Let n be a positive integer such that log _2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the inequality: $\log _2 \log _2 \log _2 \log _2 \log _2(n) < 0 < \log _2 \log _2 \log _2 \log _2(n)$.First,consider $\log _2 \log _2 \log _

Vedclass · Accepted Answer

$5$ and $16$

Vedclass · Answer

(A) Given the inequality: $\log _2 \log _2 \log _2 \log _2 \log _2(n) First,consider $\log _2 \log _2 \log _2 \log _2(n) > 0$:$\log _2 \log _2 \log _2 \log _2(n) > 0 \implies \log _2 \log _2 \log _2(n) > 1 \implies \log _2 \log _2(n) > 2 \implies \log _2(n) > 4 \implies n > 2^4 = 16$.Next,consider $\log _2 \log _2 \log _2 \log _2 \log _2(n) $\log _2 \log _2 \log _2 \log _2 \log _2(n) Thus,$16 The number of digits $l$ in the binary expansion of $n$ is given by $\lfloor \log_2(n) \rfloor + 1$.For $n > 16$,the minimum value is $\lfloor \log_2(16 + 1) \rfloor + 1 = 4 + 1 = 5$.For $n Therefore,the minimum and maximum values of $l$ are $5$ and $16$.

Let $n$ be a positive integer such that $\log _2 \log _2 \log _2 \log _2 \log _2(n) < 0 < \log _2 \log _2 \log _2 \log _2(n)$. Let $l$ be the number of digits in the binary expansion of $n$. Then the minimum and the maximum possible values of $l$ are

Similar Questions

If ${a^x} = b$,${b^y} = c$,and ${c^z} = a$,then the value of $xyz$ is:

The number of solutions to the equation $\log _2(x + 5) = 6 - x$ is

If $n = 1983!$,then the value of the expression $\frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \dots + \frac{1}{\log_{1983} n}$ is equal to

The interval of $x$ in which the inequality $5^{\frac{1}{4}(\log_5 x)^2} \geq 5x^{\frac{1}{5}(\log_5 x)}$ holds is:

$\log_{7} \log_{7} \sqrt{7 \sqrt{7 \sqrt{7}}} = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module