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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 ´ = log 9 ³³ x p x 2 63 ´ ³ x + p x 2 + 63 ´´ = log 9 ¡ x 2 ¡ x 2 63 ¢¢ = log 9 63 = log 9 (9 · 7) = log 9 9 + log 9 7 = 1 + log 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 ´ 5 ln x = e 5 ln 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 6 log 5 ( 8 + 9) = 6 6 log 5 1 = 6 0 = 6 ( 1 ) f ( 8) = 6 log 4 ( 8 + 9) 6 = 6 log 4 1 6 = 0 6 = 6 ( 3 ) f ( 8) = 7 log 4 ( 8 + 9) 7 = 7 log 4 1 7 = 0 7 = 7 ( 5 ) Since P = ( 8 , 6) , we see that f ( x ) = 6 log 4 ( x + 9) 6 .
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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 fi 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 , and then let 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 ) = ln 7 x ln e = ln 7 x = ln 7 . 6. We see that lim x →∞ (ln (5 + 2 x ) ln (7 + 5 x )) = lim x →∞ μ ln μ 5 + 2 x 7 + 5 x ¶¶ = ln μ lim x →∞ μ 5 + 2 x 7 + 5 x ¶¶ = ln μ 2 5 .
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