Is it possible to solve this exponential equation algebratically, if so, how?

$\begin{aligned} 1+\ln(x) &=x \\ \ln(x) &= x-1 \\ x &= e^{x-1} &&\color{#3D99F6}\text{Exponentiate} \\ x-e^{x-1} &= 0 \end{aligned}$ Let $f(x)=x-e^{x-1}$ . Then $f'(x) = 1-e^{x-1}$ . Solving $f'(x)=0$ yields $x=1$ as the only critical point. Evaluating $f''(1) = -e^0 = -1$ tells us that $f(1)=0$ is the maximum value of $f(x)$ . Thus (since $f(x)$ is everywhere differentiable) $x=\boxed{1}$ it is the only root of $f(x)$ and, therefore, the only solution to our original equation.