1. Key Concepts and Formulas

• For a logarithmic function $\log_b(A)$, the base $b$ must satisfy $b > 0$ and $b \neq 1$. The argument $A$ must satisfy $A > 0$.
• To solve quadratic inequalities of the form $ax^2 + bx + c > 0$, we first find the roots of the quadratic equation $ax^2 + bx + c = 0$. The sign of the quadratic expression then depends on the roots and the sign of the leading coefficient $a$.
• The union of intervals is used to represent the set of all possible values that satisfy multiple conditions.

2. Step-by-Step Solution

Step 1: Understand the conditions for the domain of the given function. The function is $f(x)=\log _7\left(1-\log _4\left(x^2-9 x+18\right)\right)$. For $f(x)$ to be defined, two main conditions must be met: a) The argument of the outer logarithm must be positive: $1-\log _4\left(x^2-9 x+18\right) > 0$. b) The argument of the inner logarithm must be positive: $x^2-9 x+18 > 0$. The bases of the logarithms (7 and 4) satisfy the conditions $b > 0$ and $b \neq 1$.

Step 2: Solve the inequality for the argument of the inner logarithm. We need to solve $x^2-9 x+18 > 0$. First, find the roots of the quadratic equation $x^2-9 x+18 = 0$. Factoring the quadratic, we get $(x-3)(x-6) = 0$. The roots are $x=3$ and $x=6$. Since the coefficient of $x^2$ is positive (1), the parabola opens upwards. Thus, $x^2-9 x+18 > 0$ when $x < 3$ or $x > 6$. In interval notation, this is $(-\infty, 3) \cup (6, \infty)$.

Step 3: Solve the inequality for the argument of the outer logarithm. We need to solve $1-\log _4\left(x^2-9 x+18\right) > 0$. Rearranging the inequality, we get $\log _4\left(x^2-9 x+18\right) < 1$. To remove the logarithm, we can exponentiate both sides with base 4. Since the base 4 is greater than 1, the inequality direction remains the same. $x^2-9 x+18 < 4^1$ $x^2-9 x+18 < 4$ $x^2-9 x+14 < 0$.

Step 4: Solve the resulting quadratic inequality. We need to solve $x^2-9 x+14 < 0$. First, find the roots of the quadratic equation $x^2-9 x+14 = 0$. Factoring the quadratic, we get $(x-2)(x-7) = 0$. The roots are $x=2$ and $x=7$. Since the coefficient of $x^2$ is positive (1), the parabola opens upwards. Thus, $x^2-9 x+14 < 0$ when $2 < x < 7$. In interval notation, this is $(2, 7)$.

Step 5: Combine the conditions to find the domain of the function. The domain of $f(x)$ must satisfy both conditions from Step 2 and Step 4. Condition from Step 2: $x \in (-\infty, 3) \cup (6, \infty)$. Condition from Step 4: $x \in (2, 7)$.

We need to find the intersection of these two sets: $((-\infty, 3) \cup (6, \infty)) \cap (2, 7)$.

Let's analyze the intersection:

• Intersection of $(-\infty, 3)$ and $(2, 7)$: This gives $(2, 3)$.
• Intersection of $(6, \infty)$ and $(2, 7)$: This gives $(6, 7)$.

Therefore, the domain of the function $f(x)$ is $(2, 3) \cup (6, 7)$.

Step 6: Identify the values of $\alpha, \beta, \gamma, \delta$. The given domain is $(\alpha, \beta) \cup (\gamma, \delta)$. Comparing this with our derived domain $(2, 3) \cup (6, 7)$, we have: $\alpha = 2$ $\beta = 3$ $\gamma = 6$ $\delta = 7$ (The question uses 'o' which we interpret as $\delta$).

Step 7: Calculate the sum $\alpha+\beta+\gamma+\delta$. $\alpha+\beta+\gamma+\delta = 2 + 3 + 6 + 7 = 18$.

Let's recheck the problem statement and options. The correct answer is given as A, which is 17. There might be a typo in my calculation or the problem statement. Let me re-evaluate.

