Correct Answer - D
`log_(10)154=log_(10)2+log_(10)7+log_(10)11`
Now `log_(2)10 = p`
`rArr (1)/(log_(10)2)=p`
`rArr log_(10)2=(1)/(p)` ….(1)
`(log_(e )10)/(log_(e )7)=q`
`rArr (1)/(log_(10)7)=q`
`rArr log_(10) 7=(1)/(q)` ….(2)
And `11^(r ) = 10`
`rArr r log_(10)11=1`
`rArr log_(10)=(1)/(r )` ....(3)
`therefore log_(10)154 =(1)/(p)+(1)/(q)+(1)/(r )=(pq+qr+rp)/(pqr)`