If n and m are integers, is bn bm?

Question

If n and m are integers, is bn > bm? 
 
(1) b^2*n > b^2*m 
(2) n > m

vivek2k70 · Answer

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?

AR15J · Answer

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

max0010 · Answer

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!

rohit8865 · Answer

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

AR15J · Answer

Did I understand the question incorrectly?
b^2*n is not same as b^(2*n)?

Bunuel · Answer

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).