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Sub 505 (Easy)|   Exponents|      
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Bunuel
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If n = 2^3, then n^n = 16 May 2018, 04:36
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D
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92% (00:35) correct 8% (00:35) wrong based on 132 sessions

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If \(n=2^3\), then \(n^n=\)

A \(2^6\)

B \(2^{11}\)

C \(2^{18}\)

D \(2^{24}\)

E \(2^{27}\)
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D
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If n = 2^3, then n^n = 16 May 2018, 10:14
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Bunuel
If \(n=2^3\), then \(n^n=\)

A \(2^6\)

B \(2^{11}\)

C \(2^{18}\)

D \(2^{24}\)

E \(2^{27}\)
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We know that \(2^3 = 8\)

Therefore, \(n^n = (2^3)^8 = 2^{3*8} = 2^{24}\) (Option D)
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If n = 2^3, then n^n = 18 May 2018, 11:25
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Bunuel
If \(n=2^3\), then \(n^n=\)

A \(2^6\)

B \(2^{11}\)

C \(2^{18}\)

D \(2^{24}\)

E \(2^{27}\)
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Since n = 8, n^n = 8^8 = (2^3)^8 = 2^24

Answer: D
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If n = 2^3, then n^n = 18 May 2018, 18:13
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If n=2^3, then n^n =

N = 8

\(N^N\) = \(8^8\) = \((2^3)^8\) = \(2^2^4\)

Ans: D
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If n = 2^3, then n^n = 12 Nov 2025, 05:38
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