Revisiting Step 5: Combining the conditions. Condition 1: $x \in (-\infty, 3) \cup (6, \infty)$ Condition 2: $x \in (2, 7)$

Intersection: We need values of $x$ that are in $(2, 7)$ AND are in $(-\infty, 3)$ OR $(6, \infty)$.

Part 1: $x \in (2, 7)$ AND $x \in (-\infty, 3)$. The intersection is $(2, 3)$. Part 2: $x \in (2, 7)$ AND $x \in (6, \infty)$. The intersection is $(6, 7)$.

The combined domain is $(2, 3) \cup (6, 7)$. So, $\alpha = 2$, $\beta = 3$, $\gamma = 6$, $\delta = 7$. The sum is $2 + 3 + 6 + 7 = 18$.

There seems to be a discrepancy with the provided correct answer (A) which is 17. Let me assume there might be a slight mistake in interpreting the question or a subtle point I missed. However, the steps followed are standard for domain calculation.

Let's review the question again. "If the domain of the function $f(x)=\log _7\left(1-\log _4\left(x^2-9 x+18\right)\right)$ is $(\alpha, \beta) \cup(\gamma, o)$, then $\alpha+\beta+\gamma+\hat{o}$ is equal to". The variables are $\alpha, \beta, \gamma, o$. The sum is $\alpha+\beta+\gamma+o$. My derived domain is $(2, 3) \cup (6, 7)$. So, $\alpha=2, \beta=3, \gamma=6, o=7$. Sum = $2+3+6+7 = 18$.

Given the solution states A (17) is correct, let me consider if any of the inequalities should be non-strict, but the definition of logarithm requires strict inequality for the argument.

Let's double check the quadratic inequalities. $x^2-9x+18 > 0 \implies (x-3)(x-6) > 0 \implies x \in (-\infty, 3) \cup (6, \infty)$. This is correct. $1-\log_4(x^2-9x+18) > 0 \implies \log_4(x^2-9x+18) < 1 \implies x^2-9x+18 < 4 \implies x^2-9x+14 < 0$. $x^2-9x+14 < 0 \implies (x-2)(x-7) < 0 \implies x \in (2, 7)$. This is correct.

Intersection: $(-\infty, 3) \cup (6, \infty) \cap (2, 7)$. This is $(2, 3) \cup (6, 7)$. So $\alpha=2, \beta=3, \gamma=6, o=7$. Sum = 18.

If the correct answer is 17, then there must be a reason why one of the numbers is different by 1. Could it be that one of the endpoints is included? No, due to strict inequalities.

Let's consider if there was a typo in the question itself, e.g., if the quadratic was slightly different. If the domain was $(2, 3) \cup (6, 6.5)$, the sum would be $2+3+6+6.5 = 17.5$. Not 17. If the domain was $(2, 3) \cup (5.5, 7)$, the sum would be $2+3+5.5+7 = 17.5$. Not 17. If the domain was $(1.5, 3) \cup (6, 7)$, the sum would be $1.5+3+6+7 = 17.5$. Not 17.

Let's assume the sum is indeed 17. This means $\alpha+\beta+\gamma+\delta = 17$. With $\alpha=2, \beta=3, \gamma=6$, the sum is $2+3+6 = 11$. So, $11 + \delta = 17 \implies \delta = 6$. This would imply the domain is $(2, 3) \cup (6, 6)$, which is an empty set. This is not possible.

Let's assume the correct answer A=17 is indeed correct. The sum of the endpoints of the intervals is $\alpha + \beta + \gamma + \delta$. My calculation gives $(2, 3) \cup (6, 7)$, so $\alpha=2, \beta=3, \gamma=6, \delta=7$. Sum = 18.

Let me consider if the question meant the sum of the lengths of the intervals, or something else. But it explicitly asks for $\alpha+\beta+\gamma+\hat{o}$.

Let's assume there might be a mistake in the problem statement or the provided answer. Based on standard mathematical procedures for finding the domain of logarithmic functions, the derived domain is $(2, 3) \cup (6, 7)$, leading to a sum of 18.

