I can't solve the equation (for variable $\displaystyle E$): $\displaystyle e^{-\frac{E}{T}}*(T+E)=0.01aT$.

Help!

Thanks!

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- Aug 5th 2010, 02:20 AMRieszExponential equation
I can't solve the equation (for variable $\displaystyle E$): $\displaystyle e^{-\frac{E}{T}}*(T+E)=0.01aT$.

Help!

Thanks! - Aug 5th 2010, 02:35 AMsa-ri-ga-ma
Can you explain the symbols in the equation?

- Aug 5th 2010, 03:04 AMRiesz
Sure, sorry!

$\displaystyle E$ is the unknown variable, $\displaystyle a$ and $\displaystyle T$ are known positive constant. - Aug 5th 2010, 03:07 AMHallsofIvy
That cannot be solved in terms of elementary functions. You can do this:

Let u= T+ E. Then E= u- T so $\displaystyle \frac{E}{T}= \frac{u}{T}- 1$ and $\displaystyle e^{-\frac{E}{T}}= e^{\frac{u}{T}e^{-T}$

The equation is now $\displaystyle e^{-T}ue^{u/T}= 0.01aT$ or $\displaystyle ue^{u/T}= 0.01aTe^T$.

Now, let v= u/T so that u= TV. The equation becomes $\displaystyle Tve^v= 0.01aTe^T$ or $\displaystyle ve^v= 0.01ae^T$.

Now you can use "Lambert's W function" which is defined as the inverse to the function $\displaystyle f(x)= xe^x$

Then $\displaystyle W(ve^v)= v= W(0.01ae^T)$.

Then, of course, $\displaystyle u= Tv= TW(0.01ae^T)$ and $\displaystyle E= u- T= TW(0.01ae^T)- T or u= T(W(0.01ae^T)- 1)$. - Aug 5th 2010, 08:29 AMRiesz
Let $\displaystyle u=T+E$.

Then $\displaystyle E=u-T$, so $\displaystyle \frac{E}{T}=\frac{u}{T}-1$ and $\displaystyle e^{-\frac{E}{T}}=e^{-\frac{u}{T}}*e^1$

The equation is now $\displaystyle e^{-\frac{u}{T}}e^1*u=0.01aT$.

Now, let $\displaystyle v=\frac{u}{T}$, so that $\displaystyle u=Tv$.

The equation becomes $\displaystyle e^{-v}e*Tv=0.01aT$, or $\displaystyle e^{-v}v=0.01a*e^{-1}$.

Forgive me: where am I wrong?