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

Answer: (B)

$7$

$N=6{\log }_{10}2+{\log }_{10}31$

$={\log }_{10}{2}^6+lo{g}_{10}\left(31\right)$

$={\log }_{10}\left(2^6.31\right)$

$={\log }_{10}(64\times 31)$

$={\log }_{10}\left(1984\right)$

Now

${\log }_{10}\left({10}^3\right)={\log }_{10}\left(1000\right)=3$

Similarly

${\log }_{10}\left({10}^4\right)=4$

Since

${10}^3<1984<{10}^4$

$3<{\log }_{10}\left(1984\right)<4$

Hence it lies between $3$ and $4$.

The sum of these two integers is $7$.