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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!

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  • $\begingroup$ I haven't read the paper but this seems to be really basic calculation. Have you considered the possibility of a typo if you struggle to get the solution? Typos appear in almost every math-heavy paper. $\endgroup$ Jul 7 at 0:11
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    $\begingroup$ Looks like you just have to set equation 2 and 3 equal and isolate $\tilde R$ ... tedious most likely but simple. Then backinsert $\tilde R$ and solve for $k := k_s = k_d$. $\endgroup$ Jul 7 at 0:47
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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 $\tilde R$.

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