# Solve exponential inequality

I've come across the following exponential inequality and, unfortunately, I encountered some difficulties trying to solve it.

$$\left | x \right |^{2x^2 - 3x + 1} \leq 1, x \in \mathbb{R}$$

I tried to solve it by rewriting $1$ as $\left | x \right | ^ 0$. After this substitution I got the equivalent inequation $2x^2 - 3x + 1 \leq 1$, which led to $x \in \left [ \frac{1}{2}, 1 \right ]$. I am not sure about whether this way of solving it is correct or not.

The solution provided by the book approaches two cases, one with $\left | x \right | \leq 1$ and the other one with $\left | x \right | > 1$. Now, here is where I encounter difficulties. I know how to solve the second case, but I don't really know what I should do in the first case, where $\left | x \right | \leq 1$.

So, I think I need some help from you.

• When $|x|<1$, the exponential is not a growing function and the direction of the inequality reverses itself. – Yves Daoust Jul 4 '16 at 8:32

$$\left | x \right |^{2x^2 - 3x + 1} \leq 1\tag1$$

I tried to solve it by rewriting $1$ as $\left | x \right | ^ 0$. After this substitution I got the equivalent inequation $2x^2 - 3x + 1 \leq 1$, which led to $x \in \left [ \frac{1}{2}, 1 \right ]$. I am not sure about whether this way of solving it is correct or not.

(you should mean $2x^2-3x+1\leq \color{red}{0}$) This is not correct. This is not equivalent to $(1)$ because you assume that $|x|\gt 1$. In other words, this is equivalent to $(1)$ only when $|x|\gt 1$. (by the way, there is no $x$ such that $\frac 12\le x\le 1$ and $|x|\gt 1$.)

The solution provided by the book approaches two cases, one with $\left | x \right | \leq 1$ and the other one with $\left | x \right | > 1$. Now, here is where I encounter difficulties. I know how to solve the second case, but I don't really know what I should do in the first case, where $\left | x \right | \leq 1$.

For $|x|=1$ or $x=0$, the inequality $(1)$ holds.

For $0\lt |x|\lt 1$, note that $(1)\iff 2x^2-3x+1\color{red}{\ge}0$.

The answer is that $\color{red}{-1\le x\le\frac 12\quad\text{or}\quad x=1}$.

hint: Take log we have: $(2x^2-3x+1)\log |x| \leq 0$. You can have two cases here to continue, but simple.