Question

If ${\log }_{10}2=0.3010$ and ${\log }_{10}3=0.4771$, then find the approximate value of ${\log }_{10}192$

• A. $1.2857$
• B. $2.2831$
• C. $0.2642$
• D. $3.7854$

Answer 1

The correct option is B $2.2831$

We can write ${\log }_{10}192={\log }_{10}(3\times 64)$
Or, ${\log }_{10}3+{\log }_{10}64\Rightarrow {\log }_{10}3+{\log }_{10}{2}^6$
$\Rightarrow {\log }_{10}3+6{\log }_{10}2=0.4771+6\times 0.3010=2.2831$