3

I'm stuck trying to solve this: $$e^x=1.25 (1+x). $$

I've looked at simple questions that use Lambert's $W,$ but it's not helping me in this case.

By trial and error I get to $0.824$, but I'd really like to know if there's a way to get an exact solution. Thanks!

Didgeridoo.
  • 37
  • 5
Kael
  • 33
  • 4
  • Why hasn't the Lambert W-function worked for you here? It seems like the best you can do is to express $x$ as a function of W. – Theoretical Economist Feb 10 '18 at 17:33
  • WA can't do better than that: https://www.wolframalpha.com/input/?i=e%5Ex%3D1.25(1%2Bx) – Arnaud Mortier Feb 10 '18 at 17:36
  • The Lambert-W function do work, but to get the number 0.8243883090329847, you need to use the value on the branch typically labeled as -1. On WA, you can get that branch using the command -ProductLog[-1,-\frac{4}{5e}]-1. There is another solution at $-0.5283280902564782$ which corresponds to principal branch (the branch $0$) of Lambert W function. – achille hui Feb 10 '18 at 18:11

1 Answers1

4

The equation in question is $$e^{x} = \frac{5}{4} \, (1 + x).$$ Multiply both sides by $e$ to obtain the following: \begin{align} e^{1 + x} &= \frac{5 e}{4} \, (1 + x) \\ \frac{4}{5 \, e} &= (1+x) \, e^{-(1+x)} \\ -(1+x) \, e^{-(1+x)} &= - \frac{4}{5 \, e} \\ -(1+x) &= W\left(- \frac{4}{5 \, e}\right) \end{align} which leads to the result $$x = - 1 - W\left(- \frac{4}{5 \, e}\right).$$

Leucippus
  • 26,329