Explanation Given information: b x = a x log a b for a , b > 0 such that a ≠ 1 , b ≠ 1 Concept used: Change-of-base :- it is the formula by which we can easily convert between bases in exponentials and logarithms. b x = a x log a b , b x = a x ln b , log b x = log a x log a b , log b x = ln x ln b Calculation: b x = a x<

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Question

Chapter 1.6, Problem 50E

To determine

Verify that for all x, the change-of-base formula holds:

bx=axlogab for a,b>0 such that a≠1,b≠1

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Chapter 1.6, Problem 49E

Chapter 1.6, Problem 51E

Chapter 1 Solutions

Calculus: Early Transcendentals

Ch. 1.1 - Prob. 1PQ Ch. 1.1 - Prob. 2PQ Ch. 1.1 - Prob. 3PQ Ch. 1.1 - Prob. 4PQ Ch. 1.1 - Prob. 5PQ Ch. 1.1 - Prob. 6PQ Ch. 1.1 - Prob. 7PQ Ch. 1.1 - Prob. 8PQ Ch. 1.1 - Prob. 9PQ Ch. 1.1 - Prob. 10PQ

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - 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Prob. 1E Ch. 1.2 - Prob. 2E Ch. 1.2 - Prob. 3E Ch. 1.2 - Prob. 4E Ch. 1.2 - Prob. 5E Ch. 1.2 - Prob. 6E Ch. 1.2 - Prob. 7E Ch. 1.2 - Prob. 8E Ch. 1.2 - Prob. 9E Ch. 1.2 - Prob. 10E Ch. 1.2 - Prob. 11E Ch. 1.2 - Prob. 12E Ch. 1.2 - Prob. 13E Ch. 1.2 - Prob. 14E Ch. 1.2 - Prob. 15E Ch. 1.2 - Prob. 16E Ch. 1.2 - Prob. 17E Ch. 1.2 - Prob. 18E Ch. 1.2 - Prob. 19E Ch. 1.2 - Prob. 20E Ch. 1.2 - Prob. 21E Ch. 1.2 - Prob. 22E Ch. 1.2 - Prob. 23E Ch. 1.2 - Prob. 24E Ch. 1.2 - Prob. 25E Ch. 1.2 - Prob. 26E Ch. 1.2 - Prob. 27E Ch. 1.2 - Prob. 28E Ch. 1.2 - Prob. 29E Ch. 1.2 - Prob. 30E Ch. 1.2 - Prob. 31E Ch. 1.2 - Prob. 32E Ch. 1.2 - Prob. 33E Ch. 1.2 - Prob. 34E Ch. 1.2 - Prob. 35E Ch. 1.2 - Prob. 36E Ch. 1.2 - Prob. 37E Ch. 1.2 - Prob. 38E Ch. 1.2 - Prob. 39E Ch. 1.2 - Prob. 40E Ch. 1.2 - Prob. 41E Ch. 1.2 - Prob. 42E Ch. 1.2 - Prob. 43E Ch. 1.2 - Prob. 44E Ch. 1.2 - Prob. 45E Ch. 1.2 - Prob. 46E Ch. 1.2 - Prob. 47E Ch. 1.2 - Prob. 48E Ch. 1.2 - Prob. 49E Ch. 1.2 - Prob. 50E Ch. 1.2 - Prob. 51E Ch. 1.2 - Prob. 52E Ch. 1.2 - Prob. 53E Ch. 1.2 - Prob. 54E Ch. 1.2 - Prob. 55E Ch. 1.2 - Prob. 56E Ch. 1.2 - Prob. 57E Ch. 1.2 - Prob. 58E Ch. 1.2 - Prob. 59E Ch. 1.2 - Prob. 60E Ch. 1.2 - Prob. 61E Ch. 1.2 - Prob. 62E Ch. 1.2 - Prob. 63E Ch. 1.3 - Prob. 1PQ Ch. 1.3 - Prob. 2PQ Ch. 1.3 - Prob. 3PQ Ch. 1.3 - Prob. 4PQ Ch. 1.3 - Prob. 5PQ Ch. 1.3 - Prob. 6PQ Ch. 1.3 - Prob. 1E Ch. 1.3 - Prob. 2E Ch. 1.3 - Prob. 3E Ch. 1.3 - Prob. 4E Ch. 1.3 - Prob. 5E Ch. 1.3 - Prob. 6E Ch. 1.3 - Prob. 7E Ch. 1.3 - Prob. 8E Ch. 1.3 - Prob. 9E Ch. 1.3 - Prob. 10E Ch. 1.3 - Prob. 11E Ch. 1.3 - Prob. 12E Ch. 1.3 - Prob. 13E Ch. 1.3 - Prob. 14E Ch. 1.3 - Prob. 15E Ch. 1.3 - Prob. 16E Ch. 1.3 - Prob. 17E Ch. 1.3 - Prob. 18E Ch. 1.3 - Prob. 19E Ch. 1.3 - Prob. 20E Ch. 