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 $
Trouble simplifying a tough equation
7
$\begingroup$
recreational-mathematics
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0OH. Now I see (after looking at the answers)! – 2013-01-01
3 Answers
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$ \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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0O.K. THats sounds better – 2013-01-01
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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$1=HappY(Ne^\omega)^{-1}Ye^{ar}$ to be more accurate. – 2013-01-01
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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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0Technically correct. – 2013-01-01