Q: IF M AND N ARE INTEGERS, AND X= 3^N AND Y= 3^M .IS X >2Y ?
A: N= M+1
B: N= 2M
DS INTEGER
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Consider A:[email protected] wrote:Q: IF M AND N ARE INTEGERS, AND X= 3^N AND Y= 3^M .IS X >2Y ?
A: N= M+1
B: N= 2M
x>2y=> 3^n > 2.3^(n-1) => 3^n > 2/3 * 3^n ---- sufficient.
Consider B:
x>2y => 3^2m > 2.3^m => insufficient.
So answer is A.
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X= 3^N AND Y= 3^M . IS X >2Y ??
3^n > 2*3^m ??
A) n=m+1. thus lhs = 3*3^m. as 3>2 sufficient
B) n=2M.
3^(2m) > 2* 3^m ?? or 3^m * 3^m > 2*3^m or 3^m>2???
if m=0, 1>2 false
if m=1, 3>2 true
Insufficient
IMO A
3^n > 2*3^m ??
A) n=m+1. thus lhs = 3*3^m. as 3>2 sufficient
B) n=2M.
3^(2m) > 2* 3^m ?? or 3^m * 3^m > 2*3^m or 3^m>2???
if m=0, 1>2 false
if m=1, 3>2 true
Insufficient
IMO A
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Cans!!
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THANKS FOR YOUR POST.cans wrote:X= 3^N AND Y= 3^M . IS X >2Y ??
3^n > 2*3^m ??
A) n=m+1. thus lhs = 3*3^m. as 3>2 sufficient
B) n=2M.
3^(2m) > 2* 3^m ?? or 3^m * 3^m > 2*3^m or 3^m>2???
if m=0, 1>2 false
if m=1, 3>2 true
Insufficient
IMO A
BUT WE DON'T KNOW THAT BOTH M AND N ARE POSITIVE OR NEGATIVE INTEGER. IF BOTH ARE NEGATIVE THAN I THINK THE ANSWER WOULD BE NOT SAME AS WHEN BOTH ARE POSITIVE
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If you plug-in negative values for M in the equality 3^m*3>2/3 *3^m you will realize that the equality works out both for negative numbers as well
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