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 $\alpha + \beta + \gamma + \delta$.

Therefore, the value is 18.