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Question

The sum of the roots of the equation x+12log2(2x+3)+2log4(102x)=0 is

A
log211
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B
log212
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C
log213
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D
log214
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Solution

The correct option is A log211
x+12log2(2x+3)+2log4(102x)=0
x+1=log2(2x+3)2log2(102x)
x+1=log2(2x+3)2(102x)
22x=(2x+3)2(102x)

Assuming 2x=t
2t=(t+3)2(10t1)
20t2=(t+3)2
t214t+11=0 (1)

If t1=2x1, t2=2x2 are roots of the equation (1), then
t1t2=2x12x2=11
x1+x2=log211

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