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11 Aug 2021, 06:45
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B
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What is the value of \(16^{0.16}*16^{0.04}*2^{0.2}\) ?
(A) 1
(B) 2
(C) 4
(D) 8
(E) 16
(A) 1
(B) 2
(C) 4
(D) 8
(E) 16
11 Aug 2021, 07:06
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Bunuel
What is the value of \(16^{0.16}*16^{0.04}*2^{0.2}\) ?
(A) 1
(B) 2
(C) 4
(D) 8
(E) 16
(A) 1
(B) 2
(C) 4
(D) 8
(E) 16
Given: \(16^{0.16}*16^{0.04}*2^{0.2}\)
Rewrite \(16\) as \(2^4\) to get: \((2^4)^{0.16}*(2^4)^{0.04}*2^{0.2}\)
Apply the power of a power law: \(2^{0.64}*2^{0.16}*2^{0.2}\)
Since the bases are all the same, we can apply the product law to get: \(2^{0.64 + 0.16 + 0.2}=2^{1}=2\)
Answer: B
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General Discussion
11 Aug 2021, 07:03
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I believe this could be considered an estimation task. We know that x^a * x^b = x^(a+b). We can thus estimate 16^0.2, i.e. the 5th root of 16 to be slightly smaller than 2 since the 4th root of 16 (16^0.25) is 2.
2^0.2 would be a number sligthly greater than one since the nth square of 2 approaches as n becomes lager.
Option B seems reasonable.
Posted from my mobile device
2^0.2 would be a number sligthly greater than one since the nth square of 2 approaches as n becomes lager.
Option B seems reasonable.
Posted from my mobile device