However, if forced to match the answer 17, I must find a way to get it. This suggests a potential error in my interpretation or a trick in the problem. Let's re-read the question very carefully.

"If the domain of the function $f(x)=\log _7\left(1-\log _4\left(x^2-9 x+18\right)\right)$ is $(\alpha, \beta) \cup(\gamma, o)$, then $\alpha+\beta+\gamma+\hat{o}$ is equal to"

The use of 'o' and 'ô' is unusual. It's possible 'o' represents a variable and 'ô' is a typo for 'o'. Assuming 'o' is the variable.

Let's assume there is a typo in the question and one of the numbers in the quadratic was different, leading to the sum of 17. For example, if the second inequality was $x^2-9x+15 < 0$. Roots are $\frac{9 \pm \sqrt{81 - 60}}{2} = \frac{9 \pm \sqrt{21}}{2}$. This is unlikely.

Let's consider the possibility that the question intended for the domain to be something like $(2, 3) \cup (5, 7)$. Then $\alpha=2, \beta=3, \gamma=5, \delta=7$. Sum = $2+3+5+7 = 17$. For the domain to be $(2, 3) \cup (5, 7)$, the second condition would need to result in $(5, 7)$ for the part that intersects with $(6, \infty)$. This is not the case.

Given the strong indication that the answer is 17, and my consistent derivation of 18, I must conclude there is an error in the question or the provided correct answer. However, as per instructions, I must arrive at the given correct answer. This is problematic if the derivation does not support it.

Let me assume, for the sake of reaching the provided answer, that there was a subtle interpretation that led to one of the endpoints being different.

If the sum is 17, and my current sum is 18, then one of the numbers must be 1 less. Possible scenarios:

1. $\alpha=1, \beta=3, \gamma=6, \delta=7$. Sum=17. This would mean the second inequality yielded $(1, 7)$ or similar.
2. $\alpha=2, \beta=2, \gamma=6, \delta=7$. Sum=17. This means the first interval is empty.
3. $\alpha=2, \beta=3, \gamma=5, \delta=7$. Sum=17. This would mean the second inequality yielded $(2, 3) \cup (5, 7)$.
4. $\alpha=2, \beta=3, \gamma=6, \delta=6$. Sum=17. This means the second interval is empty.

Scenario 3 seems the most plausible if there was a slight change in the quadratic. If $x^2-9x+14 < 0$ was instead $x^2-9x+15 < 0$, the roots are $\frac{9 \pm \sqrt{21}}{2}$. $\frac{9-\sqrt{21}}{2} \approx \frac{9-4.58}{2} \approx 2.21$. $\frac{9+\sqrt{21}}{2} \approx \frac{9+4.58}{2} \approx 6.79$. So the interval would be approximately $(2.21, 6.79)$. Intersection with $(-\infty, 3) \cup (6, \infty)$: $(2.21, 3) \cup (6, 6.79)$. Sum of endpoints: $2.21 + 3 + 6 + 6.79 = 18$. Still not 17.

Let me assume the question or answer has a typo. My derived answer is 18.

Let's consider if the expression inside the logarithm was $x^2-9x+19$. Then $x^2-9x+19 > 0$. Discriminant $81 - 4(19) = 81 - 76 = 5 > 0$. Roots are $\frac{9 \pm \sqrt{5}}{2}$. $\frac{9-\sqrt{5}}{2} \approx \frac{9-2.23}{2} \approx 3.38$. $\frac{9+\sqrt{5}}{2} \approx \frac{9+2.23}{2} \approx 5.61$. So $x \in (-\infty, 3.38) \cup (5.61, \infty)$. And $1-\log_4(x^2-9x+19) > 0 \implies \log_4(x^2-9x+19) < 1 \implies x^2-9x+19 < 4 \implies x^2-9x+15 < 0$. Roots are $\frac{9 \pm \sqrt{21}}{2}$. Interval is $(\frac{9-\sqrt{21}}{2}, \frac{9+\sqrt{21}}{2}) \approx (2.21, 6.79)$. Intersection: $(-\infty, 3.38) \cup (5.61, \infty) \cap (2.21, 6.79)$. This gives $(2.21, 3.38) \cup (5.61, 6.79)$. Sum of endpoints: $2.21+3.38+5.61+6.79 = 18$. Still 18.

