I need to solve the following equation: $$t\, e^t=t+2\, ,$$ for $t>0$. Do I need to use Lambert W function, or there is some other method?


PS. I know that any equation of the form $A+Bt+C\log(D+Et)=0$ has solutions which, if they exist, can be expressed using Lambert function. But I do not think that $t\, e^t=t+2$ can be rewrite in this way... am I wrong?

  • $\begingroup$ Numerical methods I guess. You are correct (to me) when you say that you cannot rewrite as $A+Bt+C\log(D+Et)=0$. For $t>0$, I bet that Newton method will converge in very few iterations using $x_0=1$. $\endgroup$ Jun 17, 2017 at 15:22
  • $\begingroup$ you can also use a numerical method, eg Newton-Raphson method $\endgroup$ Jun 17, 2017 at 15:23
  • $\begingroup$ WolframAlpha does not give an answer using Lambert's function so I'm guessing this cannot be solve that way. $\endgroup$ Jun 17, 2017 at 15:24
  • $\begingroup$ @T.Gunn. WolframAlpha is not the $\alpha$ and $\omega$ in mathematics. I am "almost" sure that there is a solution in terms of the generalized Lambert function. $\endgroup$ Jun 17, 2017 at 15:29

3 Answers 3


I do not believe that this can be solved using the Lambert $W$ function. When one attempts to put it in the form that you specified, one ends up with $$\ln\bigg(\frac{t}{t+2}\bigg)+t=0$$ If we let $\frac{t}{t+2}=u$, then we have $$\ln(u)+\frac{2u}{1-u}=0$$ $$(1-u)\ln(u)+2u=0$$ So if you can find some way of solving equations in the form $$A+Bt+(C+Dt)\ln(E+Ft)=0$$ then this would be possible... however, I am afraid that you may have to attack this problem without the Lambert $W$.

If you would like to approximate, I would recommend using iteration.


As I said in a comment, there is a solution involving the generalized Lambert function. If you look here, C. B. Corcino and R. B. Corcino proved that in the asymptotic expression of the generalized Bell numbers one needs to solve the equation $$x\,e^x + r x=n$$ which is exactly your case. The remaining of the paper addresses the more general case of $$e^{-cx}=a \frac{ \prod_{i=1}^n (x-t_i)} {\prod_{i=1}^m (x-s_i) }$$

Loooking at the posted equation, from a purely numerical point of view, Newton method will converge quite fast using $t_0=1$ observed by inspection. The successive iterates would be $$\left( \begin{array}{cc} 0 & 1.000000000 \\ 1 & 1.063499184 \\ 2 & 1.060100663 \\ 3 & 1.060090320 \end{array} \right)$$ which is the solution for ten significant figures.

We also could use $[1,n]$ Padé approximants built around $t=1$ and get explicit approximations of the solution. For example $$t_{(1)}=1+\frac{2 (3-e) (2 e-1)}{2+e+5 e^2}\approx 1.0599978$$ $$t_{(2)}=1+\frac{3 (3-e) \left(2+e+5 e^2\right)}{2 \left(-3+9 e+15 e^2+8 e^3\right)}\approx 1.0600921$$ $$t_{(3)}=1+\frac{8 (3-e) \left(-3+9 e+15 e^2+8 e^3\right)}{24-21 e+291 e^2+265 e^3+65 e^4}\approx 1.0600903$$

  • $\begingroup$ The paper has defined an inverse function of $x e^x+r x$ called $W_r(x)$ to solve this equation. As the op's question, the answer is $t=W_{-1}(2)$ $\endgroup$
    – mayi
    Jan 7 at 19:44

Here is a plot:


The solutions to $$t{e^t} = t + 2$$ are $$t \approx -2.23865$$ and $$t \approx 1.06009.$$

Credits: I used WolframAlpha.


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