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If you wanted to see the following (price $S_t$, log return $r$, simple return $R$) then $$ r = \log(S_{t+1}) - \log(S_t) = \log(S_{t+1}/S_{t}), $$ and $$ R = S_{t+1}/S_{t}-1, $$ thus $$ R = \exp(r)-1 $$ and $$r = \log(1+R).$$ Was this the question?



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