Is log₁ n = O(log16 n)? What about log16 n O(log₁ n)? Why or why not?
Is log₁ n = O(log16 n)? What about log16 n O(log₁ n)? Why or why not?
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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![**Is \(\log_4 n = O(\log_{16} n)\)? What about \(\log_{16} n = O(\log_4 n)\)? Why or why not?**
This question involves examining the growth rates of logarithmic functions with different bases using Big O notation.
1. **Examining \(\log_4 n = O(\log_{16} n)\):**
- To determine if \(\log_4 n\) is \(O(\log_{16} n)\), we use the change of base formula:
\[
\log_4 n = \frac{\log_{16} n}{\log_{16} 4}
\]
- Since \(\log_{16} 4\) is a constant, \(\log_4 n\) is indeed \(O(\log_{16} n)\) because dividing by a constant does not affect the asymptotic growth rate.
2. **Examining \(\log_{16} n = O(\log_4 n)\):**
- Similarly, using the change of base:
\[
\log_{16} n = \frac{\log_4 n}{\log_4 16}
\]
- Since \(\log_4 16\) is a constant, \(\log_{16} n\) is \(O(\log_4 n)\) for the same reason that dividing by a constant does not change the growth rate.
In conclusion, both \(\log_4 n = O(\log_{16} n)\) and \(\log_{16} n = O(\log_4 n)\) are true because they essentially represent the same logarithmic growth, just scaled by constant factors due to the change of base formula.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F153bfb9c-b05e-4b91-94e3-9e151aaf7f28%2F30c213eb-cc43-4b2d-b723-02f4a9a8d7da%2Fdhgafe_processed.png&w=3840&q=75)
Transcribed Image Text:**Is \(\log_4 n = O(\log_{16} n)\)? What about \(\log_{16} n = O(\log_4 n)\)? Why or why not?**
This question involves examining the growth rates of logarithmic functions with different bases using Big O notation.
1. **Examining \(\log_4 n = O(\log_{16} n)\):**
- To determine if \(\log_4 n\) is \(O(\log_{16} n)\), we use the change of base formula:
\[
\log_4 n = \frac{\log_{16} n}{\log_{16} 4}
\]
- Since \(\log_{16} 4\) is a constant, \(\log_4 n\) is indeed \(O(\log_{16} n)\) because dividing by a constant does not affect the asymptotic growth rate.
2. **Examining \(\log_{16} n = O(\log_4 n)\):**
- Similarly, using the change of base:
\[
\log_{16} n = \frac{\log_4 n}{\log_4 16}
\]
- Since \(\log_4 16\) is a constant, \(\log_{16} n\) is \(O(\log_4 n)\) for the same reason that dividing by a constant does not change the growth rate.
In conclusion, both \(\log_4 n = O(\log_{16} n)\) and \(\log_{16} n = O(\log_4 n)\) are true because they essentially represent the same logarithmic growth, just scaled by constant factors due to the change of base formula.
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