# Trouble simplifying a tough equation

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

## 3 Answers

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

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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

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

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