$$
Now we know from $(1)$ that:
$$
1 - A k h^{1/\varepsilon} = \frac{L h}{\alpha},
$$
So:
$$
\frac{\dot L}{L} = - \dfrac{1-\dfrac{L h}{\alpha}}{\dfrac{Lh}{\alpha}} \frac{\dot k}{k} - \frac{1}{\varepsilon} \frac{1-\dfrac{Lh}{\alpha}}{\dfrac{Lh}{\alpha}} \frac{\dot h}{h}- \frac{\dot h}{h},\\
= - \left(\frac{\alpha}{Lh} - 1\right)\frac{\dot k}{k} - \frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1\right)\frac{\dot h}{h} - \frac{\dot h}{h},\\
= \left(1- \frac{\alpha}{Lh}\right)\frac{\dot k}{k} - \frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{\dot h}{h}
$$
Next, $h = \dfrac{1 + \alpha(1 -g)}{g}$ so taking logs gives:
$$
\ln h = \ln (1 + \alpha(1 - g)) - \ln g
$$
taking the derivative with respect to time gives:
$$
\frac{\dot h}{h} = \frac{-\alpha g}{1 + \alpha(1 - g)}\frac{\dot g}{g} - \frac{\dot g}{g},\\
= - \frac{1 + \alpha}{1 + \alpha( 1 - g)}\frac{\dot g}{g},\\
= -\frac{(1+\alpha)}{hg} \frac{\dot g}{g},\\
= -\frac{hg + \alpha g}{hg} \frac{\dot g}{g},\\
= \frac{h + \alpha}{h} \frac{\dot g}{g}
$$
Finally, $g = f^\varepsilon s$ so:
$$
\frac{\dot g}{g} = \varepsilon \frac{\dot f}{f} + \frac{\dot s}{s}
$$
Together this gives:
$$
\begin{align\*}
\frac{\dot L}{L} &= \left(1 - \frac{\alpha}{Lh}\right) \frac{\dot k}{k} ,\\
&+ \left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{h + \alpha}{h}\frac{\dot f}{f},\\
&+\frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{h + \alpha}{h}\frac{\dot s}{s}
\end{align\*}
$$


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### Answer by username_2 on 2021-11-22. Score: 0

What are you trying to achieve?
if your aim is simply log linearizing the equation, the best you can get is

$$
ln(L) = ln(a) + {\epsilon}ln(f)+ ln(s)+ ln[1-Ak(1-\alpha+\alpha f^{\epsilon}s)^{\frac{1}{\epsilon}} f^{-1} s^\frac {-1}{\epsilon}] - ln[ 1-\alpha(1+f^{\epsilon}s)]
$$


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