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16 Apr 2010, 05:58
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If log3 2, log3 (2x – 5), log3 (2x – 7/2) are in arithmetic progression, then the value of x is equal to
1. 5
2. 4
3. 2
4. 3
1. 5
2. 4
3. 2
4. 3
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16 Apr 2010, 07:55
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ykaiim
Not a gmat question. Logs are not tested in GMAT.
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16 Apr 2010, 11:36
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log3 2, log3 (2x – 5), log3 (2x – 7/2) are in arithmetic progression
=> 2(log3 (2x – 5)) = log3 (2x – 7/2) + log3 2
=> (2x-5)^2 = 2*(2x-7/2)
=> 4x^2 + 25 - 20x = 4x -7
=> 4x^2 -24x + 32
=> x^2 - 6x + 8
=> (x-4)(x-2) = 0
=> x=4 or x =2
but 2x-5 >0 => x =2 is not possible. Hence x = 4
=> 2(log3 (2x – 5)) = log3 (2x – 7/2) + log3 2
=> (2x-5)^2 = 2*(2x-7/2)
=> 4x^2 + 25 - 20x = 4x -7
=> 4x^2 -24x + 32
=> x^2 - 6x + 8
=> (x-4)(x-2) = 0
=> x=4 or x =2
but 2x-5 >0 => x =2 is not possible. Hence x = 4
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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