Question

$\log_{18}(x^2-4x+3)<1$

• A. $(2 - \sqrt{19}, 1) \cup (3, 2 + \sqrt{19})$
• B. $x \in (-\infty, 1) \cup (3, \infty)$
• C. $x \in (2 - \sqrt{19}, 2 + \sqrt{19})$
• D. $(1, 3)$

Answer 1

Answer: (A)

$(2 - \sqrt{19}, 1) \cup (3, 2 + \sqrt{19})$

Answer 2

To solve the logarithmic inequality $\log_{18}(x^2-4x+3)<1$, we must consider two conditions:

1. Domain of the logarithm: The argument of the logarithm must be strictly positive. $x^2 - 4x + 3 > 0$ Factoring the quadratic, we get: $(x-1)(x-3) > 0$ This inequality holds when both factors are positive or both are negative.

   • If $x-1 > 0$ and $x-3 > 0$, then $x > 1$ and $x > 3$, which means $x > 3$.
   • If $x-1 < 0$ and $x-3 < 0$, then $x < 1$ and $x < 3$, which means $x < 1$. So, the domain is $x \in (-\infty, 1) \cup (3, \infty)$.

2. Solving the inequality: The base of the logarithm is $18$, which is greater than $1$. Therefore, when we remove the logarithm, the direction of the inequality remains the same. $x^2 - 4x + 3 < 18^1$ $x^2 - 4x + 3 < 18$ Subtract $18$ from both sides to get a standard quadratic inequality: $x^2 - 4x - 15 < 0$ To find the roots of the quadratic equation $x^2 - 4x - 15 = 0$, we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-15)}}{2(1)}$ $x = \frac{4 \pm \sqrt{16 + 60}}{2}$ $x = \frac{4 \pm \sqrt{76}}{2}$ $x = \frac{4 \pm \sqrt{4 \times 19}}{2}$ $x = \frac{4 \pm 2\sqrt{19}}{2}$ $x = 2 \pm \sqrt{19}$ Since the quadratic $x^2 - 4x - 15$ has a positive leading coefficient, its parabola opens upwards. Thus, the inequality $x^2 - 4x - 15 < 0$ holds for values of $x$ between its roots: $x \in (2 - \sqrt{19}, 2 + \sqrt{19})$.

3. Intersection of conditions: We must find the values of $x$ that satisfy both the domain condition and the inequality condition. Domain: $(-\infty, 1) \cup (3, \infty)$ Solution interval: $(2 - \sqrt{19}, 2 + \sqrt{19})$ We need to approximate the values of the roots to determine the intersection. $\sqrt{16} = 4$ and $\sqrt{25} = 5$, so $\sqrt{19}$ is between $4$ and $5$. Approximately, $\sqrt{19} \approx 4.36$. $2 - \sqrt{19} \approx 2 - 4.36 = -2.36$ $2 + \sqrt{19} \approx 2 + 4.36 = 6.36$ So, the solution interval is approximately $(-2.36, 6.36)$. Now we find the intersection: $((-\infty, 1) \cup (3, \infty)) \cap (2 - \sqrt{19}, 2 + \sqrt{19})$ This intersection can be broken down as: $(-\infty, 1) \cap (2 - \sqrt{19}, 2 + \sqrt{19})$ which is $(2 - \sqrt{19}, 1)$ because $2 - \sqrt{19} < 1$. AND $(3, \infty) \cap (2 - \sqrt{19}, 2 + \sqrt{19})$ which is $(3, 2 + \sqrt{19})$ because $3 < 2 + \sqrt{19}$. Combining these two parts, the final solution is $(2 - \sqrt{19}, 1) \cup (3, 2 + \sqrt{19})$.