/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Assuming the existence of roots,... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 25

Assuming the existence of roots, show that if \(c>1\), then \(c^{1 / m}n\).

Short Answer

Expert verified
We have shown that if \(c>1\), then \(c^{1/m}n\).

Step by step solution

01

Rewrite the Given Inequalities

Firstly, look at the expression \(c^{1/m} < c^{1/n}\). We can see that both are exponents of the same base \(c\). To make comparison easier rewrite the inequality using the properties of exponents: \(c^{1/m} < c^{1/n}\) is equivalent to \(c^{n/m} < c\), because if we raise both sides of the inequalities by the power of \(n\), the inequality sign will not change.
02

Introduce the Natural Logarithm

Once the expression is simplified, introduce the natural logarithm to both sides of the inequality. We get \(\ln(c^{n/m}) < \ln(c)\). According to the properties of logarithms, we can transfer the exponent in front of the logarithm. Hence, \((n/m)*\ln(c) < \ln(c)\).
03

Reduction of Common Terms

In this step, eliminate the common term present on both sides of the inequality (in this case \(\ln(c)\)) giving \((n/m) < 1\). We are allowed to do this because \(\ln(c)> 0\), as \(c > 1\) and the logarithm for values greater than one is positive.
04

Final Rearrangement

Finally, rearrange the inequality \((n/m) < 1\) to get \(n < m\), hence \(m > n\), which is the condition given in the problem.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) If \(a

Express \(\frac{1}{7}\) and \(\frac{2}{19}\) as periodic decimals.

Determine and sketch the set of pairs \((x, y)\) in \(\mathbb{R} \times \mathbb{R}\) that satisfy: (a) \(|x|=|y|\). (b) \(|x|+|y|=1\), (c) \(|x y|=2\), (d) \(|x|-|y|=2\).

Find all \(x \in \mathbb{R}\) that satisfy the inequality \(4<|x+2|+|x-1|<5\).

If \(x, y, z \in \mathbb{R}\) and \(x \leq z\), show that \(x \leq y \leq z\) if and only if \(|x-y|+|y-z|=|x-z|\). Interpret this geometrically.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.