Question

If $x-\log 48+3\log 2=\frac{1}{3}\log 125-\log 3$ and $x={\log }_{m}20$, then the value of $m$ will be

• A. $6$
• B. $5$
• C. $10$
• D. $20$

Answer 1

The correct option is D $20$

$x-\log 48+3\log 2=\frac{1}{3}\log 125-\log 3x=\log {125}^{\frac{1}{3}}-\log 3+\log 48-3\log 2x=\log 5-\log 3+\log 48-\log 2^3=\log \left(\frac{5\times 48}{3\times 8}\right)=\log 10=\log x\therefore x=1$