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

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Manager
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I came across this one and I kind of liked it, so I thought [#permalink]

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New post 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 :-D )

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

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

>= is greater than or equal to

enjoy!
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Re: DS8 [#permalink]

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New post 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
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Re: DS8 [#permalink]

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New post 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!
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New post 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
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New post 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.
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Keep on asking, and it will be given you;
keep on seeking, and you will find;
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  [#permalink] 11 Aug 2005, 14:28
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I came across this one and I kind of liked it, so I thought

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