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Manager  Joined: 02 Dec 2012
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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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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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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.

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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

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 in-equality 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 Sufficient

Therefore the answer is
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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.

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58381
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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.

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.
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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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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!
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Joined: 02 Sep 2009
Posts: 58381
Re: If n is a positive integer, is the value of b - a at least twice the  [#permalink]

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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.
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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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Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?
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Joined: 02 Sep 2009
Posts: 58381
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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.
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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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Thanks a lot Bunuel. The explanation couldn't have been more lucid.
Manager  B
Joined: 13 Feb 2011
Posts: 79
Re: If n is a positive integer, is the value of b - a at least twice the  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58381
Re: If n is a positive integer, is the value of b - a at least twice the  [#permalink]

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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.
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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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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)
Intern  Joined: 14 Oct 2015
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GMAT 1: 640 Q45 V33 Re: If n is a positive integer, is the value of b - a at least twice the  [#permalink]

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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.

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

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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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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 :-)

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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susheelh wrote:
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 :-)

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

_______________
Done. Thank you.
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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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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.
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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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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.

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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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.

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 Re: If n is a positive integer, is the value of b - a at least twice the   [#permalink] 13 Sep 2017, 19:43

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