# Exponential equation

• Aug 5th 2010, 03:20 AM
Riesz
Exponential equation
I can't solve the equation (for variable $E$): $e^{-\frac{E}{T}}*(T+E)=0.01aT$.

Help!

Thanks!
• Aug 5th 2010, 03:35 AM
sa-ri-ga-ma
Can you explain the symbols in the equation?
• Aug 5th 2010, 04:04 AM
Riesz
Sure, sorry!

$E$ is the unknown variable, $a$ and $T$ are known positive constant.
• Aug 5th 2010, 04:07 AM
HallsofIvy
Quote:

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

Help!

Thanks!

That cannot be solved in terms of elementary functions. You can do this:

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

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

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

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

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

Then, of course, $u= Tv= TW(0.01ae^T)$ and $E= u- T= TW(0.01ae^T)- T or u= T(W(0.01ae^T)- 1)$.
• Aug 5th 2010, 09:29 AM
Riesz
Let $u=T+E$.

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

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

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

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

Forgive me: where am I wrong?