Question

The least integral value of $n$ such that $7^n>{10}^5$, given that ${\log }_{10}343=2.5353$ is:

• A. $5$
• B. $6$
• C. $7$
• D. $10$

Answer 1

The correct option is B $6$

$7^n>{10}^5$ and ${\log }_{10}\left(343\right)=2.5353$

${\log }^{}(7{)}^3=2.5353$

$\Rightarrow {\log }_{10}\left(7\right)=\frac{2.5353}{3}=0.8451$

Take $\log$ on both sides of $7^n>{10}^5$

$\Rightarrow {\log }_{10}{7}^n>{\log }_{10}\left({10}^5\right)$

$\Rightarrow n{\log }_{10}7>5{\log }_{10}10$

$\Rightarrow n>5-916\Rightarrow n=6$