Exponential equals rational fraction $2^{-3x^3+5x^2-x}=\frac{x^2+1}{x}$

Question

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.

Answer 1

It's obvious that $x>0$.

Thus, by AM-GM $$2^{-3x^3+5x^2-x}=x+\frac{1}{x}\geq2,$$ which gives $$-3x^3+5x^2-x\geq1$$ or $$(x-1)^2(3x+1)\leq0.$$ The equality occurs for $x=1$ only, which gives that $1$ is an unique root.

Answer 2

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 $x=1$.