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let $P0,P1 \leq0$ $s.t$ $P0+P1=1$ (I am not sure if this assumption is required to prove the following equality. But for my application this holds). How does following hold true

\begin{equation} e^{\log(\frac{P1}{P0}+1)}=1+e^{\log(\frac{P1}{P0})} \end{equation}

And how can this be generalized for $n$ positive real numbers $s.t$ $P0+P1+…+Pn=1$ where $Px\leq0$ for $x=1:n$

NAASI
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    $$e^{\ln(\frac{P_1}{P_0}+1)}=\frac{P_1}{P_0}+1=1+\frac{P_1}{P_0}=1+e^{\ln( \frac{P_1}{P_0})} $$ – anon Mar 30 '15 at 17:57

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