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11 Jan 2023, 10:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
67% (01:03) correct
33%
(01:00)
wrong
based on 82
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What is the largest integer n for which \(2^n < 16^{-2}\) ?
A. -9
B. -8
C. -7
D. -4
E. -2
A. -9
B. -8
C. -7
D. -4
E. -2
11 Jan 2023, 20:43
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Given that \(2^n < 2^{-8}\)
If we plot the values on the number line, the values can be arranged in this fashion:
<———0————0.25————0.5—————1—————2————>
So, 1/4 is less than 1/2.
Similarly, \( \frac{1}{2^{n+1}}<\frac{1}{2^n}\)
So, the power of 9 will be less than the power of 8.
Why is the power of 9 largest in this case, if higher powers (10, 11, ……) also work with the logic?
Because, -9 > -10 > -11 ………
Hence, (A) is the answer.
If we plot the values on the number line, the values can be arranged in this fashion:
<———0————0.25————0.5—————1—————2————>
So, 1/4 is less than 1/2.
Similarly, \( \frac{1}{2^{n+1}}<\frac{1}{2^n}\)
So, the power of 9 will be less than the power of 8.
Why is the power of 9 largest in this case, if higher powers (10, 11, ……) also work with the logic?
Because, -9 > -10 > -11 ………
Hence, (A) is the answer.
12 Jan 2023, 01:10
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Bunuel
What is the largest integer n for which \(2^n < 16^{-2}\) ?
A. -9
B. -8
C. -7
D. -4
E. -2
A. -9
B. -8
C. -7
D. -4
E. -2
\(2^n < 16^{-2}\)
\(2^n < 2^{-8}\)
\(\frac{1}{2}^{-n} < \frac{1}{2}^{8}\)
-n = 9
n = -9
Option A
