If the domain of the function f(x) = log_7(1 | vedclass

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(A) For the function $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$ to be defined,we must have $1 - \log_4(x^2 - 9x + 18) > 0$ and $x^2 - 9x + 18 > 0$. St

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$18$

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(A) For the function $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$ to be defined,we must have $1 - \log_4(x^2 - 9x + 18) > 0$ and $x^2 - 9x + 18 > 0$. Step $1$: Solve $x^2 - 9x + 18 > 0$. $(x - 3)(x - 6) > 0 \implies x \in (-\infty, 3) \cup (6, \infty)$. Step $2$: Solve $1 - \log_4(x^2 - 9x + 18) > 0$. $\log_4(x^2 - 9x + 18) $x^2 - 9x + 14 Step $3$: Find the intersection of the two conditions. $x \in ((-\infty, 3) \cup (6, \infty)) \cap (2, 7) = (2, 3) \cup (6, 7)$. Thus,$(\alpha, \beta) = (2, 3)$ and $(\gamma, \delta) = (6, 7)$. Therefore,$\alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18$.

If the domain of the function $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$ is $(\alpha, \beta) \cup (\gamma, \delta)$,then $\alpha + \beta + \gamma + \delta$ is equal to

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