If `p >1a n dq >1` are such that `log(p+q)=logp+logq ,` then the value of `log(p-1)+"log"(q-1)` is equal to 0 (b) 1 (c) 2 (d) none of these
If `p >1a n dq >1` are such that `log(p+q)=logp+logq ,` then the value of `log(p-1)+"log"(q-1)` is equal to 0 (b) 1 (c) 2 (d) none of these
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Correct Answer - A
`log(p+q) =log p+log q`
` rArr p+q=pq`
` rArr pq-p-q = 0`
` rArr pq-p-q+1 = 1`
` rArr (p-1)(q-1) = 1`
` rArr log(p-1)+log(q-1) = log 1=0`
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