December 14th, 2016, 06:37 AM  #1 
Newbie Joined: Dec 2016 From: Germany, Magdeburg Posts: 3 Thanks: 0  Solving e^x  x = a
What's the mechanism for solving an equation of this kind?
Last edited by skipjack; December 17th, 2016 at 03:37 AM. 
December 14th, 2016, 07:00 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372 
You're going to have to solve this numerically. There is a closed form solution $x = W\left(e^{a}\right)a$ where $W(x)$ is the Lambert W function but $W(x)$ doesn't generally show up on your calculator. There are no lack of numerical algorithms to solve equations like this. Newton's method is a good place to start. 
December 14th, 2016, 07:13 AM  #3 
Newbie Joined: Dec 2016 From: Germany, Magdeburg Posts: 3 Thanks: 0 
Thanks a lot! I'll start with learning about the Lambert function and the Newton method.
Last edited by skipjack; December 14th, 2016 at 10:35 AM. 
December 16th, 2016, 04:59 PM  #4 
Member Joined: Dec 2016 From:  Posts: 62 Thanks: 10 
This is a trascendental equation, where the function on the left equals the function on the right: \begin{eqnarray} e^{x}=x+a \end{eqnarray} The solution is calculated graphically, by plotting both functions as a function of $x$. All possible solutions to the equation are then those values of $x$ where both graphs cross. 
December 17th, 2016, 03:48 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,831 Thanks: 2160 
If $a$ < 1, there is no real solution. If $a$ = 1, $x$ = 0 is the only real solution. If $a$ > 1, there are two real solutions.


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equation, euler, solving 
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