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.
For a logarithm $\log_b(y)$ to be defined, the argument $y$ must be greater than 0 ($y > 0$).
We solve the two inequalities derived above.
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 $(2, 3) \cup (6, 7)$.
The domain is given in the form $(\alpha, \beta) \cup (\gamma, \delta)$.
By comparing, we have:
We need to calculate $\alpha + \beta + \gamma + \delta$.
$ \alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18 $Therefore, the value is 18.
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.