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!
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 $$x = - 1 - W\left(- \frac{4}{5 \, e}\right).$$