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Take logs of both sides, i.e. $$\log Y=\log A+ a \log K +(1-a)\log L$$ This gives: $$\Delta\log Y = \Delta\log A + a \Delta\log K +(1-a) \Delta\log L$$ Then use that $\frac{d}{dx}\log x= 1/x$, which yields $\Delta\log x=\Delta x/x$. Apply that to each log-diff above.



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