Help with derivation

Question

I am trying to understand how the authors jumped from these two equations:

$k^d=k^s$

• $k^d=ln[\bar{l}\alpha^{\frac{1}{1-\alpha}}]-\frac{1}{1-\alpha}\tilde{R}$
• $k^s=ln[\bar{l}(1-\tau)(1-\alpha)\alpha^{\frac{\alpha}{1-\alpha}}]+ln(\frac{\beta_{R,\sigma}}{1+\beta_{R,\sigma}})-\frac{\alpha}{1-\alpha}\tilde{R}+\tilde{A}$

If $\sigma=1$ then $\beta_{R,\sigma} = β$ and $\tilde{R}$ and $k$ has the following closed form expression:

• $\tilde{R}=-\tilde{A}+ln[(\frac{1+\beta}{\beta})(\frac{\alpha}{1-\alpha})(\frac{1}{1-\tau})]$
• $k=-\frac{1}{1-\alpha}\tilde{A}+\frac{1}{1-\alpha}ln[(\frac{1+\beta}{\beta})(\frac{\alpha}{1-\alpha})(\frac{1}{1-\tau})]+ln[\bar{l}\alpha^{\frac{1}{1-\alpha}}]$

I am attaching the appendix of the paper here, equations 30-34. I'm just not following how it is derived but I'm sure it must be something so simple but I'm just not seeing it. If it doesn't meet the page's standard just let me know. Thanks in advance!

Answer 1

According to the paper (equations $(31)$ and $(32)$) we have: $$ \begin{align*} &k^d = \ln(\overline{l} \alpha^{\frac{1}{1-\alpha}})- \frac{1}{1-\alpha}\tilde{R},\\ &k^s = \ln(\bar l(1-\tau)(1-\alpha)\alpha^{\frac{\alpha}{1-\alpha}}) + \ln\left(\frac{\beta}{1 + \beta}\right) - \frac{\alpha}{1- \alpha}\tilde R + \tilde A \end{align*} $$ You made 2 typo's for $k^s$. First the exponent on $\alpha$ in the $\ln$ function is $\alpha/(1-\alpha)$ and not $1/(1-\alpha)$. Second the singn on the term $\dfrac{\alpha}{1 - \alpha} \tilde R$ has to be reversed.

The solution is obtained by simply setting $k^d = k^s$ and computing for $\tilde R$.