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### test03

Course: MAC 1105, Fall 2009
School: nwfsc.edu
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Algebra College Spring, 2007 Test 3 Name: Dr. D. P. Story Global Instructions: (100 points) Solve each problem and box in your final answer . (2pts ) ea. 1. True/False questions. Enter `T' for True and `F' for False. (a) (b) (c) (d) (e) F T T F T The domain of a logarithm function is (-, ). The range of the inverse of a function f is the same as the domain of f . If f is invertible, then f (f -1 (w)) = w, for...

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Algebra College Spring, 2007 Test 3 Name: Dr. D. P. Story Global Instructions: (100 points) Solve each problem and box in your final answer . (2pts ) ea. 1. True/False questions. Enter `T' for True and `F' for False. (a) (b) (c) (d) (e) F T T F T The domain of a logarithm function is (-, ). The range of the inverse of a function f is the same as the domain of f . If f is invertible, then f (f -1 (w)) = w, for all w in the domain of f -1 . To determine whether a function is one-to-one, we use the vertical line test. Suppose a > 0 and a = 1. If au = av , then it follows that u = v. In the questions that follow, fill in the blank. (f) Let a > 0 and a = 1. Then w = loga (t) if and only if t = aw (h) The inverse to the function f (x) = log3 (x) is f (i) For all x, loga a = x x -1 . . . . (g) The natural logarithm is defined to be the logarithm with a base of a = e (x) = 3 =x x and for all x > 0, a 2 loga (x) . (j) Let f (x) = 3x + 1 and g(x) = 2x - 1. Then (f g)(-2) = 22 (7pts ) 2. Let f (x) = 2 3 and g(x) = . Find the function (f g) and find its domain. Simply! x-1 x Solution: We use standard methods: f (g(x)) = f (2/x) 3 = (2/x) - 1 3x = 2-x x=0 substitution simplify, note x = 2 (f g)(x) = 3x 2-x Dom(f g) = { x | x = 0 and x = 2 } (5pts ) 3. Indicate which of the following functions are one-to-one, hence invertible. y y y y 1. x 2. x 3. x 4. x The functions depicted in graphs 2, 3 is one-to-one, separating each with a comma.) are one-to-one. (List the number of each graph that CA/T3 (7pts ) Page 2 of 4 2 . 1+x Name: 4. Find the inverse function to f (x) = 2-x x f -1 (x) = Solution: We follow the procedure: 2 2 2 f (x) = = y = = x = = x(1 + y) = 2 = x + xy = 2 1+x 1+x 1+y 2-x = xy = 2 - x = y = x 2-x = f -1 (x) = x (4pts ) ea. 5. Solve each of the following equations. (Box in your final answer.) (a) 23x-1 = 8 Solution: (c) log5 (2x) = 2 23x-1 = 8 = 23x-1 = 23 Solution: If log5 (2x) = 2, then 2x = 52 . We = 3x - 1 = 3 deduce that x = 25/2 . = x = 4/3 (b) e4x ex = e12 Solution: e4x ex = e12 = e4x+x = e12 = 4x + x2 = 12 = x2 + 4x - 12 = 0 = (x + 6)(x - 2) = 0 = x = -6, 2 (6pts ) 6. Find the domain of the function f (x) = log 1 . Use interval notation. x+1 Solution: There is a natural restraint that we can only take logarithms of positive numbers. Dom(f ) = (-1, ) (3pts ) ea. We require that x + 1 > 0 or x > -1. In interval notation, the domain is (-1, ). 2 2 2 (d) ln e4x = 12 Solution: Using the properties of logarithms, we deduce that 4x ln e = 12, or that 4x = 12, since ln e = 1. Thus, x = 3 . 7. Calculate each of the following exactly, using the various properties of logarithms and exponentials. 1 (a) log7 = 13 -13 (b) 2log2 (3x) = 3x 7 1 Solution: Immediate! Solution: log7 13 = log7 7-13 = -13 7 CA/T3 (6pts ) Page 3 of 4 Name: (x + 3)2 (2x - 1)(x2 + 1)3 8. Use the properties of logarithms to expand the expression ln ln (x + 3)2 = 2 ln(x + 3) - ln(2x - 1) - 3 ln(x2 + 1) (2x - 1)(x2 + 1)3 Solution: We use the properties of logarithms ln (x + 3)2 = ln(x + 3)2 - ln(2x - 1) - ln(x2 + 1)3 (2x - 1)(x2 + 1)3 = 2 ln(x + 3) - ln(2x - 1) - 3 ln(x2 + 1) (3pts ) ea. 9. Elementary. Solve each of the following equations for x using the various properties of exponentials and logarithms. Leave your solutions in list form, for example, x = 1, 3 , boxed in, of course. (a) log3 (2x - 1) = 4 Solution: log3 (2x - 1) = 4 = 2x - 1 = 3 = 81 = 2x = 82 4 (b) 2 log5 x = 3 log5 4 Solution: Note that x > 0. 2 log5 x = 3 log5 4 = log5 x2 = log5 43 = x2 = 43 = x = 8 = x = 8 = x = 41 (5pts ) ea. 10. Advanced. Solve each of the following equations for x using the various properties of exponentials and logarithms. Leave your solutions in list form, for example, x = 1, 3 , boxed in, of course. (a) log4 x + log4 (x - 3) = 1 Solution: After noting that x > 0, we apply (b) 5 23x = 8 (round the answer to 3 decimal the properties of logs: places) Solution: log4 x + log4 (x - 3) = 1 = log4 x(x - 3) = 1 = x(x - 3) = 4 = x2 - 3x - 4 = 0 = (x - 4)(x + 1) = 0 = x = -1, 4 = x = 4 5 23x = 8 = 23x = 8/5 = ln 23x = ln(8/5) = 3x ln 2 = ln(8/5) = x = ln(8/5) 3 ln 2 = x 0.226 CA/T3 Page 4 of 4 Name: Compound Interest: Recall the following formulas r -nt r nt P =A 1+ A=P 1+ n n A = P ert P = Ae-rt (3pts ) ea. 11. A...

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