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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)
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16 Jan 2023, 01:38
16 Jan 2023, 01:38
Expert Reply
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Question Stats:
80% (01:25) correct
20% (03:30) wrong based on 5 sessions
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If \((2^x)(3^y) = 288\), where x and y are positive integers, then \((2^{x-1})(3^{y-2})\) equals:
A. 16
B. 24
C. 48
D. 96
E. 144
A. 16
B. 24
C. 48
D. 96
E. 144
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)
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18 Jan 2023, 06:47
18 Jan 2023, 06:47
Expert Reply
Algebraically we can approach the problem in another way as well.
If (2^x)(3^y) = 288
then (2^x-1)(3^y-2) = 288/(2[*]3[*]3)
The reason being the exponent of 2 is reduced by 1 and exponent of 3 is reduced by 2.
Hence, the answer would be 288/18 = 16.
Answer: A
If (2^x)(3^y) = 288
then (2^x-1)(3^y-2) = 288/(2[*]3[*]3)
The reason being the exponent of 2 is reduced by 1 and exponent of 3 is reduced by 2.
Hence, the answer would be 288/18 = 16.
Answer: A
Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)
[#permalink]
23 Oct 2024, 11:28
23 Oct 2024, 11:28
1
2 ^x* 3^y =288
288= 2*144
288=2*12^2
288=2*(2*6)^2
288=2*(2^2*3)^2
288=2*2^4*3^2
288=2^5 * 3^2
2 ^x* 3^y =2^5 * 3^2
x=5, y=2
2^(5-1)*3^(2-2)=16
Answer is : A
288= 2*144
288=2*12^2
288=2*(2*6)^2
288=2*(2^2*3)^2
288=2*2^4*3^2
288=2^5 * 3^2
2 ^x* 3^y =2^5 * 3^2
x=5, y=2
2^(5-1)*3^(2-2)=16
Answer is : A
