My Math Forum Solving e^x - x = a

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 December 14th, 2016, 07:37 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 04:37 AM.
 December 14th, 2016, 08:00 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,632 Thanks: 1472 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. Thanks from topsquark and knownothing1994
 December 14th, 2016, 08:13 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 11:35 AM.
 December 16th, 2016, 05:59 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, 04:48 AM #5 Global Moderator   Joined: Dec 2006 Posts: 21,105 Thanks: 2324 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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