Exponential and ln function

Question

It seems quite simple but how would I find the exact solution for:

$$\exp(x) = -\ln(x) $$

I'm not too sure where to start?

Answer 1

I wouldn't bet on a closed form here and use Newton-Raphson iterations to get :

\begin{align} x_0&:=1\\ x_{n+1}&:=x_n-\frac{f(x_n)}{f'(x_n)}\\ x_{n+1}&:=x_n-\frac{e^{x_n}+\log(x_n)}{e^{x_n}+\dfrac 1{x_n}}\;\\ \end{align} \begin{align} x_0&=1\\ x_1& = 0.2689414213699951207488407582\\ x_2& = 0.2698730572541564559160264039\\ x_3& = 0.2698741375720040468246780600\\ x_4& = 0.2698741375734492238773798650\\ x_5& = 0.2698741375734492238773824512\\ \end{align}

Answer 2

$$e^x=\log \frac{1}{x}$$ take the e for both side $$e^{e^x}=\frac{1}{x}$$ hence $$x=\frac{1}{e^{e^x}}$$ $$x=\frac{1}{e^{e^{\frac{1}{e^{e^{...}}}}}}$$