   Chapter 11.4, Problem 47ES ### 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
1 views

# For each positive real number u ,   log 2   u < u Use this fact and the result of exercise 21 in Section 11.1 to prove the following: For every integer n ≥ 1 , if x is any real number with x > ( 2 n ) 2 n then log 2 x < x 1 / n .

To determine

To prove:

For all integers n = 1, log2x < x1/nfor all real numbers x >(2n)2n.

Explanation

Given information:

For all positive real numbers u, log2u < u.

Proof:

Let n be a positive integer and let x be a positive real number such that x<(2n)2n.

Since x is a positive real number, x1/2n is also a positive real number.

We know that the inequality log2u<u holds for all positive real numbers.

Let u=x1/2n

log2x1/2n<x1/2n

Multiply each side of the equation by 2n (note that n > 0)

2nlog2x1/2n<2nx1/2n

Property logarithm: logbac=clogba

log2x=2n12nlog2x=2nlog2x1/2n<2nx1/2n

However, we also know that x>(2n)2n. Let us take the 1/2th power of each side (note that xm/n is an increasing function for all positive integersm and n ).

x1/2>(2n)n

Multiply both sides by x1/2

x1/2x1/2>(2n)nx1/2

Note that x1/2x1/2=x

x>(2n)nx1/2

Take the 1/ nth power of each side (note that xm/n is an increasing function for all positive integersm and n )

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 