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if log(x-1) +log (x+1) = $\log_{2} 1$

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Correct option is B

log(x-1) +log(x+1)= log21 log[(x-1)(x+1)]= log 1 Since log 1=0 ∴ x2-1=1 x= 2

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Find for x: if log(x-1) +log (x+1) =log21

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Find for x:
if log(x-1) +log (x+1) = $\log_{2} 1$