<p>According to the paper (equations <span class="math-container">$(31)$</span> and <span class="math-container">$(32)$</span>) we have:
<span class="math-container">$$
\begin{align*}
&amp;k^d = \ln(\overline{l} \alpha^{\frac{1}{1-\alpha}})- \frac{1}{1-\alpha}\tilde{R},\\
&amp;k^s = \ln(\bar l(1-\tau)(1-\alpha)\alpha^{\frac{\alpha}{1-\alpha}}) + \ln\left(\frac{\beta}{1 + \beta}\right) - \frac{\alpha}{1- \alpha}\tilde R + \tilde A
\end{align*}
$$</span>
You made 2 typo's for <span class="math-container">$k^s$</span>. First the exponent on <span class="math-container">$\alpha$</span> in the <span class="math-container">$\ln$</span> function is <span class="math-container">$\alpha/(1-\alpha)$</span> and not <span class="math-container">$1/(1-\alpha)$</span>. Second the singn on the term <span class="math-container">$\dfrac{\alpha}{1 - \alpha} \tilde R$</span> has to be reversed.</p>
<p>The solution is obtained by simply setting <span class="math-container">$k^d = k^s$</span> and computing for <span class="math-container">$\tilde R$</span>.</p>

CC BY-SA 4.0
2021-07-07T06:31:09.410Z
2021-07-07T06:31:09.410Z
tdm
<p>I am trying to understand how the authors jumped from these two equations:</p>
<p><span class="math-container">$k^d=k^s$</span></p>
<ul>
<li><span class="math-container">$k^d=ln[\bar{l}\alpha^{\frac{1}{1-\alpha}}]-\frac{1}{1-\alpha}\tilde{R}$</span></li>
<li><span class="math-container">$k^s=ln[\bar{l}(1-\tau)(1-\alpha)\alpha^{\frac{\alpha}{1-\alpha}}]+ln(\frac{\beta_{R,\sigma}}{1+\beta_{R,\sigma}})-\frac{\alpha}{1-\alpha}\tilde{R}+\tilde{A}$</span></li>
</ul>
<p>If <span class="math-container">$\sigma=1$</span>  then <span class="math-container">$\beta_{R,\sigma} = β$</span> and <span class="math-container">$\tilde{R}$</span> and <span class="math-container">$k$</span> has the following closed form expression:</p>
<ul>
<li><span class="math-container">$\tilde{R}=-\tilde{A}+ln[(\frac{1+\beta}{\beta})(\frac{\alpha}{1-\alpha})(\frac{1}{1-\tau})]$</span></li>
<li><span class="math-container">$k=-\frac{1}{1-\alpha}\tilde{A}+\frac{1}{1-\alpha}ln[(\frac{1+\beta}{\beta})(\frac{\alpha}{1-\alpha})(\frac{1}{1-\tau})]+ln[\bar{l}\alpha^{\frac{1}{1-\alpha}}]$</span></li>
</ul>
<p><a href="https://www.aeaweb.org/content/file?id=11085" rel="nofollow noreferrer">I am attaching the appendix of the paper here</a>, 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!</p>

CC BY-SA 4.0
2021-07-07T00:11:38.430Z
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.
Alalalalaki
CC BY-SA 4.0
2021-07-07T00:47:19.553Z
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 $k := k_s = k_d$.
Jesper Hybel
<macroeconomics>
Help with derivation
Maybeline Lee
# Help with derivation

(Asked by: username_0 on 2021-07-06)

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](https://www.aeaweb.org/content/file?id=11085), 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!

**Tags:** macroeconomics
> **Comments:**
> **username_1**: 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.  
> **username_2**: 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 $k := k_s = k_d$.  

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## 1 answers

### Answer by username_3 on 2021-07-07. Score: 4

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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