Question

If $n$ is a natural number such that $n=p_1^{a_1}.p_2^{a_2}\cdots p_k^{a_k}$ and $p_1,p_2,p_3,\cdots ,p_k$ are distrinct prime numbers, then ${\log }_{e}n\ge$

• A. $k\log 2$
• B. $k\log 3$
• C. $k\log 4$
• D. none of these

Answer 1

The correct option is A $k\log 2$

We have
$n=p_1^{a_1}.p_2^{a_2}...p_k^{a_k}$
Since, $p_i^{\prime }s$ are primes and $n$ is a natural number. Therefore, each $a_i$ is a natural number.

Also, $p_i\ge 2$ for $i=1,2,\cdots ,k$
[$\because 2$ is the smallest prime]
Now,
$n=p_1^{a_1}.p_2^{a_2}\cdots p_k^{a_k}$
$\Rightarrow \log n=a_1\log p_1+a_2\log p_2+\cdots +a_k\log p_k$
$\Rightarrow \log n\ge a_1\log 2+a_2\log 2+\cdots +a_k\log 2$
$\Rightarrow \log n\ge (a_1+a_2+\cdots +a_k)\log 2$
$\Rightarrow \log n\ge k\log 2\left[\begin{array}{c}\because a_1\ge 1\text{for}i=1,2,\cdots ,k\\ \therefore a_1+a_2+\cdots +a_k\ge k\end{array}\right]$