Question

The number $N=6{\log }_{10}2+{\log }_{10}31$ lies between two successive integers whose sum is equal to:

• A. 5
• B. 7
• C. 9
• D. 10

Answer 1

The correct option is B 7

$N={\log }_{10}64+{\log }_{10}31={\log }_{10}1984$

Now, we know that ${\log }_{10}1000=3$ and ${\log }_{10}10000=4$.

So, $1000<1984<10000$

$\Rightarrow {\log }_{10}1000<{\log }_{10}1984<{\log }_{10}10000$

$\Rightarrow 3<{\log }_{10}1984<4$

$\Rightarrow 3<{\log }_{10}64+{\log }_{10}31<4$

So, it lies between two successive integers $3$ and $4$.

So, their sum is equal to $3+4=7$

Hence. option $B$ is correct.