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QOTD #1 81^3 + 27^4 is equivalent to which
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27 Jun 2016, 08:42
27 Jun 2016, 08:42
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\(81^3\) + \(27^4\) is equivalent to which of the following expressions?
Indicate all such expressions.
I) (\(3^7\))(\(2\))
II) (\(3^{12}\))(\(2\))
III) (\(9^6\))(\(2\))
IV) \(9^{12}\)
V) \(3^{24}\)
Indicate all such expressions.
I) (\(3^7\))(\(2\))
II) (\(3^{12}\))(\(2\))
III) (\(9^6\))(\(2\))
IV) \(9^{12}\)
V) \(3^{24}\)
Show: :: OA
II and III
Re: QOTD #1 81^3 + 27^4 is equivalent to which
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04 Jul 2016, 21:20
04 Jul 2016, 21:20
1
The original expression can be written as 3^24; Choice ii 3^12(2) = 3^24 and Choice iii 9^6(2) = 3^12(2) = 3^24 are equivalent to original expression,hence answers
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Re: QOTD #1 81^3 + 27^4 is equivalent to which
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05 Apr 2017, 12:07
05 Apr 2017, 12:07
1
Carcass wrote:
\(81^3\) + \(27^4\) is equivalent to which of the following expressions?
Indicate all such expressions.
I) (\(3^7\))(\(2\))
II) (\(3^{12}\))(\(2\))
III) (\(9^6\))(\(2\))
IV) \(9^{12}\)
V) \(3^{24}\)
Indicate all such expressions.
I) (\(3^7\))(\(2\))
II) (\(3^{12}\))(\(2\))
III) (\(9^6\))(\(2\))
IV) \(9^{12}\)
V) \(3^{24}\)
Show: :: OA
II and III
Scanning the answer choices, I can see that I should try to rewrite the original expression with base 3 and base 9.
BASE 3
\(81^3\) + \(27^4\) = \((3^4)^3\) + \((3^3)^4\)
= \(3^{12}\) + \(3^{12}\)
= (\(2\))(\(3^{12}\))
= II
BASE 9
Take the result from above:
\(81^3\) + \(27^4\) = (\(2\))(\(3^{12}\))
Notice that \(3^{12}\) can be rewritten as \((3^2)^6\), and also notice that \(3^2 = 9\)
So, we get: (\(2\))(\(3^{12}\)) = \((2)\)\((3^2)^6\)
= (\(2\))(\(9^6\))
= III
Answer:
Show: ::
II and III
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