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by shashank.ism » Mon Feb 08, 2010 11:22 am
If log 2, log (2x -1) and log (2x + 3) are in A.P., then x is equal to

a) 5/2
b) log25
c) log32
d) 3/2
e) 1/2
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by harsh.champ » Mon Feb 08, 2010 11:23 am
shashank.ism wrote:If log 2, log (2x -1) and log (2x + 3) are in A.P., then x is equal to

a) 5/2
b) log25
c) log32
d) 3/2
e) 1/2
2 log (2x - 1) = log 2 + log (2x + 3)
⇒ log (2x - 1) 2 = log 2(2x + 3)
Removing logarithm from both sides of the equation, we get
(2x - 1) 2 = 2 (2x + 3) ⇒ 4x 2 - 4x + 1 = 4x + 6
4x 2 - 8x - 5 = 0 ⇒ (2x - 5)(2x + 1) = 0.
As 2x cannot be negative, [spoiler]2x = 5 ⇒ x = 5/2. A is[/spoiler] the answer.
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by ajith » Mon Feb 08, 2010 11:29 am
shashank.ism wrote:If log 2, log (2x -1) and log (2x + 3) are in A.P., then x is equal to

a) 5/2
b) log25
c) log32
d) 3/2
e) 1/2
Since the numbers are in AP => 2b = a+c

2 log(2x-1) = log 2 + log (2x+3)

(2x-1)^2 = 2(2x+3)

4x^2 - 4x+1 = 4x +6
4x^2 -8x -5 = 0

(2x-5) (2x+1) =0

x cannot be negative (log doesnt exist for negative numbers)

2x-5 =0
x= 5/2, A
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by Ian Stewart » Mon Feb 08, 2010 1:20 pm
Just in case this thread causes any test takers to panic, logarithms are certainly not tested on the GMAT, ever.
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