Question

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

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

Answer 1

The correct option is A $a+b+1$

Let, $x=\log 60$
$\therefore x=\log (2^2\cdot 3\cdot 5)$
$\therefore x=\log 2^2+\log 3+\log \frac{10}{2}$ $(\log a.b=\log a+\log b)$
$\therefore x=2\log 2+\log 3+1-\log 2$ ...$(\log a^b=b\log a)$
$\therefore x=\log 2+\log 3+1$
$\therefore x=a+b+1$