Math31B Midterm1

Math31B Midterm1 - Math 31B Lecture 4 Fall 2011 Exam#1 17...

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Math 31B, Lecture 4, Fall 2011 Exam #1, 17 October 2011 Name: ID Number: Section and TA: You have 50 minutes for the exam. No calculators, phones, notes, or books allowed. You must show your work for credit. Question: 1 2 3 4 5 6 7 Total Points: 8 13 14 24 14 19 8 100 Score:
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1. Prove the change of base formula for logarithms: log a ( x ) = log b ( x ) log b ( a ) with a 6 = 1 and b 6 = 1 (8 points) . x = b log b ( x ) , a = b log b ( a ) x = ( a ) log a ( x ) = ( b log b ( a ) ) log a ( x ) b log b ( x ) = b log b ( a ) log a ( x ) log b ( x ) = log b ( a ) log a ( x ) log a ( x ) = log b ( x ) log b ( a )
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2. Using the chain rule, find d d x ln( f ( x )) in terms of d d x f ( x ) and f ( x ) (4 points). Then, using logarithmic differentiation, evaluate d d x x ( x + 1) 3 (3 x - 1) 2 (9 points) . Do not find a common denominator. d d x ln( f ( x )) = d d x f ( x ) f ( x ) , so for the second part d d x f ( x ) = f ( x ) d d x ln( f ( x )) . Using this formula,
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This note was uploaded on 04/06/2012 for the course MATH 31B taught by Professor Valdimarsson during the Fall '08 term at UCLA.

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Math31B Midterm1 - Math 31B Lecture 4 Fall 2011 Exam#1 17...

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