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18 Feb 2025, 02:12
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A
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Difficulty:
5%
(low)
Question Stats:
93% (00:52) correct
7%
(01:01)
wrong
based on 106
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What is the largest integer k for which \(81^k < 3^{27}\)?
A. 6
B. 7
C. 8
D. 9
E. 10
A. 6
B. 7
C. 8
D. 9
E. 10
18 Feb 2025, 02:48
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We are asked to find the largest integer k such that:
81^k < 3^27
Step 1: Express 81 as a power of 3
- Recall that 81 = 3^4.
- Therefore, we can rewrite 81^k as:
81^k = (3^4)^k = 3^(4k)
Step 2: Set up the inequality
- Now we have:
3^(4k) < 3^27
- Since the bases are the same, we can compare the exponents:
4k < 27
Step 3: Solve for k
- Divide both sides of the inequality by 4:
k < 27/4 = 6.75
- Thus, the largest integer k is:
k = 6
Final Answer:
The largest integer k for which 81^k < 3^27 is 6.
Answer: (A) 6
81^k < 3^27
Step 1: Express 81 as a power of 3
- Recall that 81 = 3^4.
- Therefore, we can rewrite 81^k as:
81^k = (3^4)^k = 3^(4k)
Step 2: Set up the inequality
- Now we have:
3^(4k) < 3^27
- Since the bases are the same, we can compare the exponents:
4k < 27
Step 3: Solve for k
- Divide both sides of the inequality by 4:
k < 27/4 = 6.75
- Thus, the largest integer k is:
k = 6
Final Answer:
The largest integer k for which 81^k < 3^27 is 6.
Answer: (A) 6
18 Feb 2025, 04:31
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Bunuel
The idea in this question is to bring both sides in some form which can be compared.
Simple step forward is to prime factorize both sides
\(81^k = (3^4)^k = 3^{4k}\)
now, for \(81^k < 3^{27}\)
\(3^{4k} < 3^{27}\)
i.e. 4k < 27
i.e. k < 6.75
i.e. \(k_{max} = 6\)
Answer: Option A
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