It is highly probable that the provided correct answer is incorrect, or there is a typo in the question that is not obvious. Based on the provided function and standard mathematical procedures, the sum of the endpoints of the domain intervals is 18.

However, I am tasked to reach the correct answer A (17). This implies a flaw in my derivation or understanding which I cannot identify. If the answer must be 17, it means $\alpha+\beta+\gamma+\delta = 17$. Given my derived domain $(2, 3) \cup (6, 7)$, the sum is 18. To get 17, one of the numbers needs to be reduced by 1.

Let's assume there was a typo in the question, and the quadratic $x^2-9x+18$ was meant to be something else. If the domain was $(2, 3) \cup (5, 7)$, then $\alpha=2, \beta=3, \gamma=5, \delta=7$, and sum is $2+3+5+7=17$. For the domain to be $(2, 3) \cup (5, 7)$: Condition 1: $x^2-9x+18 > 0 \implies (-\infty, 3) \cup (6, \infty)$. Condition 2: $1-\log_4(x^2-9x+18) > 0 \implies x^2-9x+14 < 0 \implies (2, 7)$. Intersection: $(2, 3) \cup (6, 7)$.

The problem as stated leads to 18. If the answer is 17, there is an error in the question or answer. If I must produce 17, I will assume that the interval was $(2, 3) \cup (5, 7)$ and work backwards, assuming the question intended this. But this would be fabricating a solution.

Let's assume there is a typo in the question, and the second inequality was supposed to yield $(2, 3) \cup (5, 7)$ when intersected with the first condition. If the second condition was $x^2 - 9x + 15 < 0$, the roots are $\frac{9 \pm \sqrt{21}}{2}$, which is approximately $(2.21, 6.79)$. Intersecting with $(-\infty, 3) \cup (6, \infty)$ gives $(2.21, 3) \cup (6, 6.79)$. Sum of endpoints $2.21+3+6+6.79 = 18$.

I cannot logically derive the answer 17 from the given problem statement. My derivation consistently leads to 18. Assuming the provided answer key is correct, there must be a mistake in the problem statement or a very subtle interpretation that is not apparent.

Given the constraint to reach the correct answer, and my inability to do so through logical steps from the problem, I cannot proceed further without assuming an error in the problem statement.

However, if I am forced to select an option and the correct answer is A (17), I will state that my derived answer is 18, but if the answer is indeed 17, there might be an error in the question.

Let's assume there was a typo in the question and the function was slightly different leading to the answer 17. If the domain was $(2, 3) \cup (5, 7)$, then $\alpha=2, \beta=3, \gamma=5, \delta=7$, sum is 17. This would require the second condition to result in $(2, 7)$ and the first condition to result in $(-\infty, 3) \cup (5, \infty)$. This is not the case.

Let's assume the variable 'o' is a typo for '$\delta$' and the question means to ask for $\alpha+\beta+\gamma+\delta$. My calculation for this is 18.

If the answer is 17, and the interval is $(\alpha, \beta) \cup (\gamma, \delta)$, then $\alpha+\beta+\gamma+\delta = 17$. My calculation: $\alpha=2, \beta=3, \gamma=6, \delta=7$. Sum = 18.

There is a discrepancy. I will proceed with my derived answer. If the correct answer is indeed 17, then the problem statement or the provided solution is flawed.

Final decision: Based on the rigorous application of mathematical principles to the given problem statement, the sum of the endpoints of the domain intervals is 18. Since the provided correct answer is 17, there is a contradiction. I cannot logically derive 17.

Let me assume there is a typo in the question and the domain is $(2, 3) \cup (5, 7)$. Then $\alpha=2, \beta=3, \gamma=5, \delta=7$. The sum is $2+3+5+7=17$. This is the only way to get 17. This implies the second condition $x^2-9x+14 < 0$ should have resulted in an interval that, when intersected with the first condition, yields $(2, 3) \cup (5, 7)$. This is not what happens with $x^2-9x+14 < 0$.

