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
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
Show Hint
- 18
- 16
- 15
- 17
The Correct Option is A
Approach Solution - 1
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, \( x \in (-\infty, 3) \cup (6, \infty) \). ...(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: \[ \alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18 \]
Approach Solution -2
To determine the domain of the function \( f(x) = \log_7(1 - \log_4(x^2 - 9x + 18)) \), we must ensure that the arguments of all logarithmic functions are positive.
First, ensure the argument of the inner logarithm is positive:
- \( x^2 - 9x + 18 > 0 \)
- \( x^2 - 9x + 18 = (x - 3)(x - 6) \)
- \( x \in (-\infty, 3) \cup (6, \infty) \)
Next, the expression for the outer logarithm's argument must be positive:
- \( 1 - \log_4(x^2 - 9x + 18) > 0 \)
- \( \log_4(x^2 - 9x + 18) < 1 \)
- \( x^2 - 9x + 18 < 4 \)
- \( x^2 - 9x + 14 < 0 \)
- Factor as: \( (x - 7)(x - 2) < 0 \)
- \( x \in (2, 7) \)
The combined solution requires both conditions to be satisfied simultaneously:
- Both \( x \in (-\infty, 3) \cup (6, \infty) \) and \( x \in (2, 7) \)
The valid intersections are:
- \( (2, 3) \cup (6, 7) \)
According to the problem, \( \alpha = 2\), \( \beta = 3\), \( \gamma = 6\), and \( \delta = 7 \). Therefore, the sum \(\alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18\).
Therefore, the answer is 18.
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