HW 14 Solutions - Mathematics 408K: Homework 14 Gagan Tara...

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Mathematics 408 K: Homework 14 Gagan Tara Nanda 23 rd November, 2004 HW 14 : # s 1 23 1. Using properties of the log function, we see that log 9 ³ x p x 2 63 ´ +log 9 ³ x + p x 2 63 ´ =l o g 9 ³³ x p x 2 63 ´³ x + p x 2 +63 ´´ =l o g 9 ¡ x 2 ¡ x 2 63 ¢¢ =l o g 9 63 =l o g 9 (9 · 7) =l o g 9 9+log 9 7 =1+ l o g 9 7 . 2. Using the identity a log a b = b , we see that 6 5(log 6 e )ln x = ³ 6 log 6 e ´ 5ln x = e 5ln x = ³ e ln x ´ 5 = x 5 . 3. Since the graph has a vertical asymptote at x = 9 , the expression for f ( x ) has the form f ( x )= a log c ( x +9) a . This eliminates choices (2) and (4) . Now evaluate f ( 8) for the remaining expressions: f ( 8) = 6 6log 5 ( 8+9)=6 6log 5 1=6 0=6 ( 1 ) f ( 8) = 6 log 4 ( 8+9) 6=6log 4 1 6=0 6= 6 ( 3 ) f ( 8) = 7 log 4 ( 8+9) 7=7log 4 1 7=0 7= 7 ( 5 ) Since P =( 8 ,
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Gagan Tara Nanda 2 4. Combine the left-hand side to get ln ( x ( x + 1)) = ln (30) x ( x +1) = 30 x 2 + x 30 = 0 ( x +6)( x 5) = 0 x = 6 , 5 However, ln ( 6) is not de f ned, so x 6 = 6 x =5 . 5. Multiply both sides of the given equation by e x to get e x · e x 21 e x · e x =4 e x , andthenlet y = e x to get y 2 21 = 4 y y 2 4 y 21 = 0 ( y 7) ( y +3) = 0 y =7 , 3 So this implies e x =7 , 3 , but since e x > 0 for all x , we deduce that e x =7 .Then ln ( e x )=l n 7 x ln e =l
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This note was uploaded on 02/28/2010 for the course M 56200 taught by Professor Radin during the Fall '09 term at University of Texas at Austin.

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HW 14 Solutions - Mathematics 408K: Homework 14 Gagan Tara...

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