   Chapter 11.4, Problem 27ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# Use Theorems 11.2.7-11.2.9 and properties 11.4.11, 11.4.12, and 11.4.13 to derive each statement in 27-30.27. 2 n + log 2 n is Θ ( n )

To determine

Derive the statement: 2n+log2n is Θ(n).

Explanation

Given information:

2n+log2n is Θ(n).

Calculation:

2n+log2n

For all real numbers b and r with b > 1 and r > 0, logbxxr for all sufficiently large real numbers x. thus there exists a real number k such that for all real numbers n > k ( b = 2 and r = 1):

log2nn

Note that it is safe to assume that k1. Adding 2x to both sides of the inequality, we obtain

2n+log2n3n

However, 2n+log2n is positive when n >1 (as n > 0 and log2n>0 ) and 3n is nonnegative when n > 0

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