HW 13 Solutions - 3 e 5 x 3 x 2 + C . Since y (0) = 1 , we...

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Mathematics 408 K: Homework 13 Gagan Tara Nanda 16 th November, 2004 HW 13 : # s 7 , 19 , 20 7. Here, we should f rst convert both sides of the equation to a common base. So 4 x 2 =16 11 2 x 15 = ¡ 4 2 ¢ 11 2 x 15 =4 11 x 30 . This implies x 2 =11 x 30 ,so x 2 11 x +30=0 . From here, it is tempting to factor x 2 11 x +30= ( x 5) ( x 6) , but this is unnecessary. The goal is to f nd the value of a + b , and this is ( 11) = 11 , since the sum of the roots of the quadratic equation x 2 + cx + d =0 is c . 19. We shall f rst f nd the general form for anti-derivatives of the given function. Since dy dx = y 0 =15 e 5 x 6 x , we see that y = μ 15 5 e 5 x 3 x 2 + C =
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Unformatted text preview: 3 e 5 x 3 x 2 + C . Since y (0) = 1 , we have 3 e 5(0) 3 2 + C = 1 3 + C = 1 C = 4 Hence y (1) = 3 e 5(1) 3 1 2 + 4 = 3 e 5 + 1 . 20. We see that 3 log 6 4 6 3 6 log 3 3 3 4 = 3 log 6 6 3 1 / 4 6 log 3 3 4 1 / 3 = 3 log 6 6 3 / 4 6 log 3 3 4 / 3 = 3 3 4 log 6 6 6 4 3 log 3 3 = 9 4 log 6 6 8 log 3 3 = 9 4 8 = 23 4 ....
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