1-e^(-x)=5e^(-2x)+3x^(-x)
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I don't think there is an analytical way to solve because of the $\displaystyle x^{-x}$ term. Should that read $\displaystyle e^{-x}$ per chance?
3x^(-x) is correct
If$\displaystyle f(x)=1-e^{-x}-5e^{-2x}-3x^{-x}$ Then, $\displaystyle f(1)<0$ and $\displaystyle f(3)>0$, so there is at least one root in $\displaystyle [1,3]$.
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