Log linearization of Tobin Q equation (SW, 2003)

Hi everybody,
I am struggling with the following log linearization (it is the FOC for K(t) in SW, 2003); in particular, given that functional form for the capital utilization function, I was wondering whether I have to take first order taylor expansion of the entire function when log linearizing(in that case, I use the entire function evaluated at utilization rate=1 which is the stady steady state value of capital utilization as a pivotal point) or just of the exponential term (in that case, I just use utilization rate = 1 as a pivotal point). I’ve tried both ways, but I am not able to obtain equation 30. Would anyone be so helpful to give me some insights? Thank everybody for the attention
tobin q log linearization.pdf (443.3 KB)

You need to do a full Taylor of the whole equation.

What do you mean exactly by doing a full Taylor expansion of the whole equation? I have always done log linearization on the single terms (ie. x’=log(x(t)) - log(x) ==>x(t)=xexp(x’)==>x(t)=x(1+x’)) and I have obtained the rest of the equations in the SW paper this way. In this particular case, I do not kwon how to treat the exponential term in the variable capital utilization function, as if I applied the aforementioned method, I wouldn’t get rid of the exponential. Thanks for being helpful.

I case like this it’s easier to do a Taylor approximation than the Uhlig trick.
{e^{\psi \left( {{u_{t + 1}} - 1} \right)}} \approx \psi {e^{\psi \left( {u - 1} \right)}}\left( {{u_{t + 1}} - u} \right) = \psi {e^{\psi \left( {u - 1} \right)}}u\frac{{\left( {{u_{t + 1}} - u} \right)}}{u} = u\psi {e^{\psi \left( {u - 1} \right)}}{\hat u_{t + 1}}

Since the steady state value of capital utilization rate is 1 by assumption, is it correct to claim that the resulting value for that expression is u(hat)(t+1)? If so, can I procceed with the Uhlig trick in the rest of equation? I have tried so but I get something very different from the SW paper.
Thank you in advance

Given that the steady state is 1, percentage deviations and absolute deviations are the same. Thus, the answer should be yes.