Question

$\text{What is the value of}m\text{if}3\log 4+2\log 5-\frac{1}{3}\log 64-\frac{1}{2}\log 16=m$

Answer 1

Given that:
$3\log 4+2\log 5-\frac{1}{3}\log 64-\frac{1}{2}\log 16=m$

Using ${\log }_b\left(x^m\right)=m{\log }_{b}x$
$\Rightarrow \log 4^3+\log 5^2-\log (64{)}^{\frac{1}{3}}-\log (16{)}^{\frac{1}{2}}=m$
$\Rightarrow \log 64+\log 25-\log 4-\log 4=m$

Using ${\log }_{b}x+{\log }_{b}y={\log }_{b}xy$
$\Rightarrow \log (64\times 25)-(\log 4+\log 4)=m$
$\Rightarrow \log (64\times 25)-\log 16=m$

Using ${\log }_{b}x-{\log }_{b}y={\log }_b\frac{x}{y}$
$\Rightarrow \log \left(\frac{64\times 25}{16}\right)=m$
$\Rightarrow \log 100=m$
$\Rightarrow m=2$

Hence the correct answer is option (b).