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If n and m are integers, is bn > bm?

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Bunuel
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If n and m are integers, is bn > bm? [#permalink] New post   11 Mar 2017, 03:44
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C
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E

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If n and m are integers, is bn > bm?

(1) b^2*n > b^2*m
(2) n > m
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E
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vivek2k70
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Re: If n and m are integers, is bn > bm? [#permalink] New post   11 Mar 2017, 04:09
If we consider the positive number then condition are met but when considering negative number sign changes. So two solution.
So I think its E. Am I right?
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Re: If n and m are integers, is bn > bm? [#permalink] New post   11 Mar 2017, 15:20
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We are given that m and n has to be integer
statement 1: b^(2*n)> b^(2*m)

case 1: b=2
n=3
m=1
2^6>2^2 and bn>bm(6>2)

case 2: b=1/2
n=1
m=2

(1/2)^2>(1/2)^4

1/4>1/16 but bn< bm(1/2<1)

so statement1 is not suffiecient


Statement 2: n> m


Case 1: n=3
m=2
b=1


bn> bm(3>2)

Case 2: n=3
m=2
b=-1
bn<bm (-3<-2)


Statement 2 is not sufficient


Combining both statements
Case 2 of statement 1 does not exist when we combine 2 statements
so correct answer should be C

+1 Kudos if you like the post :)
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Re: If n and m are integers, is bn > bm? [#permalink] New post   11 Mar 2017, 18:49
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Don't cancel terms on either side of the equality without knowing the terms' sign.

With the above,
A: We can cancel b^2 on both sides safely (the square is always positive). We are left with n > m, this is not sufficient, since the we still don't know what is b, so options A and D are out
B: as discussed above. this is essentially the simplified version of A, so even option B is out
Option C is out simply because of the fact that both the statements A and B convey the exact same incomplete information (typical GMAT option trap)

E wins!
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Re: If n and m are integers, is bn > bm? [#permalink] New post   11 Mar 2017, 19:43
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Bunuel wrote:
If n and m are integers, is bn > bm?

(1) b^2*n > b^2*m
(2) n > m



is bn > bm
or b(n-m)>0
means if b>0 then n-m>0

or

if b<0 then n-m<0

(1) as b^2 is >0 always we can cancel from both sides
leaving n-m>0
but no info of b ..(if b>0then YES but if b<0 then NO)

(2) no new information (similar to (1) )

Combining no info on b ..thus
Ans E
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Re: If n and m are integers, is bn > bm? [#permalink] New post   12 Mar 2017, 03:37
Did I understand the question incorrectly?
b^2*n is not same as b^(2*n)?
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Bunuel
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Re: If n and m are integers, is bn > bm? [#permalink] New post   12 Mar 2017, 04:02
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AR15J wrote:
Did I understand the question incorrectly?
b^2*n is not same as b^(2*n)?


Mathematically b^2*n CANNOT mean b^(2*n) it can only mean b^2 multiplied by n. If it were b^(2*n) it would be written as b^(2*n).
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If n and m are integers, is bn > bm? [#permalink] New post   13 Mar 2017, 03:47
If n and m are integers, is bn > bm?


Rephrasing: Is bn - bm>0? or is b(n-m)>0

(1) b^2*n > b^2*m

b^2*n - b^2*m> 0.......b^2(n - m)> 0

b^2 is always positive....then n-m>0

but b could be positive or negative

If b is positive .........Answer is Yes

If b is negative........Answer is NO

Insufficient

(2) n > m

n-m>0........No info about b

Insufficient


Combining 1 & 2

n - m> 0 but still no info about b

Answer: E
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Re: If n and m are integers, is bn > bm? [#permalink] New post   13 Mar 2017, 12:35
1 is insufficient because b can be a negative or positive number.
2 says n is greater than m but we do not know the value of b so 2 is insufficient

combining 1 & 2, still insufficient because we do not know if b is negative or positive
therefore, the answer is E.
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Re: If n and m are integers, is bn > bm? [#permalink] New post   29 Jul 2021, 16:28
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Re: If n and m are integers, is bn > bm? [#permalink]
29 Jul 2021, 16:28
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