Question:
If \(\log_2 \log_7 (x^2 - x + 37) = 1\), then what could be the value of 'x'?
If \(\log_2 \log_7 (x^2 - x + 37) = 1\), then what could be the value of 'x'?
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Undo the outer logarithm first: \(\log_2 y = 1\) gives \(y = 2\), then undo the base-7 logarithm.
Updated On: Nov 4, 2026
- 4
- 5
- 3
- None of these
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The Correct Option is A
Solution and Explanation
Step 1: Remove the outer logarithm:
\(\log_2 \left[\log_7 (x^2 - x + 37)\right] = 1 \Rightarrow \log_7 (x^2 - x + 37) = 2^1 = 2\).
Step 2: Remove the inner logarithm:
\(x^2 - x + 37 = 7^2 = 49 \Rightarrow x^2 - x - 12 = 0 \Rightarrow (x-4)(x+3) = 0\).
Step 3: Read off x:
x = 4 or x = -3. Among the options only 4 is a possible value (x = 3 and x = 5 give \(x^2-x+37 = 43\) and 57, neither equal to 49).
Final Answer:
The correct option is (1): 4
\(\log_2 \left[\log_7 (x^2 - x + 37)\right] = 1 \Rightarrow \log_7 (x^2 - x + 37) = 2^1 = 2\).
Step 2: Remove the inner logarithm:
\(x^2 - x + 37 = 7^2 = 49 \Rightarrow x^2 - x - 12 = 0 \Rightarrow (x-4)(x+3) = 0\).
Step 3: Read off x:
x = 4 or x = -3. Among the options only 4 is a possible value (x = 3 and x = 5 give \(x^2-x+37 = 43\) and 57, neither equal to 49).
Final Answer:
The correct option is (1): 4
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