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# If n and m are integers, and x=3^n, and y=3^m, is the value

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If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]  12 Sep 2012, 18:09
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If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y?

(1) n=m+1
(2) n=2m
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Sep 2012, 23:36, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Data Sufficiency [#permalink]  12 Sep 2012, 19:58
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saurabhsingh24 wrote:
If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y?
(1) n=m+1
(2) n=2m

There will be three cases:
when M is positive: option 1 is sufficient to answer. and option 2 also give answer
When M is negative: Option 1 is sufficient to answer. But option 2 is not sufficient to answer.
When M is Zero: option Option 1 is sufficient to answer But Option 2 is not sufficient to answer.

Try with some nos. for these three cases.
If you like the post, kindly give me 1 kudos.
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Re: Data Sufficiency [#permalink]  12 Sep 2012, 23:10
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If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y?
(1) n=m+1
(2) n=2m

we have to find if x > 2y
i.e. 3^n > 2*3^m

STAT1:
n = m+1
so we have to prove that
3^(m+1) > 2* 3^m
=> 3 * 3^m > 2*3^m
=> 3 * 3^m - 2*3^m > 0
=> 3^m > 0

So ALL integer value of m 3^m will be greater that 0
So, SUFFICIENT

STAT2:
n=2m
so we have to prove that
3^(2m) > 2* 3^m
3^(2m) - 2* 3^m > 0
3^m* (3^m - 2) > 0

for ALL values of m 3^m will be greater than 0
So, 3^m* (3^m - 2) > 0 if (3^m - 2) is > 0

(3^m - 2) > 0 for all positive values of m
(3^m - 2) < 0 for 0 and all negative values of m

Hope it helps!
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Re: If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]  13 Sep 2012, 00:32
3
KUDOS
saurabhsingh24 wrote:
If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y?

(1) n=m+1
(2) n=2m

The question is in fact "is $$3^n>2\cdot{3^m}$$?"

(1) The given inequality becomes $$3^{m+1}>2\cdot{3^m}$$. Dividing through by $$3^m$$, which is for sure a positive number, we obtain $$3 > 2,$$ obviously true.
Sufficient.

(2) Now the given inequality becomes $$3^{2m}>2\cdot{3^m}$$. Dividing through again by $$3^m$$, we obtain $$3^m>2$$. This inequality holds only for $$m>0$$.
Not sufficient.

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Re: If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]  05 Sep 2013, 07:42
saurabhsingh24 wrote:
If n and m are integers, and x=3^n, and y=3^m, is the value of x greater than the value of 2y?

(1) n=m+1
(2) n=2m

Basically the Q asks whether 3^n> 2*3^m

St 1 substituting for n we get

3^(m+1) - 2*3^m > 0
3* 3^m -2 *3^m>0
3^m>0 ------> m can be 0 or 1 or 2 and so one
so n= m+1 (1 or 2 or 3 for corresponding values of m)
We see that 3^n > 2*3^m

So A is sufficient so ruling out B,C and E

From st 2 we have n= 2m

We get 3^2m -2*3^m>0
9*3^m - 2*3^m> 0 or 7*3^m>0

So m =0, n=0 ---->putting in the above equation we get (3^2m -2*3^m>0) --> 1-2 =-1 not greater than zero
But if m=1,n=2 then we get 3^2m -2*3^m>0 ------>9-6>0 3>0 so 2 ans choices possible so D ruled out

Ans is A
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Re: If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]  18 Sep 2013, 17:35
Here is how I solved it,
x = 3^n and y =3^m
1. n = m +1; therefore, x = 3^m+1 => 3 * 3^m, now 3^m = y, therefore x = 3y and hence x > 2y Sufficient
2. n =2m; x = 3^2m; x = y^2; now nothing is given about x and y; therefore for y = 0; x =0 and hence the x is not greater than 2y
whereas if y =-1, x = 1 again not satisfied, if y = -2, x = 4 ; x = 2y; if y = 3, x = 9; hence NS

Please let me know in case my line of reasoning is not correct, especially for the second statement.
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Re: If n and m are integers, and x=3^n, and y=3^m, is the value [#permalink]  30 Sep 2014, 22:23
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Re: If n and m are integers, and x=3^n, and y=3^m, is the value   [#permalink] 30 Sep 2014, 22:23
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