Key Concepts and Formulas

• Domain of a Logarithmic Function $\log_b(a)$: For the function $\log_b(a)$ to be defined in real numbers, the following conditions must hold:
  1. Argument must be positive: $a > 0$
  2. Base must be positive: $b > 0$
  3. Base must not be equal to 1: $b \neq 1$
• Solving Quadratic Inequalities: For a quadratic $ax^2 + bx + c$ with roots $r_1$ and $r_2$ ($r_1 < r_2$):
  • If $a > 0$, $ax^2 + bx + c < 0 \implies x \in (r_1, r_2)$.
  • If $a > 0$, $ax^2 + bx + c > 0 \implies x \in (-\infty, r_1) \cup (r_2, \infty)$.
• Solving Rational Inequalities: Use the Wavy Curve Method by finding the roots of the numerator and denominator, ordering them, and testing intervals to determine where the expression is positive or negative.

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Step-by-Step Solution

Part 1: Finding the domain of $f_1(x) = \log_5(18x - x^2 - 77)$

The base of the logarithm is $5$, which is positive and not equal to $1$. Thus, we only need to ensure the argument is positive.

• Step 1: Set the argument of the logarithm greater than zero. The argument is $18x - x^2 - 77$. We require: $18x - x^2 - 77 > 0$

• Step 2: Rearrange the quadratic inequality. To make solving easier, multiply the inequality by $-1$ and reverse the inequality sign: $x^2 - 18x + 77 < 0$

• Step 3: Factor the quadratic expression. We look for two numbers that multiply to $77$ and add to $-18$. These numbers are $-7$ and $-11$. $(x - 7)(x - 11) < 0$

• Step 4: Solve the quadratic inequality. The roots are $x=7$ and $x=11$. Since the coefficient of $x^2$ is positive ($1 > 0$) and the inequality is $< 0$, the solution is the interval between the roots. $7 < x < 11$ Thus, the domain of the first function is $(7, 11)$.

• Step 5: Identify $\alpha$ and $\beta$. Given that the domain is $(\alpha, \beta)$, we have: $\alpha = 7, \quad \beta = 11$

Part 2: Finding the domain of $f_2(x) = \log_{(x-1)} \left( \frac{2x^2 + 3x - 2}{x^2 - 3x - 4} \right)$

For this function, we must satisfy three conditions for the base $b = x-1$ and the argument $a = \frac{2x^2 + 3x - 2}{x^2 - 3x - 4}$.

• Step 1: Apply the base conditions.

  • Base must be positive: $x - 1 > 0 \implies x > 1$. Let this be Condition A.
  • Base must not be equal to 1: $x - 1 \neq 1 \implies x \neq 2$. Let this be Condition B.

• Step 2: Apply the argument condition. The argument must be positive: $\frac{2x^2 + 3x - 2}{x^2 - 3x - 4} > 0$

• Step 3: Factorize the numerator and the denominator.

  • Numerator: $2x^2 + 3x - 2 = (2x - 1)(x + 2)$
  • Denominator: $x^2 - 3x - 4 = (x - 4)(x + 1)$ The inequality becomes: $\frac{(2x - 1)(x + 2)}{(x - 4)(x + 1)} > 0$

• Step 4: Solve the rational inequality using the Wavy Curve Method. The critical points are the roots of the numerator and denominator: $x = 1/2, x = -2, x = 4, x = -1$. Ordering these points: $-2, -1, 1/2, 4$. We analyze the sign of the expression in the intervals defined by these points:

  • $x < -2$: $\frac{(-)(-)}{(-)(-)} = +$
  • $-2 < x < -1$: $\frac{(-)(+)}{(-)(-)} = -$
  • $-1 < x < 1/2$: $\frac{(-)(+)}{(-)(+)} = +$
  • $1/2 < x < 4$: $\frac{(+)(+)}{(-)(+)} = -$
  • $x > 4$: $\frac{(+)(+)}{(+)(+)} = +$ Since the inequality is $> 0$, the solution is: $x \in (-\infty, -2) \cup (-1, 1/2) \cup (4, \infty)$ Let this be Condition C.

• Step 5: Combine all conditions to find the domain of $f_2(x)$. We need to find the intersection of Condition A ($x > 1$), Condition B ($x \neq 2$), and Condition C ($x \in (-\infty, -2) \cup (-1, 1/2) \cup (4, \infty)$).

  • Condition A ($x > 1$) restricts our search to the interval $(1, \infty)$.
  • Condition B ($x \neq 2$) means we must exclude $2$ from $(1, \infty)$, giving $(1, 2) \cup (2, \infty)$.
  • Now, we intersect $(1, 2) \cup (2, \infty)$ with Condition C.
    • The interval $(1, 2)$ has no overlap with Condition C.
    • The interval $(2, \infty)$ overlaps with $(4, \infty)$ from Condition C. The intersection is $(4, \infty)$. Therefore, the domain of the second function is $(4, \infty)$.

• Step 6: Identify $\gamma$ and $\delta$. Given that the domain is $(\gamma, \delta)$, we have: $\gamma = 4 \quad \text{and} \quad \delta = \infty$ We only need $\gamma$ for the final calculation.

Part 3: Calculate $\alpha^2 + \beta^2 + \gamma^2$

We have found $\alpha = 7$, $\beta = 11$, and $\gamma = 4$.

• Step 1: Substitute the values into the expression. $\alpha^2 + \beta^2 + \gamma^2 = (7)^2 + (11)^2 + (4)^2$

• Step 2: Compute the squares. $= 49 + 121 + 16$

• Step 3: Sum the results. $= 170 + 16$ $= 186$

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Common Mistakes & Tips

• Logarithm Domain Conditions: Always remember all three conditions for the base ($b>0, b\neq 1$) and argument ($a>0$). For logarithmic functions with variable bases, all three are crucial.
• Rational Inequalities: When solving rational inequalities, the roots of the denominator are always excluded from the solution set, as they lead to division by zero.
• Combining Intervals: Carefully find the intersection of all derived conditions. A number line or visual representation can be very helpful.

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Summary

The problem involves finding the domains of two logarithmic functions by applying the fundamental rules for logarithms and solving quadratic and rational inequalities. For the first function, $\log_5(18x - x^2 - 77)$, the domain was found to be $(7, 11)$, yielding $\alpha = 7$ and $\beta = 11$. For the second function, $\log_{(x-1)} \left( \frac{2x^2 + 3x - 2}{x^2 - 3x - 4} \right)$, after considering the base and argument conditions and solving the rational inequality, the domain was determined to be $(4, \infty)$, giving $\gamma = 4$. Finally, the expression $\alpha^2 + \beta^2 + \gamma^2$ was computed using these values.

The final answer is $\boxed{186}$.