Concept:Use logarithm laws to split each term into powers of prime factors and combine them.Explanation:Let log2=a, log3=b, and log5=c.Then 7log(1516)=7(4a−b−c)=28a−7b−7c.5log(2425)=5(2c−3a−b)=−15a−5b+10c.3log(8081)=3(4b−4a−c)=−12a+12b−3c.Adding all terms:(28a−15a−12a)+(−7b−5b+12b)+(−7c+10c−3c)=a+0+0=a.Since a=log2, the whole expression simplifies to log2.Answer:log2 (Option C)