Exam Solutions for Appm 1345, Spring 2010: Logarithmic Functions and L'Hopital's Rule, Exams of Calculus

Download Exam Solutions for Appm 1345, Spring 2010: Logarithmic Functions and L'Hopital's Rule and more Exams Calculus in PDF only on Docsity! APPM 1345 Exam 3 Solutions Spring 2010 1. (a) ln(lnx) = 1 lnx = e x = ee . (b) 25log5 2x = 400 (52)log5 2x = 400 5log5(2x) 2 = 400 (2x)2 = 400 2x = 20 x = 10 . 2. Let y = f(x) = 3 2ex . (a) We interchange x and y to find the inverse function. x = 3 2ey 2xey = 3 ey = 3 2x y = ln ( 3 2x ) . The inverse function is f−1(x) = ln ( 3 2x ) . (b) The domain of f is (−∞,∞). The range of f is (0,∞). The domain of f−1 is (0,∞).The range of f−1 is (−∞,∞). (c) Note that f(0) = 3/2. f(x) = 3 2ex = 3 2 e−x df dx = −3 2 e−x df dx ∣∣∣∣ x=0 = −3 2 . f−1(x) = ln ( 3 2x ) = ln 3− ln 2− lnx df−1 dx = −1 x df−1 dx ∣∣∣∣ x=3/2 = −2 3 . Each slope is the reciprocal of the other. 3. (a) We convert log6 x to (lnx)/(ln 6). Let u = ln x, du = dx x . Then the u-limits are ln 6 to ln 36.∫ 36 6 dx x(log6 x) 2 = (ln 6)2 ∫ 36 6 dx x(lnx)2 = (ln 6)2 ∫ ln 36 ln 6 u−2 du = (ln 6)2 −1 u ]ln 36 ln 6 = (ln 6)2 ( − 1 ln 36 + 1 ln 6 ) = (ln 6)2 ( − 1 2 ln 6 + 1 ln 6 ) = ln 6 2 . (b) Let u = 3t − 1, du = 3t(ln 3) dt. Then the u-limits are 2 to 8. ∫ 2 1 3t 3t − 1 dt = 1 ln 3 ∫ 8 2 du u = 1 ln 3 ln |u| ]8 2 = 1 ln 3 (ln 8− ln 2) = 1 ln 3 (3 ln 2− ln 2) = 2 ln 2 ln 3 = 2 log3 2 . (c) Let u = e3t − 2, du = 3e3t dt. Then y = ∫ e3t√ e3t − 2 dt = 1 3 ∫ u−1/2 du = 1 3 2u1/2 + C = 2 3 √ e3t − 2 + C Next we use the initial value y(ln 3) = 1/3 to find the value of C. 1 3 = 2 3 √ e3 ln 3 − 2 + C = 2 3 √ eln 27 − 2 + C = 2 3 √ 27− 2 + C = 10 3 + C C = −3. The solution is y = 2 3 √ e3t − 2− 3 .