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If n is a positive integer, is the value of b  a at least twice the [#permalink]
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18 Dec 2012, 09:00
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If n is a positive integer, is the value of b  a at least twice the value of \(3^n  2^n\)? (1) \(a= 2^{(n+1)}\) and \(b= 3^{(n+1)}\) (2) \(n = 3\)
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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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18 Dec 2012, 09:03



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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29 Sep 2013, 02:59
Bunuel wrote: If n is a positive integer, is the value of b  a at least twice the value of 3^n  2^n?
Question: is \(ba\geq{2(3^n  2^n)}\)?
(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is \(3^{n+1}2^{n+1}\geq{2*(3^n  2^n)}\)? > is \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\)? > is \(3^{n}\geq{0}\)? 3^n is always more than zero, so this statement is sufficient.
(2) n = 3. Clearly insufficient.
Answer: A. For statement (1), I understand how you arrived at \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\) but how does this equation tell you whether ba is at least twice the value of 3^n2^n? For example, if i sub in n=1 into the equation, it is true that ba is at least twice the value (ba= 5 and 3^n2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as ba then becomes 19 and the right side of the equation then becomes 10, which is not at least twice the value. What am i missing here?



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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29 Sep 2013, 13:01
bulletpoint wrote: Bunuel wrote: If n is a positive integer, is the value of b  a at least twice the value of 3^n  2^n?
Question: is \(ba\geq{2(3^n  2^n)}\)?
(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is \(3^{n+1}2^{n+1}\geq{2*(3^n  2^n)}\)? > is \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\)? > is \(3^{n}\geq{0}\)? 3^n is always more than zero, so this statement is sufficient.
(2) n = 3. Clearly insufficient.
Answer: A. For statement (1), I understand how you arrived at \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\) but how does this equation tell you whether ba is at least twice the value of 3^n2^n? For example, if i sub in n=1 into the equation, it is true that ba is at least twice the value (ba= 5 and 3^n2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as ba then becomes 19 and the right side of the equation then becomes 10, which is not at least twice the value. What am i missing here? For the first statement after some manipulation the question becomes: is \(3^{n}\geq{0}\)? Irrespective of the actual value of n, 3^n will always be greater than zero, thus the answer to the question is YES.
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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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30 Sep 2013, 18:49
Walkabout wrote: If n is a positive integer, is the value of b  a at least twice the value of 3^n  2^n?
(1) a= 2^(n+1) and b= 3^(n+1) (2) n = 3 The Question asks: b  a > 2*(3^n  2^n)The answer should be a definitive yes/no Statement 1:a= 2^(n+1) and b= 3^(n+1) Lets take the smallest possible value of a Positive Integer i.e 1 and put it in for "n" 3^(1+1)  2^ (1+1) > 2*(3^1  2^1) 3^2  2^2 > 2* (3  2) 9  4 > 2 *1 5>2 (a definitive answer) Lets test another number (just to be on the Safe Side), lets test n=2 3^(2+1)  2^ (2+1) > 2*(3^2  2^2) 3^3  2^3 > 2* (9  4) 27  8 > 2* 5 19> 10 (a definitive answer) Thus, Sufficient.Statement 2:
n=3 Let's Put it in the inequality b  a > 2*(3^n  2^n) b  a> 2*(3^3  2^3) b  a> 2*(27  8) b  a> 2*(19) b  a> 38 If, b=100 and a=10, than definitive answer If, b= 2 and a= 10, than definitive answer But since it doesn't provide any value for either "a" or "b" Thus, Not SufficientTherefore the answer is



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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20 Jul 2014, 00:12
Do you think that it's better to approach this problem by plugging in numbers for n (as long as you're sure you can do it fast for cases n=1 and n=2), or to do the problem algebraically (which you know will take ~4590seconds to crunch)?
Would be particularly interested in getting Bunuel's take...Thanks!



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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20 Jul 2014, 06:06



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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20 Jul 2014, 09:17
Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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20 Jul 2014, 09:43



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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21 Jul 2014, 05:50
Thanks a lot Bunuel. The explanation couldn't have been more lucid.



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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03 Aug 2014, 10:09
Bunuel wrote: suhaschan wrote: Can someone please show me the simplification as to how the equation becomes whether 3^n > 0? \(3^{n+1}2^{n+1}\geq{2*(3^n  2^n)}\); \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\); Cancel \(2*2^{n}\): \(3*3^{n}\geq{2*3^n}\); Subtract 2*3^n from both sides: \(3*3^{n}2*3^n\geq{0}\); \(3^{n}\geq{0}\) Hope it's clear. Hi Bunuel, In the end instead of subtracting \(2*3^n\) from both sides, can we cancel \(3^n\) from both sides (as it's always positive) and reach \(3>2\) making the statement sufficient? Is that also a correct approach? Thanks.



