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If x=(3^8)(81^5) and y = (9^7)(27^12), then xy=
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10 Aug 2020, 10:06
10 Aug 2020, 10:06
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Question Stats:
89% (00:54) correct
10% (02:32) wrong based on 49 sessions
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If \(x=(3^8)(81^5)\) and \(y = (9^7)(27^{12})\), then \(xy=\)
A. \(3^{48}\)
B. \(3^{54}\)
C. \(3^{61}
\)
D. \(3^{68}\)
E. \(3^{78}\)
A. \(3^{48}\)
B. \(3^{54}\)
C. \(3^{61}
\)
D. \(3^{68}\)
E. \(3^{78}\)
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Re: If x=(3^8)(81^5) and y = (9^7)(27^12), then xy=
[#permalink]
11 Aug 2020, 08:14
11 Aug 2020, 08:14
1
Carcass wrote:
If \(x=(3^8)(81^5)\) and \(y = (9^7)(27^{12})\), then \(xy=\)
A. \(3^{48}\)
B. \(3^{54}\)
C. \(3^{61}\)
D. \(3^{68}\)
E. \(3^{78}\)
A. \(3^{48}\)
B. \(3^{54}\)
C. \(3^{61}\)
D. \(3^{68}\)
E. \(3^{78}\)
When we scan the answer choices (ALWAYS scan the answer choices before solving the question), I see that all of the answers have base 3.
So let's first convert all powers to base 3.
Given: \(x=(3^8)(81^5)\)
Rewrite as: \(x=(3^8)[(3^4)^5]\)
Apply the Power of a Power law to get: \(x=(3^8)(3^{20})\)
Apply the Product law to get: \(x=3^{28}\)
Given: \(y = (9^7)(27^{12})\)
Rewrite as: \(y = [(3^2)^7][(3^3)^{12}]\)
Apply the Power of a Power law to get: \(y=(3^{14})(3^{36})\)
Apply the Product law to get: \(y=3^{50}\)
So, \(xy = (3^{28})(3^{50}) = 3^{78}\)
Answer: E
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Re: If x=(3^8)(81^5) and y = (9^7)(27^12), then xy= [#permalink]
11 Aug 2020, 08:14
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