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I'm having trouble simplifying the following equation. I've tried grouping terms in different ways, but it's not looking any more joyful. Can someone please help with its resolution? Hopefully by midnight?? $$ \omega - \ln Y=\ln\left(H p^2 a + \exp(ra)\right) - \ln N $$

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What do you want to do with the equation ? –  Amr Dec 31 '12 at 23:38
    
OH. Now I see (after looking at the answers)! –  Amr Jan 1 '13 at 0:13
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3 Answers 3

up vote 5 down vote accepted

$$ \omega - \ln Y=\ln\left(H p^2 a + \exp(ra)\right) - \ln N $$ $$ \omega =\ln\left(H p^2 a + \exp(ra)\right)+ \ln Y - \ln N $$ $$ \omega =\ln\left(H p^2 a + \exp(ra)\right)+ \ln (Y/N ) $$ $$ \omega =\ln\left({YH p^2 a + Y\exp(ra)\over N}\right) $$ $$ \omega =\ln\left({Ha p^2Y + Ye^{ar})\over N}\right) $$ $$e^{\omega}=\frac{Ha p^2Y + Ye^{ar}}{N}$$ $$0=Ha ppY-Ne^{\omega} + Ye^{ar}$$

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Or just $HappY = Ne^{\omega} - Ye^{ar}$ :). –  mjqxxxx Jan 1 '13 at 0:14
    
O.K. THats sounds better –  Adi Dani Jan 1 '13 at 0:16
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Assuming everything there are numbers (or at least, everything commutes), may be you are wishing the community "$HappY Ne^\omega Ye^{ar}$".

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$1=HappY(Ne^\omega)^{-1}Ye^{ar}$ to be more accurate. –  c.p. Jan 1 '13 at 0:01
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Here, I solved for $a$ for you:

$$ a = \frac{N r e^\omega-H p^2 Y \cdot \mathrm{W}\left(\frac{e^{\frac{e^\omega N r}{H p^2 Y}} r}{H p^2}\right)}{H p^2 r Y} $$

Where $\mathrm{W}(x)$ is the Lambert-W function.

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Technically correct. –  Thomas Jan 1 '13 at 3:55
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