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This question refers to the Capital stock in macroeconomics, so K is the capital stock.

Hi everyone. I have this problem where I don't actually understand why you invert the exponent. You still have to remove the exponent in the 4th step so why invert it in the first place. Is this because of the negative exponent it will give because of alpha being so low?

Kevin Abdul
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They want to go from the second line to fifth line (and then probably to $K = \dots$) and are just taking baby steps:

  • 2 to 3: invert both sides;
  • 3 to 4: simplify both sides;
  • 4 to 5: raise both sides to the power $1/(1-\alpha)$.

They might as well have taken all those steps as one; the authors probably believe this is easier to understand for their readers.

Magdiragdag
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  • Is it because of EN:s exponent on the right hand side that they would want to reverse the K^alpha-1 exponent to K^1-alpha? – Kevin Abdul Feb 24 '21 at 12:52
  • I understand the steps, but just now saw why they would want to reverse the K^alpha-1 to K^1-alpha.It is because of EN on the other side having the same exponent. Thus more readability and more easier way to simplify the function. – Kevin Abdul Feb 24 '21 at 12:54
  • It might be that usually $\alpha<1$, so they would prefer to use the positive exponent of $1-\alpha$ rather than of $\alpha-1$. – Jaap Scherphuis Feb 24 '21 at 12:54
  • This is actually quite right. In statistics alpha was always lower than one. That was why I was wondering. If it was because of alpha always being lower than 1 that the person who solved the question reverses K:s exponent or if it was because of it being easier to simplify regards to wanting to have the same exponent as EN. – Kevin Abdul Feb 24 '21 at 12:56
  • So your question is in fact not really about the steps taken, but about the final expression for $K$. Apparently, the authors prefer this version over the version with the fraction the other way around and $\frac{1}{\alpha - 1}$ in the exponent. This is an understandable preference if, as @JaapScherphuis suggests, $\alpha < 1$ usually. Also, it is also nicer that the final expression now has two occurrences of $1 - \alpha$ instead of one occurrence of $1 - \alpha$ and one of $\alpha - 1$. – Magdiragdag Feb 24 '21 at 13:02