Question

Solve the logarithmic inequality $log_{\sqrt{8}}(x^2-4x+3)\leq 1$.

• A. $[2 - \sqrt{1 + 2\sqrt{2}}, 1) \cup (3, 2 + \sqrt{1 + 2\sqrt{2}}]$
• B. $(-\infty, 1) \cup (3, \infty)$
• C. $[2 - \sqrt{1 + 2\sqrt{2}}, 2 + \sqrt{1 + 2\sqrt{2}}]$
• D. $(1, 3)$

Answer 1

Answer: (A)

$[2 - \sqrt{1 + 2\sqrt{2}}, 1) \cup (3, 2 + \sqrt{1 + 2\sqrt{2}}]$

Answer 2

To solve the logarithmic inequality $log_{\sqrt{8}}(x^2-4x+3)\leq 1$, we must satisfy two conditions:

1. Domain of the logarithm: The argument of the logarithm must be positive. $x^2-4x+3 > 0$ $(x-1)(x-3) > 0$ This inequality holds for $x \in (-\infty, 1) \cup (3, \infty)$.

2. Solving the inequality: Since the base of the logarithm, $\sqrt{8}$, is greater than 1, we can remove the logarithm by raising the base to the power of the right side and maintaining the inequality direction. $x^2-4x+3 \leq (\sqrt{8})^1$ $x^2-4x+3 \leq \sqrt{8}$ $x^2-4x+(3-2\sqrt{2}) \leq 0$

   To find the roots of the quadratic equation $x^2-4x+(3-2\sqrt{2}) = 0$, we use the quadratic formula: $x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(3-2\sqrt{2})}}{2}$ $x = \frac{4 \pm \sqrt{16 - 12 + 8\sqrt{2}}}{2}$ $x = \frac{4 \pm \sqrt{4 + 8\sqrt{2}}}{2}$ $x = \frac{4 \pm 2\sqrt{1 + 2\sqrt{2}}}{2}$ $x = 2 \pm \sqrt{1 + 2\sqrt{2}}$

   Let $\alpha = 2 - \sqrt{1 + 2\sqrt{2}}$ and $\beta = 2 + \sqrt{1 + 2\sqrt{2}}$. The inequality $x^2-4x+(3-2\sqrt{2}) \leq 0$ holds for $x \in [\alpha, \beta]$.

3. Intersection of conditions: We need to find the intersection of the domain $(-\infty, 1) \cup (3, \infty)$ and the solution interval $[2 - \sqrt{1 + 2\sqrt{2}}, 2 + \sqrt{1 + 2\sqrt{2}}]$. We note that $2 - \sqrt{1 + 2\sqrt{2}} < 1$ and $2 + \sqrt{1 + 2\sqrt{2}} > 3$. Therefore, the intersection is $[2 - \sqrt{1 + 2\sqrt{2}}, 1) \cup (3, 2 + \sqrt{1 + 2\sqrt{2}}]$.