If log32,log3(2x−5) and log3(2x−72) are in A.P. then x is equal to
A
2
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B
3
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C
4
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D
8
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Solution
The correct option is B 3 We have 2log3(2x−5)=log3(2x−72)+log32 ⇒log3(2x−5)2=log32(2x−72)⇒(2x−5)2=2x+1−7⇒22x−10.2x+25=2×2x−7⇒(2x)2−12.2x+32=0⇒(2x−8)(2x−4)=0 ⇒2x=8,2x=4 But log3(2x−5) is not defined if 2x<5.