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Question

If 3(log9x)292log9x+5=33, then possible value(s) of x is/are

A
9
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B
37/2
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C
37
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D
35/2
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Solution

The correct options are
A 9
C 37
3(log9x)292log9x+5=33
Taking log on both sides to the base 3, we get
(log9x)292log9x+5=32
Put y=log9x
y292y+5=322y29y+7=0
(2y7)(y1)=0
y=72,1
log9x=1 or log9x=72
x=9 or x=97/2=37

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