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How do I rearrange the steady state function in the beginning of b into the k* and y* equations? (function from part a is Akα)

 

 

Guest Apr 20, 2018

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 #1
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If we start from \(sAk^\alpha=(\delta+n+g)k\) then divide both sides by \(k^\alpha(\delta+n+g)\) we get \(\frac{sA}{\delta+n+g}=\frac{k}{k^\alpha}\)

 

This is \(\frac{sA}{\delta+n+g}=k^{1-\alpha}\)  Take the \(1-\alpha\) root of both sides to get:  \((\frac{sA}{\delta+n+g})^\frac{1}{1-\alpha}=k\)  which isn't quite the same as your expression for \(k^*\), unless \(k^*=k^\alpha\).

Alan  Apr 20, 2018
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 #1
avatar+26689 
+2
Best Answer

If we start from \(sAk^\alpha=(\delta+n+g)k\) then divide both sides by \(k^\alpha(\delta+n+g)\) we get \(\frac{sA}{\delta+n+g}=\frac{k}{k^\alpha}\)

 

This is \(\frac{sA}{\delta+n+g}=k^{1-\alpha}\)  Take the \(1-\alpha\) root of both sides to get:  \((\frac{sA}{\delta+n+g})^\frac{1}{1-\alpha}=k\)  which isn't quite the same as your expression for \(k^*\), unless \(k^*=k^\alpha\).

Alan  Apr 20, 2018

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