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# I came across this one and I kind of liked it, so I thought

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Manager
Joined: 28 Aug 2004
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I came across this one and I kind of liked it, so I thought [#permalink]

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10 Aug 2005, 07:59
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I came across this one and I kind of liked it, so I thought of sharing it with you guys! (God how good I am )

x,y>0. Is 2^x>3^y?

(1) x>2y
(2) x>=y+3

>= is greater than or equal to

enjoy!
SVP
Joined: 03 Jan 2005
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10 Aug 2005, 08:10
x,y>0. Is 2^x>3^y?

(1) x>2y
2^x>2^(2y)=4^y>3^y
Sufficient

(2) x>=y+3
2^x>=2^(y+3)=8*2^y
May or may not be greater than 3^y
Insufficient

A
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Manager
Joined: 31 Aug 2004
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11 Aug 2005, 11:26
HongHu wrote:
x,y>0. Is 2^x>3^y?

(1) x>2y
2^x>2^(2y)=4^y>3^y
Sufficient

(2) x>=y+3
2^x>=2^(y+3)=8*2^y
May or may not be greater than 3^y
Insufficient

A

Is 8*2^y>3^y? I cannot think of a case where 8*2^y<3^y. When y is an integer, 8*2^y>3^y; when y is a fraction (ie. 0.5), 8*2^y is still greater than 3^y. Are you able to show us under what circumstance 8*2^y<3^y?
Thanks!
Manager
Joined: 06 Aug 2005
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11 Aug 2005, 11:45
Yes for any y > (log 8 / log (3/2)) approx = 5.13

Try y=6

8 x 2^6 = 512 < 729 = 3 ^ 6
SVP
Joined: 03 Jan 2005
Posts: 2233
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Kudos [?]: 342 [0], given: 0

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11 Aug 2005, 14:28
Think about it this way. 3^y increases more rapidly than 2^y. Originally 8*2^y has a large start. But each time when y increases, the value of function 3^y increases three times the previous value, while the other function only increases twice the previous value. Some point on the way the more rapidly increasing function is due to surpass the slower increasing function.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

11 Aug 2005, 14:28
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