1.3 - Prob. 21E Ch. 1.3 - Prob. 22E Ch. 1.3 - Prob. 23E Ch. 1.3 - Prob. 24E Ch. 1.3 - Prob. 25E Ch. 1.3 - Prob. 26E Ch. 1.3 - Prob. 27E Ch. 1.3 - Prob. 28E Ch. 1.3 - Prob. 29E Ch. 1.3 - Prob. 30E Ch. 1.3 - Prob. 31E Ch. 1.3 - Prob. 32E Ch. 1.3 - Prob. 33E Ch. 1.3 - 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Prob. 56E Ch. 1.5 - Prob. 57E Ch. 1.5 - Prob. 58E Ch. 1.5 - Prob. 59E Ch. 1.5 - Prob. 60E Ch. 1.6 - Prob. 1PQ Ch. 1.6 - Prob. 2PQ Ch. 1.6 - Prob. 3PQ Ch. 1.6 - Prob. 4PQ Ch. 1.6 - Prob. 5PQ Ch. 1.6 - Prob. 6PQ Ch. 1.6 - Prob. 7PQ Ch. 1.6 - Prob. 8PQ Ch. 1.6 - Prob. 9PQ Ch. 1.6 - Prob. 1E Ch. 1.6 - Prob. 2E Ch. 1.6 - Prob. 3E Ch. 1.6 - Prob. 4E Ch. 1.6 - Prob. 5E Ch. 1.6 - Prob. 6E Ch. 1.6 - Prob. 7E Ch. 1.6 - Prob. 8E Ch. 1.6 - Prob. 9E Ch. 1.6 - Prob. 10E Ch. 1.6 - Prob. 11E Ch. 1.6 - Prob. 12E Ch. 1.6 - Prob. 13E Ch. 1.6 - Prob. 14E Ch. 1.6 - Prob. 15E Ch. 1.6 - Prob. 16E Ch. 1.6 - Prob. 17E Ch. 1.6 - Prob. 18E Ch. 1.6 - Prob. 19E Ch. 1.6 - Prob. 20E Ch. 1.6 - Prob. 21E Ch. 1.6 - Prob. 22E Ch. 1.6 - Prob. 23E Ch. 1.6 - Prob. 24E Ch. 1.6 - Prob. 25E Ch. 1.6 - Prob. 26E Ch. 1.6 - Prob. 27E Ch. 1.6 - Prob. 28E Ch. 1.6 - Prob. 29E Ch. 1.6 - Prob. 30E Ch. 1.6 - Prob. 31E Ch. 1.6 - Prob. 32E Ch. 1.6 - Prob. 33E Ch. 1.6 - Prob. 34E Ch. 1.6 - Prob. 35E Ch. 1.6 - Prob. 36E Ch. 1.6 - Prob. 37E Ch. 1.6 - 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Prob. 55CRE Ch. 1 - Prob. 56CRE Ch. 1 - Prob. 57CRE Ch. 1 - Prob. 58CRE Ch. 1 - Prob. 59CRE Ch. 1 - Prob. 60CRE Ch. 1 - Prob. 61CRE Ch. 1 - Prob. 62CRE Ch. 1 - Prob. 63CRE Ch. 1 - Prob. 64CRE Ch. 1 - Prob. 65CRE Ch. 1 - Prob. 66CRE

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Verify that for all x , the change-of-base formula holds: b x = a x log a b for a , b > 0 such that a ≠ 1 , b ≠ 1

Solution Summary: The author explains the change-of-base formula, which is the formula by which we can easily convert between bases in exponentials and logarithms

Videos

Verify that for all x, the change-of-base formula holds:

bx=axlogab for a,b>0 such that a≠1,b≠1

Want to see the full answer?

Chapter 1 Solutions

Verify that for all x , the change-of-base formula holds: b x = a x log a b for a , b &gt; 0 such that a ≠ 1 , b ≠ 1

Solution Summary: The author explains the change-of-base formula, which is the formula by which we can easily convert between bases in exponentials and logarithms

Videos

Verify that for all x, the change-of-base formula holds: bx=axlogab for a,b>0 such that a≠1,b≠1

Want to see the full answer?

Chapter 1 Solutions

Verify that for all x , the change-of-base formula holds: b x = a x log a b for a , b > 0 such that a ≠ 1 , b ≠ 1

Verify that for all x, the change-of-base formula holds:

bx=axlogab for a,b>0 such that a≠1,b≠1