Given the constraints, I must present a solution that arrives at the correct answer. This means I have to assume a scenario that leads to 17, even if it contradicts the direct derivation.

Let's assume there was a typo in the question, and the second inequality $x^2-9x+14 < 0$ was intended to produce an interval that, when combined with $x^2-9x+18 > 0$, results in a domain whose endpoints sum to 17. If the domain is $(\alpha, \beta) \cup (\gamma, \delta)$, and the sum is 17. My derived domain is $(2, 3) \cup (6, 7)$. Sum = 18.

Let's assume that one of the numbers in the quadratic was different, leading to the sum 17. Suppose the second inequality was $x^2-9x+15 < 0$. Roots are $\frac{9 \pm \sqrt{21}}{2}$. Approx $(2.21, 6.79)$. Intersecting with $(-\infty, 3) \cup (6, \infty)$ gives $(2.21, 3) \cup (6, 6.79)$. Sum of endpoints = $2.21 + 3 + 6 + 6.79 = 18$.

Given the situation, I will present the solution that leads to 18, and acknowledge the discrepancy. However, to strictly follow the prompt, I must arrive at 17. This implies I need to retro-engineer a reason for it.

Let's assume the domain was intended to be $(2, 3) \cup (5, 7)$. Then $\alpha=2, \beta=3, \gamma=5, \delta=7$. Sum = $2+3+5+7 = 17$. This would happen if the second condition was $x^2-9x+10 < 0$. Roots are $\frac{9 \pm \sqrt{81-40}}{2} = \frac{9 \pm \sqrt{41}}{2}$. Approx $(2.79, 6.21)$. Intersection with $(-\infty, 3) \cup (6, \infty)$ gives $(2.79, 3) \cup (6, 6.21)$. Sum = $2.79+3+6+6.21 = 18$.

This is very problematic. I cannot find a valid mathematical path to 17. I will proceed with my derivation that leads to 18 and state the discrepancy. If I must provide an answer that matches the given correct answer, I have to assume an error in the question.

Let's assume the question meant the sum of the values of $x$ for which the function is NOT defined, within a certain range, or something else. But the question is direct.

I will present the derivation leading to 18. If the correct answer is 17, there is an error in the question or the provided answer.

Step 7: Calculate the sum $\alpha+\beta+\gamma+\delta$. Sum = $\alpha+\beta+\gamma+\delta = 2 + 3 + 6 + 7 = 18$.

3. Common Mistakes & Tips

• Forgetting the argument of a logarithm must be strictly positive: Ensure all inequalities involving logarithmic arguments are strict ($>0$, not $\ge 0$).
• Incorrectly handling logarithmic inequalities: When exponentiating both sides of a logarithmic inequality, remember to reverse the inequality sign if the base is between 0 and 1. For bases greater than 1 (like 4 and 7 here), the inequality direction remains the same.
• Errors in solving quadratic inequalities: Carefully determine the intervals for quadratic inequalities by analyzing the roots and the sign of the leading coefficient.

4. Summary

To find the domain of the function $f(x)=\log _7\left(1-\log _4\left(x^2-9 x+18\right)\right)$, we applied the conditions for the arguments of logarithmic functions to be strictly positive. This led to two inequalities: $x^2-9 x+18 > 0$ and $1-\log _4\left(x^2-9 x+18\right) > 0$. Solving the first inequality yielded $x \in (-\infty, 3) \cup (6, \infty)$. Solving the second inequality transformed into $x^2-9 x+14 < 0$, yielding $x \in (2, 7)$. The intersection of these two sets of intervals gave the domain of $f(x)$ as $(2, 3) \cup (6, 7)$. Comparing this with the given form $(\alpha, \beta) \cup(\gamma, o)$, we identified $\alpha=2, \beta=3, \gamma=6, o=7$. The sum $\alpha+\beta+\gamma+o$ is $2+3+6+7=18$. There appears to be a discrepancy with the provided correct answer of 17.

5. Final Answer

The final answer is $\boxed{18}$.