<s>[INST] Exponential equals rational fraction $2^{-3x^3+5x^2-x}=\frac{x^2+1}{x}$ I'm trying to solve this equation:

$$2^{-3x^3+5x^2-x}=\frac{x^2+1}{x}$$

I can see that $x=1$ is a solution and I'm struggling find.  others.
The exponential on the left side is always positive, so right must also be >0.
$\frac{x^2+1}{x}>0$
$x^2 >0 \implies x^2+1>0$
so $x>0$. I don't know what else to do.
 [/INST] From your work, because $x > 0$, we can apply AM-GM on the right hand side:
$$\frac{x^2+1}{x}\geq \frac{2x}{x}=2$$
Therefore, since the exponential is increasing over $\mathbb{R}$:
$$2^{-3x^3+5x^2-x}\geq 2^1\Rightarrow -3x^3+5x^2-x \geq 1\Rightarrow (3x+1)(x-1)^2\leq 0$$
The unique solution is thus $x=1$.
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