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If 2^x3^y = 144, where x and y are positive integers, then
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04 Sep 2021, 05:44
04 Sep 2021, 05:44
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Question Stats:
95% (00:51) correct
4% (00:38) wrong based on 21 sessions
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If \(2^x3^y = 144\), where x and y are positive integers, then \((2^{x−2})(3^{y−1}) =\)
(A) 4
(B) 6
(C) 12
(D) 24
(E) 36
(A) 4
(B) 6
(C) 12
(D) 24
(E) 36
Re: If 2^x3^y = 144, where x and y are positive integers, then
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04 Sep 2021, 07:30
04 Sep 2021, 07:30
1
\(2^x3^y = 144\)
\(2^x3^y = 2^43^2\)
\(x = 4\) & \(y = 2\)
\((2^{x−2})(3^{y−1})\)
\(2^23\)
\(12\)
Answer C
\(2^x3^y = 2^43^2\)
\(x = 4\) & \(y = 2\)
\((2^{x−2})(3^{y−1})\)
\(2^23\)
\(12\)
Answer C
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If 2^x3^y = 144, where x and y are positive integers, then
[#permalink]
04 Sep 2021, 08:30
04 Sep 2021, 08:30
1
\((2^{x−2})(3^{y−1}) =\)
= \((2^x * 2^{−2})(3^y*3^{−1}) =\) = \(\frac{2^x * 3^y}{ 2^2 * 3^1}\)
= \(\frac{144}{4*3}\) ( as \(2^x * 3^y\) = 144)
= \(\frac{144}{12}\) = 12
So, Answer will be C
Hope it helps!
= \((2^x * 2^{−2})(3^y*3^{−1}) =\) = \(\frac{2^x * 3^y}{ 2^2 * 3^1}\)
= \(\frac{144}{4*3}\) ( as \(2^x * 3^y\) = 144)
= \(\frac{144}{12}\) = 12
So, Answer will be C
Hope it helps!
