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Question

If log(a+π)log2(4x+1+4x)=0,xϵR, then 64x is equal to

A
1
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B
3
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C
9
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D
12
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Solution

The correct option is C 9
log2(4x+1+4x)=1
4x+1+4x=21=2
Squaring both sides, we get
4x+1+4x+24x(4x+1)=4
8x3+24x(4x+1)=0
24x(4x+1)=38x
Squaring both sides:
4(4x(4x+1))=948x+64x2
4(16x2+4x)=64x248x+9
64x2+16x=64x248x+9
64x=9

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