soln1 - MATH 138 Calculus 2, Solutions to Assignment 1 1:...

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Unformatted text preview: MATH 138 Calculus 2, Solutions to Assignment 1 1: (a) Find all values of x such that e 2 ln( x- 1) = x + 5. Solution: Using the formula e b ln a = a b we have e 2 ln( x- 1) = ( x- 1) 2 , and so e 2 ln( x- 1) = x + 5 ( x- 1) 2 = x +5 x 2- 2 x +1 = x +5 x 2- 3 x- 4 = 0 ( x +1)( x- 4) = 0 x =- 1 or 4. Notice that when x =- 1, ln( x- 1) is not defined, and so the only solution is x = 4. (b) Find all x such that ln x- ln 2 = 1 + ln x . Solution: We have ln x- ln 2 = 1 + ln x ln x- ln x = ln 2 + 1 = ln 2 + ln e ln x x = ln(2 e ) ln x = ln 2 e x = 2 e x = 4 e 2 . (c) Let f ( x ) = log 2 ( x + 2) 3 4 . Find numbers a , b and c such that f- 1 ( x ) = a + 2 bx + c . Solution: Note that f ( x ) = log 2 ( x + 2) 3- log 2 4 = 3 log 2 ( x + 2)- 2. We have y = 3 log 2 ( x + 2)- 2 = y + 2 = 3 log 2 ( x + 2) = ( y + 2) / 3 = log 2 ( x + 2) = 2 ( y +2) / 3 = x + 2 = x =- 2 + 2 ( y +2) / 3 . Thus f- 1 ( x ) =- 2 + 2 ( x +2) / 3 and so we have a =- 2, b = 1 3 and c = 2 3 ....
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This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.

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soln1 - MATH 138 Calculus 2, Solutions to Assignment 1 1:...

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