Question

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

• A. 18
• B. 16
• C. 15
• D. 17

Hint:

For a logarithmic function \( \log_b(g(x)) \) to be defined, two conditions must be met: the base \( b \) must be positive and not equal to 1 (\( b>0, b \neq 1 \)), and the argument \( g(x) \) must be positive (\( g(x)>0 \)). When dealing with nested logarithms, apply these conditions from the outermost logarithm inwards. Finally, find the intersection of all the conditions to determine the domain of the function.

Answer 1

The Correct Option is A

For the function \( f(x) = \log_7(1 - \log_4(x^2 - 9x + 18)) \) to be defined, we need two conditions to be satisfied:

The argument of the outer logarithm must be positive: \[ 1 - \log_4(x^2 - 9x + 18)>0 \] \[ 1>\log_4(x^2 - 9x + 18) \] \[ 4^1>x^2 - 9x + 18 \] \[ 4>x^2 - 9x + 18 \] \[ 0>x^2 - 9x + 14 \] \[ x^2 - 9x + 14<0 \] Factoring the quadratic: \[ (x - 2)(x - 7)<0 \] This inequality holds for \( 2<x<7 \). So, \( x \in (2, 7) \). \quad ...(2)

The argument of the inner logarithm must be positive: \[ x^2 - 9x + 18>0 \] Factoring the quadratic: \[ (x - 3)(x - 6)>0 \] This inequality holds for \( x<3 \) or \( x>6 \). So, \( x \in (-\infty, 3) \cup (6, \infty) \). \quad ...(1)

The domain of the function is the intersection of the intervals obtained from conditions (1) and (2). Intersection of \( (-\infty, 3) \) and \( (2, 7) \) is \( (2, 3) \). Intersection of \( (6, \infty) \) and \( (2, 7) \) is \( (6, 7) \). 
Therefore, the domain of the function is \( (2, 3) \cup (6, 7) \). Given that the domain is \( (\alpha, \beta) \cup (\gamma, \delta) \), we have: \( \alpha = 2 \), \( \beta = 3 \), 
\( \gamma = 6 \), \( \delta = 7 \). The value of \( \alpha + \beta + \gamma + \delta \) is: \[ \alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18 \]