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Aug 31 '21 at 3:16 history became hot network question
Aug 30 '21 at 22:49 answer Michael Greinecker timeline score: 3
Aug 30 '21 at 19:34 comment added Jsck No it's not: $AK^{\alpha}L^{1-\alpha}=A\left(\frac{\alpha}{1-\alpha}\frac{W}{R}L\right)^{\alpha}L^{1-\alpha}=A\left(\frac{\alpha}{1-\alpha}\frac{W}{R}\right)^{\alpha}L^{\alpha+1-\alpha}=A\left(\frac{\alpha}{1-\alpha}\frac{W}{R}\right)^{\alpha}L$
Aug 30 '21 at 19:29 comment added Bertrand The term $L^{1-\alpha}$ is missing in the last eq.
Aug 30 '21 at 19:25 comment added Jsck Sorry for the typo, I put them.
Aug 30 '21 at 19:23 history edited Jsck CC BY-SA 4.0
edited title
Aug 30 '21 at 19:20 comment added Bertrand Where are the power $\alpha$ and $1-\alpha$ in the expression of $Y=...$?
S Aug 30 '21 at 19:15 review First questions
Aug 31 '21 at 7:33
S Aug 30 '21 at 19:15 history asked Jsck CC BY-SA 4.0