Given $\log 2=0.301\text{ and }\log 3=0.477$then the number of digits before decimal in $3^{12}\times 2^8$ is 

Given that , $\log 2=0.301$
                  $\log 3=0.477$
Let   $\mathrm{y}=3^{12}\times 2^8$
$\begin{array}{l}\log \mathrm{y}=12\log 3+8\log 2\\ =12\times (0.477)+8\left(0301\right)=8.132\end{array}$
$\text{ So, number of digits before decimal in }{3}^{12}\times 2^8=8+1=9$