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Re: If n is a positive integer, is the value of b - a at least [#permalink]
18 Dec 2012, 08:03

Expert's post

2

This post was BOOKMARKED

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 [#permalink]
29 Sep 2013, 01:59

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.

Re: If n is a positive integer, is the value of b - a at least [#permalink]
29 Sep 2013, 12:01

Expert's post

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?

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 [#permalink]
19 Jul 2014, 23:12

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!

Re: If n is a positive integer, is the value of b - a at least [#permalink]
20 Jul 2014, 05:06

Expert's post

warriorsquared wrote:

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. _________________

If n is a positive integer, is the value of b - a at least [#permalink]
03 Aug 2014, 09:09

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.

Re: If n is a positive integer, is the value of b - a at least [#permalink]
12 Aug 2014, 01:55

2

This post received KUDOS

Expert's post

Dienekes wrote:

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.

Yes, this also would be a correct way of solving. _________________

If n is a positive integer, is the value of b - a at least [#permalink]
04 May 2015, 05:45

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)

gmatclubot

If n is a positive integer, is the value of b - a at least
[#permalink]
04 May 2015, 05:45

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