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Solow steady state ratio using Cobb-Douglas function

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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)

Apr 20, 2018

#1
+27902
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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 $$k^*=k^\alpha$$.

Apr 20, 2018

#1
+27902
+2
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 $$k^*=k^\alpha$$.