If loga30=A,loga(53)=−B\log_{a}{30}=A,\log_{a}({\frac{5}{3}})=-Bloga30=A,loga(35)=−B and log2a=13\log_2{a}=\frac{1}{3}log2a=31, then log3a\log_3{a}log3a equals
2A+B−3\frac{2}{A+B-3}A+B−32
2A+B−3\frac{2}{A+B}-3A+B2−3
A+B2−3\frac{A+B}{2}-32A+B−3
A+B−32\frac{A+B-3}{2}2A+B−3
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