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)
+33616 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 \(k^*=k^\alpha\).
+33616 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 \(k^*=k^\alpha\).
