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If \(N = 2^{0.15}\) and \(N^b = 16,\) \(b\) must equal.

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by M7MBA » Wed Oct 07, 2020 10:44 am

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B

C

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E

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If \(N = 2^{0.15}\) and \(N^b = 16,\) \(b\) must equal.

A. \(\dfrac3{80}\)
B. \(\dfrac35\)
C. \(4\)
D. \(\dfrac53\)
E. \(\dfrac{80}3\)

Answer: E

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Re: If \(N = 2^{0.15}\) and \(N^b = 16,\) \(b\) must equal.

by Scott@TargetTestPrep » Sun Oct 11, 2020 5:40 am
M7MBA wrote: ↑
Wed Oct 07, 2020 10:44 am
 If \(N = 2^{0.15}\) and \(N^b = 16,\) \(b\) must equal.

A. \(\dfrac3{80}\)
B. \(\dfrac35\)
C. \(4\)
D. \(\dfrac53\)
E. \(\dfrac{80}3\)

Answer: E

Solution:

Substituting 2^0.15 for N in the second equation, we have:

(2^0.15)^b = 16

2^(0.15b) = 2^4

0.15b = 4

b = 4/0.15 = 400/15 = 80/3

Answer: E

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