Question

The value of $(\log 4+\frac{1}{3}\log 125-\frac{1}{5}\log 32)$ is equal to

• A. $3$
• B. $1$
• C. $2$
• D. $6$

Answer 1

Answer: (B)

$1$

$x=\log 4+\frac{1}{3}\log 125-\frac{1}{5}\log 32$

$\therefore x=\log 4+\log {125}^{\frac{1}{3}}-\log {32}^{\frac{1}{5}}$ ....($\log a^b=b\log a$)

$\therefore x=\log 10$....($\log \frac{a}{b}=\log a-\log b$, $\log a.b=\log a+\log b$)

$\therefore x=1$