Question

Find the largest integer n such that \({\log ^*}n = 5\). Determine the number of decimal digits in this number.

Short answer

The \({2^{65536}}\) is the largest integer \(n\) for which \({\log ^ * }n = 5\).

Step-by-step solution

Describe the given information and the formulas:

It is given that,

\({\log ^{\left( k \right)}}n = \;\left\{ \begin{aligned}{l}n\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{if}}\;k = 0\\\log \left( {{{\log }^{\left( {k - 1} \right)}}n} \right)\;\;\;\;\;\;\;\;{\rm{if lo}}{{\rm{g}}^{\left( {k - 1} \right)}}n{\rm{ is defined and positive}}\\{\rm{undefined}}\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{otherwise}}{\rm{.}}\end{aligned} \right.\)

Here,\({\log ^*}n\)is the smallest nonnegative integer ksuch that\({\log ^{\left( k \right)}}n \le 1\).

Determine the number of decimal digits in this number

The logarithms have base 2.

In this case \({\log ^ * }n = 5\), thus \(5\) is the smallest nonnegative integer such that

\({\log ^{\left( 5 \right)}}n \le 1\)

Use the recursive definition:

\(\log \left( {{{\log }^{\left( 4 \right)}}n} \right) \le 1\)

Take the exponential with base \(2\) of each side:

\(\begin{aligned}{c}{\log ^{\left( 4 \right)}}n \le {2^1}\\ = 2\end{aligned}\)

Use the recursive definition:

\(\log \left( {{{\log }^{\left( 3 \right)}}n} \right) \le 2\)

Take the exponential with base 2 of each side:

\(\begin{aligned}{c}{\log ^{\left( 3 \right)}}n \le {2^2}\\ = 4\end{aligned}\)

Use the recursive definition:

\(\log \left( {{{\log }^{\left( 2 \right)}}n} \right) \le 4\)

Take the exponential with base \(2\) of each side:

\(\begin{aligned}{c}{\log ^{\left( 2 \right)}}n \le {2^4}\\ = 16\end{aligned}\)

Use the recursive definition:

\(\log \left( {{{\log }^{\left( 1 \right)}}n} \right) \le 16\)

Take the exponential with base \(2\) of each side:

\(\begin{aligned}{c}{\log ^{\left( 1 \right)}}n \le {2^{16}}\\ = 65536\end{aligned}\)

Use the recursive definition:

\(\log \left( {{{\log }^{\left( 0 \right)}}n} \right) \le 65536\)

Take the exponential with base \(2\) of each side:

\({\log ^{\left( 0 \right)}}n \le {2^{65536}}\)

Use the definition of the logarithm:

\(n \le {2^{65536}}\)

Therefore, \({2^{65536}}\) is the largest integer \(n\) for which \({\log ^ * }n = 5\).