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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?
Question: is \(b-a\geq{2(3^n - 2^n)}\)?
(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 \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\)? --> is \(3^{n}\geq{0}\)? 3^n is always more than zero, so this statement is sufficient.
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 \(b-a\geq{2(3^n - 2^n)}\)?
(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 \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\)? --> 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 \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\) but how does this equation tell you whether b-a is at least twice the value of 3^n-2^n?
For example, if i sub in n=1 into the equation, it is true that b-a is at least twice the value (b-a= 5 and 3^n-2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as b-a then becomes 19 and the right side of the equation then becomes 10, which is not at least twice the value.
If n is a positive integer, is the value of b - a at least twice the value of 3^n - 2^n?
Question: is \(b-a\geq{2(3^n - 2^n)}\)?
(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 \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\)? --> 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 \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\) but how does this equation tell you whether b-a is at least twice the value of 3^n-2^n?
For example, if i sub in n=1 into the equation, it is true that b-a is at least twice the value (b-a= 5 and 3^n-2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as b-a 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.
_________________
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 ~45-90seconds to crunch)?
Would be particularly interested in getting Bunuel's take...Thanks!
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 ~45-90seconds to crunch)?
Would be particularly interested in getting Bunuel's take...Thanks!
You should take the approach which fits you the best.
_________________
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
1
This post received KUDOS
Bunuel wrote:
suhaschan wrote:
Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?
\(3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}\);
\(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\);
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.
Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?
\(3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}\);
\(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\);
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.
Yes, this also would be a correct way of solving.
_________________
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?
\(3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}\);
\(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\);
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 \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\); clearly shows that it it IS bigger. (Given that n is a positive integer)
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 \(b-a\geq{2(3^n - 2^n)}\)?
(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 \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\)? --> 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 \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\) but how does this equation tell you whether b-a is at least twice the value of 3^n-2^n?
For example, if i sub in n=1 into the equation, it is true that b-a is at least twice the value (b-a= 5 and 3^n-2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as b-a 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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11 Jan 2017, 11:21
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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 b-a= 9-4=5 which is 5 times more than 3-2=1 In fact, b-a 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 b-a. Insufficient.
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 \(b-a\geq{2(3^n - 2^n)}\)?
(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 \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\)? --> 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?
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68% (01:31) correct
