Question

Find an integer nwith$\mathit{n}\mathbf{>}\mathbf{2}$for which${\mathbf{(}\mathbf{log}\mathbf{}\mathbf{n}\mathbf{)}}^{{\mathbf{2}}^{\mathbf{100}}}\mathbf{<}\sqrt{n}$.

Short answer

The integer is$n=2^{2^{108}}$

Step-by-step solution

Determine an integer

$(\log \mathit{n}{)}^{2^{100}}<\sqrt{n}$

Take square of both sides of this inequality shows that,

$$\begin{gathered} {\left((\log \mathit{n}{)}^{2^{100}}\right)}^2<n \\ \log \mathit{n}{)}^{2^{101}}<n \end{gathered}$$

Let$n=2^a$

$a^{2^{101}}={\left(\log 2^a\right)}^{2^{101}}<2^a$

Final Solution

By substituting various values of a,$a=2^{102},2^{103},2^{104}....$to get the correct value that holds the inequality.

The value of a that holds the inequality is $a=2^{108}$.

Since$n=2^a$

$n=2^a=2^{2^{108}}$