Solve for $x$ given $2.544 = e^{-x} + e^{-2x} + e^{-3x}$

Question

Can this problem be solved algebraically? $$e^{-x}+e^{-2x} + e^{-3x} = 2.544$$

Answer 1

Hint:

Let $e^{-x}=u$ to get $$u+u^2+u^3=2.544$$ when $0<u\le 1$

Answer 2

Let $$y=e^{-x}\quad 0<y\le 1$$

then solve

$$y^3+y^2+y=2.544$$

then $$-x=\ln y \implies x=\ln \frac 1y$$

To prove rigorously that there is only a (real) solution we can consider

$$f(y)=y^3+y^2+y-2.544$$

which is continuous with $f(0)<0$, $f(1)>0$, $f'(x)>0$ and refer to IVT.

Our dear friend Wolfy can help to find the numerical value for that solution.