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If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1)
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Updated on: 18 Jun 2018, 04:31
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If \((2^x)(3^y) = 288\), where x and y are positive integers, then \((2^{x1})(3^{y2})\) equals: A. 16 B. 24 C. 48 D. 96 E. 144
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Originally posted by marcusaurelius on 21 May 2010, 11:33.
Last edited by Bunuel on 18 Jun 2018, 04:31, edited 2 times in total.
Renamed the topic, edited the question and added the OA.




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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1)
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21 May 2010, 11:51
If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1)(3^y2) equals:
16 24 48 96 144
So I would start attacking this problem by quickly performing the prime factorization of 288. With that it is easy to count the 5 twos and the 2 threes that are the prime factors. So x=5, y=2. now quickly 2^4(3^0)=16. Than answer should be number 1.
If my comments were helpful to you, please give kudos. Thanks, Skip




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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1)
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21 May 2010, 11:54
Thanks, I didn't even think to use the prime factorization.



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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1)
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15 Jun 2010, 04:37
Algebraically we can approach the problem in another way as well.
If (2^x)(3^y) = 288 then (2^x1)(3^y2) = 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.
 Please give kudos, if my comments were helpful.



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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1)
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12 Nov 2011, 18:45
skipjames wrote: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1)(3^y2) equals:
16 24 48 96 144
So I would start attacking this problem by quickly performing the prime factorization of 288. With that it is easy to count the 5 twos and the 2 threes that are the prime factors. So x=5, y=2. now quickly 2^4(3^0)=16. Than answer should be number 1.
If my comments were helpful to you, please give kudos. Thanks, Skip Hi Skip, could you please explain prime factorization and how you count the three's. I think I see how you count the two's by taking 2, 8, and 8 seperately and see how many two's making up each number but I don't quite understand how you get the three's. Thanks, Steven



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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1)
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14 Nov 2011, 14:06
Here's how I do prime factorization.
In one row you start with your number to be factored and in the other column you write the smallest possible prime that number is divisible by. Then you do the same for the (number/prime) and so on until you get to 1.
For example:
number:______288______144______72______36______18______9______3______1 (done) prime factor:_______2_________2______ 2_______2_______2______3______3
In this case you it's easy to see that 288 = 2*2*2*2*2*3*3 = 2^5 * 3^2.



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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1)
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15 Nov 2011, 09:27
I approached it by using prime factorization. you need to know how many two's and three's are there in 288. using prime factorization, i got 4 two's and 2 three's. you just need to plug in the values for x and y and solve the given equation. (2^51)(3^22)= (2^4)(3^0)= (16)(1)= 16 hope that helps
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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1)
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29 Apr 2017, 04:25
Prime factorize 288 then do your magic: \(\left( { 2 }^{ x } \right) \left( 3^{ y } \right) =288\\ \left( { 2 }^{ x } \right) \left( 3^{ y } \right) ={ 2 }^{ 5 }{ 3 }^{ 2 }\\ x=5\\ y=2\\ \\ \left( { 2 }^{ x1 } \right) \left( { 3 }^{ y2 } \right) =\left( { 2 }^{ 51 } \right) \left( { 3 }^{ 22 } \right) ={ 2 }^{ 4 }{ 3 }^{ 0 }=16\)
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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x1) &nbs
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