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.
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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

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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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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Thanks a lot Bunuel. The explanation couldn't have been more lucid.

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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Target question: Is \(b - a \geq 3^n - 2^n\)?

Statement 1: \(a= 2^{n+1}\) and \(b= 3^{n+1}\)

Substitute values into target question to get: Is \(b= 3^{n+1}- 2^{n+1} \geq 3^n - 2^n\)?

Rearrange terms: Is \(3^{n+1}- 3^n \geq 2^{n+1}-2^n \)?

Factor both sides: Is \(3^n(3- 1) \geq 2^n(2- 1)\)?

Simplify: Is \(3^n(2) \geq 2^n\)?

Divide both sides of the inequality by 2 to get: Is \(3^n \geq 2^{n-1}\)?

At this point, it's clear that the left side is greater than the right side (since the left side of the inequality is the product of n 3's, while the right side of the inequality is the product of n-1 2's)

As such we can be certain that \(3^n \geq 2^{n-1}\)
Statement 1 is SUFFICIENT


Statement 2: n = 3
Since we have no information about the variables a and b, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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Hey - can someone explain the 'cancel' step? I am lost on how the cancelling is done here, thx.

Notice that we have like terms, \(-2*2^{n}\), in both sides of \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}}\), so we can cancel them. Add \(2*2^{n}\) to both sides to get \(3*3^{n}\geq{2*3^n}\).
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Plugging in values doesn't look like a bad idea for this one.

we know n is +ve integer, what's the least n can be ?... 1, right ?

so let's give it a go..

\(a= 2^{(n+1)}\) and \(b= 3^{(n+1)}\)

if \(n=1\)----> \(b-a= 3^2-2^2= 9-4=5\). clearly this more than twice what we'll get from the equation in the stem, for n=1 :- \(2(3^n-2^n)\)----> \(2(3^1-2^1)= 2(1)= 2\). stmt is suff.

Stm 2. N>3. this doesn't do anything for us. we have no idea what b and a are.

C is an obvious trap. try harder, GMAT

Bunuel

I'm probably missing something really obvious here, but if we plug in n = 2 for statement 1:

\(3^3 - 2^3 = 19\)

\(2 (3^2 - 2^2) = 10\)

19 is not twice the value of 10. Don't we get yes and no answers?
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19 is at least twice the value of 5 (3^2 - 2^2).

Is the value of b - a at least twice the value of 3^n - 2^n?

If n = 2, then:

b - a = 19
3^n - 2^n = 5. Twice the value of 5 is 10.

19 > 10.
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Ah, yes. Silly mistake indeed.

Thank you Bunuel.
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Bunuel VeritasKarishma

how did you come from this step
Is \(3*3^{n}-2*2^{n} \geq 2*3^n-2*2^{n}\)?

to this step
Is \(3^{n}\geq{0}\)?

I hope this post might help.
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1st.
\(b= 3^{(n+1)}\)
\(a= 2^{(n+1)}\)

b-a = \(3^{(n+1)}\) - \(2^{(n+1)}\)
is \(3^{(n+1)}\) - \(2^{(n+1)}\) >= 2(\(3^n - 2^n\)) ?
\(3^{(n+1)}\) - \(2^{(n+1)}\) >= \(2^1*3^n - 2^n*2^1\) add \(2^{(n+1)}\) to both side
\(3^{(n+1)}\) >= \(2^1*3^n\) -- > \(3^n*3^1\) >= \(2^1*3^n\) divide both side by \(3^n\)
Question becomes is \(3^1\) >= \(2^1\) ? Sufficient.

2st.
does not provide any values for a and b - thus, is not sufficient