Log-linearization of the market clearing condition I am dealing with a paper of Walsh &amp; Ravenna.

www.banque-france.fr/fondation/gb/telechar/bourses_recherche/Welfare-based_Ravenna.pdf

I am kind of confused by the euqation (19) on page 33.

The market clearing condition goes as follows:

$$Y_{t} = C_{t} - w^{u}(1 - N_{t}) + \kappa\upsilon_{t}$$

Log linearization around the steady state yields

$$\hat{y}_{t} = \frac{\bar{C}}{\bar{Y}}\hat{c}_{t} - w^{u}\hat{n}_{t} + \left( \frac{\kappa\bar{\upsilon}}{\bar{Y}}\right)(\hat{\Theta}_{t} + \hat{u}_{t} )    \;\;\; \mathbf{(19)} $$

I don't know how this formula (19) is derived?
Isn't there something missing? From my basic understanding of log linearization it ought to look like this:

$$\hat{y}_{t} = \frac{\bar{C}}{\bar{Y}}\hat{c}_{t} - \left( \mathbf{\frac{\bar{N}}{\bar{Y}}} \right) w^{u}\hat{n}_{t} + \left( \frac{\kappa\bar{\upsilon}}{\bar{Y}}\right)(\hat{\Theta}_{t} + \hat{u}_{t} )$$

with
$$ Y_{t}\;\;...\;\; output $$
$$ C_{t}\;\;...\;\;  consumption $$
$$ w^{u}\;\;...\;\;  wage\;of\;unmatched\;workers $$
$$ 1-N_{t}\;\;...\;\;  unmatched\;workers $$
$$ w^{u}(1-N_{t})\;\;...\;\;home\;production$$
$$ \kappa\;\;...\;\;  cost\;of\;posting\;vacancy $$
$$ v_{t}\;\;...\;\;  vacancies $$
$$ \hat{v}_{t} = (\hat{\Theta}_{t} + \hat{u}_{t} ) $$
$$ \omega = \frac{v_{t}}{u_{t}}\;\;...\;\;measure\;of\;labour\;market\;tightness$$
$$ \hat{\cdot}\;\;...\;\;log\;deviation\;from\;steady\;state\;value$$
$$ \bar{\cdot}\;\;...\;\;steady\;state\;value$$

Small letter with a hat: log deviation of a variable arount its steady state.
Big letter with a bar: steady state value.
K: cost of posting a job vancancy.
w^u: "wage" of unemployed workers.

I've read the paper and the appendix aswell, read both the papers in the bibliography of this one aswell as later ones basing on this publication, but I could not find a helpful hint.

Is there any special relationship between N and Y in the steady state that explains why this whole term vanishs? Or do I have a wrong understanding of log-linearization?

I have to apologyze for my rusty English. I am already taking care of this problem. But for the above mentioned I would like to have your help. Does anyone has a decisive hint?
We have:

$$y_t = c_t − w^u(1 − N_t) + \kappa v_t$$

We take the log of both sides:

$$\ln y_t = \ln \left[ c_t − w^u(1 − N_t) + \kappa v_t \right]$$

And then linearize around the steady states:

$$\begin{align}
\ln \bar{y} + \frac{1}{\bar{y}}(y_t - \bar{y}) &amp; = \ln \left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right] + \frac{1}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(c_t - \bar{c}) \\
&amp; + \frac{w_u}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(N_t - \bar{N}) \\
&amp; + \frac{\kappa}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(V_t - \bar{V})
\end{align}
$$

Cancel $\ln \bar{y}$ and $\ln \left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]$

$$\begin{align}
\frac{1}{\bar{y}}(y_t - \bar{y}) &amp; = \frac{1}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(c_t - \bar{c}) \\
&amp; + \frac{w_u}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(N_t - \bar{N}) \\
&amp; + \frac{\kappa}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}(V_t - \bar{V})
\end{align}
$$

Multiply the first term on the right hand side by $\frac{\bar{c}}{\bar{c}}$, and similarly for the other terms:

$$\begin{align}
\frac{1}{\bar{y}}(y_t - \bar{y}) &amp; = \frac{\bar{c}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(c_t - \bar{c})}{\bar{c}} \\
&amp; + \frac{w_u \bar{N}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(N_t - \bar{N})}{\bar{N}} \\
&amp; + \frac{\kappa \bar{V}}{\left[ \bar{c} - w^u(1-\bar{N}) + \kappa \bar{V} \right]}\frac{(V_t - \bar{V})}{\bar{V}}
\end{align}
$$

And we simplify, taking advantage of the fact that $\hat{v}_t = \hat{\theta}_t + \hat{u}_t$, to get what you have:

$$\hat{y}_t = \frac{\bar{C}}{\bar{Y}}\hat{c}_t + w^u \frac{\bar{N}}{\bar{Y}} \hat{n}_t + \left( \frac{K\bar{V}}{\bar{Y}} \right) (\hat{\theta}_t + \hat{u}_t) $$

I don't think this derivation is wrong. But you can see that $y_t$ and $N_t$ are directly proportional to each other. It's definitely plausible that $\frac{\bar{N}}{\bar{Y}} = 1$, but I've pored over the equations in the setup for 3 hours to no avail at showing that rigorously. It's probably something silly that I don't understand. I'll look back later and if nothing else, I'll turn this answer into a community wiki so anyone who has the right idea can edit it.