Domain Calculation for Logarithmic Function
The function is given by $f(x) = \log_7 (1-\log_4 (x^2-9x + 18))$. To find the domain, we need to ensure that the arguments of both logarithms are positive.
Logarithm Argument Conditions
For a logarithm $\log_b(y)$ to be defined, the argument $y$ must be greater than 0 ($y > 0$).
- Outer Logarithm Condition: The argument of $\log_7$ must be positive: $1-\log_4 (x^2-9x + 18) > 0$ $1 > \log_4 (x^2-9x + 18)$ Since the base $4 > 1$, we can rewrite this as: $4^1 > x^2-9x + 18$ $4 > x^2-9x + 18$ $0 > x^2-9x + 14$ $x^2-9x + 14 < 0$
- Inner Logarithm Condition: The argument of $\log_4$ must be positive: $x^2-9x + 18 > 0$
Solving Quadratic Inequalities
We solve the two inequalities derived above.
- Inequality 1: $x^2-9x + 14 < 0$ Factor the quadratic: $(x-2)(x-7) < 0$. The roots are $x=2$ and $x=7$. Since the parabola opens upwards, the inequality holds for values between the roots. So, $2 < x < 7$. The interval is $(2, 7)$.
- Inequality 2: $x^2-9x + 18 > 0$ Factor the quadratic: $(x-3)(x-6) > 0$. The roots are $x=3$ and $x=6$. Since the parabola opens upwards, the inequality holds for values outside the roots. So, $x < 3$ or $x > 6$. The interval is $(-\infty, 3) \cup (6, \infty)$.
Determining the Domain
The domain of the function $f(x)$ is the intersection of the intervals obtained from both conditions.
We need $x$ values that satisfy both $(2 < x < 7)$ and $(x < 3 \text{ or } x > 6)$.
Intersection 1: $(2, 7) \cap (-\infty, 3) = (2, 3)$.
Intersection 2: $(2, 7) \cap (6, \infty) = (6, 7)$.
Combining these, the domain is $(2, 3) \cup (6, 7)$.
Calculating the Sum
The domain is given in the form $(\alpha, \beta) \cup (\gamma, \delta)$.
By comparing, we have:
- $\alpha = 2$
- $\beta = 3$
- $\gamma = 6$
- $\delta = 7$
We need to calculate $\alpha + \beta + \gamma + \delta$.
$ \alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18 $Therefore, the value is 18.
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