<p>Take the natural logarithm, rewrite</p>
<p><span class="math-container">$$ x_t - c_t = \frac{1}{1 - v}\ln\left[ 1 - \theta + \theta e^{(1-v)(l_t - c_t)}\right]$$</span></p>
<p>Notice that I'm doing nothing, since <span class="math-container">$e^{(1-v)(l_t - c_t)}= (L_t/C_t)^{1-v}$</span></p>
<p>Substitute <span class="math-container">$ y_t = l_t - c_t$</span> and take a First Order Taylor Expansion around <span class="math-container">$l - c = 0$</span> (differentiate w.r.t. <span class="math-container">$y$</span>), and notice that you can rewrite <span class="math-container">$1 = L/C$</span> and you get the expression.</p>
<p>P.S. Are you sure that the last <span class="math-container">$l_t - c_t$</span> is not a log deviation? In theory, you linearize around the steady state!</p>

CC BY-SA 4.0
2022-10-15T11:33:01.703Z
This is from Monetary policy, inflation and the business cycle (Gali, 2008), pg. 29. the book says it's doing a log linearization around a zero inflation steady state, but it also clearly states that x_t = ln(X_t)
NC520
CC BY-SA 4.0
2022-10-15T12:01:26.963Z
Ok, but what is $L_t$ and what is $C_t$, then? Initially, I thought consumption and leisure, but here? Liquidity and consumption?
Matteo Bulgarelli
CC BY-SA 4.0
2022-10-16T01:24:47.563Z
Consumption and leisure
NC520
CC BY-SA 4.0
2022-10-14T13:39:33.693Z
2022-10-14T19:20:26.533Z
2022-10-14T19:20:26.533Z
Matteo Bulgarelli
<p>Can anyone show how by applying log-linearization we can turn this equation:</p>
<p><span class="math-container">$\frac{X_t}{C_t} = \left( (1 - \vartheta) + \vartheta \left( \frac{L_t}{C_t} \right)^{1-\nu} \right)^\frac{1}{1-\nu}$</span></p>
<p>into this one:</p>
<p><span class="math-container">$x_t - c_t = \frac{1}{1 - \nu} \frac{\vartheta \left( \frac{L}{C}\right)^{1-\nu} }{(1 - \vartheta) + \vartheta \left( \frac{L}{C}\right)^{1-\nu} } (1-\nu) (l_t - c_t)$</span></p>
<p>where values without the subscript <span class="math-container">$t$</span> denote steady state values and values in smaller case denote log values.</p>

CC BY-SA 4.0
2022-10-14T11:24:09.530Z
Source for the equation?
Daneel Olivaw
CC BY-SA 4.0
2022-10-15T11:27:23.050Z
Monetary policy, inflation and the business cycle (Gali, 2008), pg. 29
NC520
<rbc><log-linearization>
Log-linearization in RBC model
NC520
# Log-linearization in RBC model

(Asked by: username_0 on 2022-10-13)

Can anyone show how by applying log-linearization we can turn this equation:

$\frac{X\_t}{C\_t} = \left( (1 - \vartheta) + \vartheta \left( \frac{L\_t}{C\_t} \right)^{1-\nu} \right)^\frac{1}{1-\nu}$

into this one:

$x\_t - c\_t = \frac{1}{1 - \nu} \frac{\vartheta \left( \frac{L}{C}\right)^{1-\nu} }{(1 - \vartheta) + \vartheta \left( \frac{L}{C}\right)^{1-\nu} } (1-\nu) (l\_t - c\_t)$

where values without the subscript $t$ denote steady state values and values in smaller case denote log values.

**Tags:** rbc, log-linearization
> **Comments:**
> **username_1**: Source for the equation?  
> **username_0**: Monetary policy, inflation and the business cycle (Gali, 2008), pg. 29  

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## 1 answers

### Answer by username_2 on 2022-10-14. Score: 1 | Accepted answer

Take the natural logarithm, rewrite

$$ x\_t - c\_t = \frac{1}{1 - v}\ln\left[ 1 - \theta + \theta e^{(1-v)(l\_t - c\_t)}\right]$$

Notice that I'm doing nothing, since $e^{(1-v)(l\_t - c\_t)}= (L\_t/C\_t)^{1-v}$

Substitute $ y\_t = l\_t - c\_t$ and take a First Order Taylor Expansion around $l - c = 0$ (differentiate w.r.t. $y$), and notice that you can rewrite $1 = L/C$ and you get the expression.

P.S. Are you sure that the last $l\_t - c\_t$ is not a log deviation? In theory, you linearize around the steady state!

> **Comments:**
> **username_0**: This is from Monetary policy, inflation and the business cycle (Gali, 2008), pg. 29. the book says it's doing a log linearization around a zero inflation steady state, but it also clearly states that x_t = ln(X_t)  
> **username_2**: Ok, but what is $L_t$ and what is $C_t$, then? Initially, I thought consumption and leisure, but here? Liquidity and consumption?  
> **username_0**: Consumption and leisure  

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