Question

If $x-\log 48+3\log 2=\frac{1}{3}\log 125-\log 3$, then value of $x$ will be

• A. $2$
• B. $5$
• C. $1$
• D. $9$

Answer 1

Answer: (C)

$1$

$x-\log 48+3\log 2=\frac{1}{3}\log 125-\log 3$
$\therefore x-\log 48+\log 2^3=\log {125}^{\frac{1}{3}}-\log 3$ ...($\log a^b=b\log a$)
$\therefore x=\log 48-\log 8+\log {125}^{\frac{1}{3}}-\log 3$
$\therefore x=\log 10$ ....($\log \frac{a}{b}=\log a-\log b,\log a.b=\log a+\log b$)
$\therefore x=1$