Question

If $n$ is a natural number $>1$, then the least value of $n$ for which ${\log }_{n}121<3.8$

• A. $4$
• B. $11$
• C. $121$
• D. $13$

Answer 1

The correct option is A $4$

Given ${\log }_{n}121<3.8$

$\Rightarrow {\log }_n(11{)}^2<3.8$

$\Rightarrow {\log }_n\left(11\right)<\left(1.9\right)$

$(\because {\log }_c{a}^b=b{\log }_{c}a)$

Using antilog
$\Rightarrow 11<n^{1.9}$

$\Rightarrow n>(11{)}^{\frac{1}{1.9}}$

$\Rightarrow n>3.53$

so, the least integral value will be

$\Rightarrow n=4$