Question

( ln x ) 2 − 3 ln x + 3 ln x − 1 < 1

Answer 1

$(e^{r_3}, 1) \cup (e^{r_1}, e^{r_2})$ where $r_1, r_2, r_3$ are roots of $y^3 - 3y^2 - 2y + 3 = 0$.

Answer 2

Let $y = \ln x$. The inequality becomes $y^2 - 3y + \frac{3}{y} - 1 < 1$. The domain requires $x > 0$ and $\ln x \neq 0$, so $x \neq 1$. Thus, $x \in (0, \infty) \setminus \{1\}$. For $y = \ln x$, the domain for $y$ is $(-\infty, \infty) \setminus \{0\}$.

The inequality is $y^2 - 3y + \frac{3}{y} - 2 < 0$.

Let's consider the original inequality again. $(\ln x)^2 - 3 \ln x + \frac{3}{\ln x} - 1 < 1$. Let $y = \ln x$. $y^2 - 3y + \frac{3}{y} - 2 < 0$.

Let's consider the function $f(y) = y^2 - 3y + \frac{3}{y} - 2$. We want to find $y$ such that $f(y) < 0$.

Let's examine the behavior of $f(y)$ as $y \to 0^+$, $y \to 0^-$, $y \to \infty$, $y \to -\infty$. As $y \to 0^+$, $f(y) \approx \frac{3}{y} \to +\infty$. As $y \to 0^-$, $f(y) \approx \frac{3}{y} \to -\infty$. As $y \to \infty$, $f(y) \approx y^2 \to +\infty$. As $y \to -\infty$, $f(y) \approx y^2 \to +\infty$.

We need to find the roots of $f(y) = 0$, which is $y^3 - 3y^2 - 2y + 3 = 0$. Let the roots be $r_1, r_2, r_3$.

The inequality is $\frac{P(y)}{y} < 0$. This holds when $P(y)$ and $y$ have opposite signs.

Case 1: $y > 0$ and $P(y) < 0$. For $y > 0$, the roots of $P(y)$ are $r_1 \in (0, 1)$ and $r_2 \in (3, 4)$. $P(y)$ is positive for $0 < y < r_1$, negative for $r_1 < y < r_2$, and positive for $y > r_2$. So, for $y > 0$, $P(y) < 0$ when $r_1 < y < r_2$.

Case 2: $y < 0$ and $P(y) > 0$. For $y < 0$, the only root of $P(y)$ is $r_3 \in (-2, -1)$. $P(y)$ is negative for $y < r_3$ and positive for $r_3 < y < 0$. So, for $y < 0$, $P(y) > 0$ when $r_3 < y < 0$.

Combining the cases, the solution for $y = \ln x$ is $y \in (r_3, 0) \cup (r_1, r_2)$. Substituting back $y = \ln x$: $r_3 < \ln x < 0 \implies e^{r_3} < x < e^0 = 1$. $r_1 < \ln x < r_2 \implies e^{r_1} < x < e^{r_2}$.

The solution for $x$ is $x \in (e^{r_3}, 1) \cup (e^{r_1}, e^{r_2})$. We know $r_3 \in (-2, -1)$, $r_1 \in (0, 1)$, $r_2 \in (3, 4)$. So $e^{r_3} \in (e^{-2}, e^{-1})$. $e^{r_1} \in (e^0, e^1) = (1, e)$. $e^{r_2} \in (e^3, e^4)$.

The solution set is $(e^{r_3}, 1) \cup (e^{r_1}, e^{r_2})$, where $r_1, r_2, r_3$ are the roots of the equation $(\ln x)^3 - 3(\ln x)^2 - 2(\ln x) + 3 = 0$. Since the roots of the cubic are not simple rational numbers, the solution must be expressed in terms of these roots.