Question

If ${\log }_{10}2=a$ and ${\log }_{10}3=b$, then express $\log 60$ in terms of $a$ and $b$

• A. $a-b-1$
• B. $a-b+1$
• C. $a+b+1$
• D. $a+b-1$

Answer 1

Answer: (C)

$a+b+1$

${\log }_{10}2=a{\log }_{10}3=b\log 60=\log (2\times 3\times 10)=\log 2+\log 3+\log 10=a+b+1$