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Question

Find the value of x for which log2(5.2x+1),log4(21x+1) and 1 form an arithmetic progression.

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Solution

log2(5.2x+1),log4(21x+1)and1 are in AP.
Common difference d=a2a1=a3a2
log4(21x+1)log2(5.2x+1)=1log4(21x+1)
log4(21x+1)+log4(21x+1)=log22+log2(5.2x+1)[logaa=1]
2log4(21x+1)=log22+log2(5.2x+1)
22log2(21x+1)=log2(2(5.2x+1))
[Since,lognba=nlogba & logca+logcb=logc(ab)]
21x+1=5.2x+1+2
212x=102x+1
Let 2x=y
2y=10y+1
10y2+y2=0
10y2+5y4y2=0
5y(2y+1)2(2y+1)=0
(5y2)(2y+1)=0
y=25ory=12
2x=25or2x=12
2x=25(Since exponent of any real number can never be a negative value).
Applying log2 on both sides,
log2(2x)=log2(25)
x=log22log25
x=1log25

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