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by maihuna » Sun Jan 30, 2011 8:50 am
Seems type what is less than 5k,

seeing the options :
(1) 4^(x+1) > 16,000
(2) 4^(x+1) > 4^(x) + 12,288

1=> 4^x*4 > 16k => x^4>4K => x>8

2=> 4^x*4 > 4^x + 12888 =? 3*4^x > 12888 => 4^x > 4296

if u put whole q, we can deduce further
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by Target2009 » Thu Feb 03, 2011 8:51 pm
IMO - D
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by MAAJ » Fri Feb 04, 2011 7:16 am
The second statement is 4^x > 4096
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by GMATGuruNY » Fri Feb 04, 2011 7:34 am
Is 4^x less than 5000?

1) 4^(x+1) > 16,000
2) 4^(x+1) = 4^x + 12,288
The correct version of the question appears above.

Statement 1: 4^(x+1) > 16,000
4^x * 4 > 16,000
4^x > 4000.
Since 4^x = 4001 and 4^x = 5001 both satisfy statement 1, insufficient.

Statement 2: 4^(x+1) = 4^x + 12,288
Since statement 2 gives us an equation (as opposed to an inequality), we can solve for the value of 4^x (and for the value of x).
Sufficient.

The correct answer is B.

Here's the math behind statement 2:
4^(x+1) = 4^x + 12,288
4^(x+1) - 4^x = 12,288
(4^x * 4) - 4^x = 12,288
4^x * (4-1) = 12,288
4^x = 4096
(since 4096 = 4^6, x=6).
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