Question:

The domain of the function \( f(x) = \log_7(\log_3(\log_5(20x - x^2 - 91))) \) is:

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When solving logarithmic domain problems, work step by step from the innermost log outwards, enforcing stricter inequalities at each level.
Updated On: Aug 23, 2025
  • (7, 13)
  • (8, 12)
  • (7, 12)
  • (12, 13)
  • None of the above
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The Correct Option is B

Solution and Explanation

Step 1: Condition for logarithm.
For \(\log_b(y)\) to exist, we require: \[ y>0 \] So, step by step we must ensure each inside expression is positive.

Step 2: Innermost logarithm.
We need: \[ \log_5(20x - x^2 - 91)>0 \] which means: \[ 20x - x^2 - 91>1 \quad \text{since } \log_5(z)>0 \Rightarrow z>1 \] So, \[ 20x - x^2 - 91>1 \] \[ 20x - x^2>92 \] \[ x^2 - 20x + 92<0 \]

Step 3: Solve quadratic inequality.
Discriminant: \[ D = (20)^2 - 4(92) = 400 - 368 = 32 \] So, roots are: \[ x = \frac{20 \pm \sqrt{32}}{2} = \frac{20 \pm 4\sqrt{2}}{2} = 10 \pm 2\sqrt{2} \] \[ \Rightarrow 10 - 2\sqrt{2}<x<10 + 2\sqrt{2} \] Approximately: \(7.17<x<12.83\).

Step 4: Middle logarithm condition.
We need: \[ \log_3(\log_5(20x - x^2 - 91))>0 \] \[ \Rightarrow \log_5(20x - x^2 - 91)>1 \] \[ \Rightarrow 20x - x^2 - 91>5^1 = 5 \] \[ 20x - x^2>96 \] \[ x^2 - 20x + 96<0 \]

Step 5: Solve second inequality.
Discriminant: \[ D = 400 - 384 = 16 \] Roots: \[ x = \frac{20 \pm 4}{2} = \{8, 12\} \] So range: \[ 8<x<12 \]

Step 6: Outermost logarithm condition.
Since \( \log_7(\cdot) \) requires only positive input, and inside is already ensured>0 by Step 4, no further restriction.

Step 7: Final domain.
Combine Step 3 and Step 5: - Step 3 gave: \(7.17<x<12.83\) - Step 5 refined it to: \(8<x<12\) So final domain: \[ (8, 12) \]

Final Answer:
\[ \boxed{(8, 12)} \]
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