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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12 Aug 2014, 02:55



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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04 May 2015, 06:45
Bunuel wrote: suhaschan wrote: Can someone please show me the simplification as to how the equation becomes whether 3^n > 0? \(3^{n+1}2^{n+1}\geq{2*(3^n  2^n)}\); \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\); Cancel \(2*2^{n}\): \(3*3^{n}\geq{2*3^n}\); Subtract 2*3^n from both sides: \(3*3^{n}2*3^n\geq{0}\); \(3^{n}\geq{0}\) Hope it's clear. You could stop at the part I have boldfaced to realize that the statement is sufficient, no? I mean \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\); clearly shows that it it IS bigger. (Given that n is a positive integer)



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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22 Oct 2015, 14:20
bulletpoint wrote: Bunuel wrote: If n is a positive integer, is the value of b  a at least twice the value of 3^n  2^n?
Question: is \(ba\geq{2(3^n  2^n)}\)?
(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is \(3^{n+1}2^{n+1}\geq{2*(3^n  2^n)}\)? > is \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\)? > is \(3^{n}\geq{0}\)? 3^n is always more than zero, so this statement is sufficient.
(2) n = 3. Clearly insufficient.
Answer: A. For statement (1), I understand how you arrived at \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\) but how does this equation tell you whether ba is at least twice the value of 3^n2^n? For example, if i sub in n=1 into the equation, it is true that ba is at least twice the value (ba= 5 and 3^n2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as ba then becomes 19 and the right side of the equation then becomes 10, which is not at least twice the value. What am i missing here? How did you get that the right side of the equation is 10?? 3^2 = 9 and 2^2 = 4 which means 5. 5 X 2 = 10. 19 > 10.



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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20 Jan 2017, 03:56
From (1), we have to find whether:
b  a > 2*[3^n  2^n] ?
Substituting a and b,
3^(n+1)  2^(n+1) > 2*[3^n  2^n] ? 3^(n+1)  2^(n+1) > 2*3^n  2^(n+1) ?
2^(n+1) gets cancelled on both sides.
3^(n+1) > 2*3^n? 3*3^n > 2*3^n?
3^n gets cancelled on both sides.
3 > 2?
Yes.



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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20 Aug 2017, 03:57
Hello Moderators, Can you please make the Stem math friendly? This is a OG question and making the stem Math friendly may help ins solving the question :) Thanks in advance! Walkabout wrote: If n is a positive integer, is the value of b  a at least twice the value of 3^n  2^n?
(1) a= 2^(n+1) and b= 3^(n+1) (2) n = 3
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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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20 Aug 2017, 04:15



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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26 Aug 2017, 09:45
It is easier if you take any positive integer value of n. Statement 1: Suppose n=1 (least value) then ba= 94=5 which is 5 times more than 32=1 In fact, ba is actually 3^(n+1)  2^(n+1) which will always be far greater than 3^n  2^n as the value of n increases. sufficient. Statement 2: there is no value for ba. Insufficient.



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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13 Sep 2017, 19:42
Bunuel wrote: If n is a positive integer, is the value of b  a at least twice the value of 3^n  2^n?
Question: is \(ba\geq{2(3^n  2^n)}\)?
(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is \(3^{n+1}2^{n+1}\geq{2*(3^n  2^n)}\)? > is \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\)? > is \(3^{n}\geq{0}\)? 3^n is always more than zero, so this statement is sufficient.
(2) n = 3. Clearly insufficient.
Answer: A. Hi Bunuel 3*3^n>=2.3^n We can divide 3^n on both sides then we will be left with 3>2 right? So, statement A is sufficient Please guide me if Im wrong



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Re: If n is a positive integer, is the value of b  a at least twice the [#permalink]
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13 Sep 2017, 19:43
Bunuel wrote: If n is a positive integer, is the value of b  a at least twice the value of 3^n  2^n?
Question: is \(ba\geq{2(3^n  2^n)}\)?
(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is \(3^{n+1}2^{n+1}\geq{2*(3^n  2^n)}\)? > is \(3*3^{n}2*2^{n}\geq{2*3^n2*2^{n}\)? > is \(3^{n}\geq{0}\)? 3^n is always more than zero, so this statement is sufficient.
(2) n = 3. Clearly insufficient.
Answer: A. Hi Bunuel 3*3^n>=2.3^n We can divide 3^n on both sides then we will be left with 3>2 right? So, statement A is sufficient Please guide me if Im wrong




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