Question

Let n be a positive integer such that ${{\log }_{2}}{{\log }_{2}}{{\log }_{2}}{{\log }_{2}}{{\log }_{2}}\left( n \right)< 0< {{\log }_{2}}{{\log }_{2}}{{\log }_{2}}{{\log }_{2}}\left( n \right)$. Let I be the number of digits in the binary expansion of n. Then the minimum and maximum possible values of I are.
A. 5 and 16
B. 5 and 17
C. 4 and 16
D. 4 and 17

Answer 1

Hint: We need to apply the theorem of logarithm to find the inequation on n. we change the logarithm portion back into indices form. The inequation will them give the number of digits in the binary expansion of n from the indices number. We know the relation between these two to find the solution of the problem.

Complete step-by-step answer:
We have been given a linear inequation ${{\log }_{2}}{{\log }_{2}}{{\log }_{2}}{{\log }_{2}}{{\log }_{2}}\left( n \right)< 0< {{\log }_{2}}{{\log }_{2}}{{\log }_{2}}{{\log }_{2}}\left( n \right)$. We need to break it into two parts and apply rules of logarithm to find the possible values of n.
We know that ${{\log }_{a}}b=x\Rightarrow {{a}^{x}}=b$. This relation holds for inequality also.
We take the first part as ${{\log }_{2}}{{\log }_{2}}{{\log }_{2}}{{\log }_{2}}{{\log }_{2}}\left( n \right)< 0$.
Applying the same formula again and again we get
\[\begin{align}
  & {{\log }_{2}}{{\log }_{2}}{{\log }_{2}}{{\log }_{2}}\left( n \right)< \left( {{2}^{0}}=1 \right) \\
 & \Rightarrow {{\log }_{2}}{{\log }_{2}}{{\log }_{2}}\left( n \right)< \left( {{2}^{1}}=2 \right) \\
 & \Rightarrow {{\log }_{2}}{{\log }_{2}}\left( n \right)< \left( {{2}^{2}}=4 \right) \\
 & \Rightarrow {{\log }_{2}}\left( n \right)< \left( {{2}^{4}}=16 \right) \\
 & \Rightarrow n< {{2}^{16}} \\
\end{align}\]
So, the value of n has to be less than 16.
The second part of the relation is ${{\log }_{2}}{{\log }_{2}}{{\log }_{2}}{{\log }_{2}}\left( n \right)> 0$
Applying the same formula of logarithm again and again we get
\[\begin{align}
  & {{\log }_{2}}{{\log }_{2}}{{\log }_{2}}\left( n \right)> \left( {{2}^{0}}=1 \right) \\
 & \Rightarrow {{\log }_{2}}{{\log }_{2}}\left( n \right)> \left( {{2}^{1}}=2 \right) \\
 & \Rightarrow {{\log }_{2}}\left( n \right)> \left( {{2}^{2}}=4 \right) \\
 & \Rightarrow n> \left( {{2}^{4}}=16 \right) \\
\end{align}\]
We got two equations and joining them together we get $16< n< {{2}^{16}}$.
Now we know that in the binary expansion of ${{2}^{n}}$, there would be $\left( n+1 \right)$ terms or digits. So, we got that ${{2}^{4}}=16< n$ which means the minimum number of terms in the expansion will be $4+1=5$.
Also, in $n< {{2}^{16}}$, the maximum number of terms in the expansion will be $16+1=17$.
So, the minimum and maximum possible values of I are 5 and 17.

So, the correct answer is “Option B”.

Note: The binary expansion is of the digits of possible values for n. the digits always have to be integer. That’s why we apply the theorem that in the binary expansion of ${{2}^{n}}$, there would be $\left( n+1 \right)$ terms or digits. Logarithmic formula of ${{\log }_{a}}b=x\Rightarrow {{a}^{x}}=b$ can be applied for both equality and inequality.