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If 60^a=3 and 60^b=5 then 12^((1-a-b)/(2...

If `60^a`=3 and `60^b=5` then `12^((1-a-b)/(2(1-b)))` is equal to

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We have
`60^(a)=3" "rArr" "a=log_(60)3`
`60^(b)=5" "rArr" "b=log_(60)5`
`So," "(1-(a+b))/(2(1-b))=(1-(log_(60)3+log_(60)5))/(2(log_(60)60-log_(60)5))`
`=(log_(60)60-log_(60)15)/(2(log_(60)60-log_(60)5))=(log_(60)4)/(2log_(60)12)`
`=(1)/(2)log_(12)4=log_(12)2`
`rArr" "12^((1-a-b)/(2(1-b)))=12^(log_(12)2)=2`
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