1. Math 140 Quiz 4 - Summer 2004 Solution Review (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

2. Problem 1 (33) Approximate each value using a calculator. Express answer rounded to three decimal places. log 9 (1/81) Without a calculator one notes, log 9 (1/81) = log 9 (9 -2 ) = -2log 9 (9) = -2. Or, evaluate one, or other, change-of-base formula: log 9 (1/81) = log(1/81)/log(9), log 9 (1/81) = ln(1/81)/ln(9).

3. Problem 2 (06) Approximate each value using a calculator. Express answer rounded to three decimal places. e 2.64 If necessary, use keystrokes 1, INV(or 2 nd ), ln to get value of e. Then, press y x key and enter 2.64 followed by equal (or enter) key. Result is 14.01320361..., which rounds to 14.013.

4. Problem 3 (28) Convert to logarithmic form: 5 3 = 125. One takes the base-5 logarithm of both sides to get log 5 (5 3 ) = log 5 (125). Then one observes that log 5 (5 3 ) = 3 log 5 (5) = 3. Thus, log 5 (125) = 3.

5. Problem 4 (06) Use a calculator to find the natural logarithm correct to four decimal places. ln(82 1/2 ) One notes, ln(82 1/2 ) = (1/2)ln(82) = 2.203359624 = 2.2034. Or, evaluate the square root first and find: ln(82 1/2 ) = ln(9.055385138) = 2.203359624 = 2.2034.

6. Problem 5 (50) If the following defines a one-to-one function, find the inverse of: f = {(-3, -7), (-2, -7), (-1, -1), (0, 2)}. Since both domain values –3 and –2 lead to –7, f fails the horizontal line test and is not 1-to-1. If one proceeded by swapping x y in f, getting: f -1 = {(-7, -3), (-7, -2), (-1, -1), (2, 0)}? But this is not a function and would not check. For example, f -1 (f(-3)) = f -1 (-7) = -3 or -2 and not just -3.

7. Problem 6 (11) If the following defines a one-to-one function, find the inverse of: f(x) = 2x + 8. The horizontal line test is satisfied since a graph of _ y = f(x) = 2x + 8 is a line. Hence, f(x) is 1-to-1. One proceeds by swapping x y and solving for y. Thus, x = 2y + 8 => y = (x - 8)/2. This means that f -1 (x) = (x - 8)/2. Check: f -1 (f(x)) = [(2x + 8) - 8]/2 = x.

8. A = P (1 + r/n ) nm = $480 (1 + 0.07/4 ) (4)(7) = $480 (1.0175 ) 28 = $780.20 Problem 7 (61) Compute the amount in m years if a principal P is invested at a nominal annual interest rate of r compounded as given. Round to the nearest cent. P = $480, m = 7, r = 7% compounded quarterly

9. Or, solve x = 4y for y to get f - -1 (x) = x/4. Then graph it. Draw the line L: y = x. Construct the inverse function line by reflection in L. Recognize f(x) as a straight line of slope m = 4 and y-intercept (0,0). Problem 8 (56) Graph f(x) as a solid line/curve & its inverse as a dashed line/curve on the same axes. f(x) = 4x. L This is answer C).

10. Problem 9 (61) Use a calculator and the base conversion formula to find the logarithm, correct to three decimal places. log 6.9 (5.5) Evaluate one, or other, change-of-base formula: log 6.9 (5.5) = log(5.5)/log(6.9) = 0.74036/0.83885 = 0.883, log 6.9 (5.5) = ln(5.5)/ln(6.9) = 1.70475/1.93152 = 0.883.

11. Problem 10 (72) Decide whether or not f(x) and g(x) are inverses of each other with: a) f(x) = 3x + 5, g(x) = (1/3)x - 5; b) f(x) = (x - 6) 2, g(x) = x 1/2 + 6. One needs to see if: (g o f )(x) = g(f(x)) = x or (f o g )(x) = f(g(x)) = x. a) f(g(x) ) = 3[(1/3)x - 5] + 5 = x - 10 => NO. b) f(g(x)) = [(x 1/2 + 6) - 6] 2 = [x 1/2 ] 2 = x => YES.

12. A = Pe rt 3P = Pe 0.0875t 3 = e 0.0875t ln(3) = ln(e 0.0875t ) = 0.0875t ln(e) = 0.0875t 0.0875t = ln(3) => t = ln(3)/0.0875 = 12.556 yr Problem 11 (83) How long will it take for an investment to triple in value if it earns 8.75% compounded continuously? Round your answer to three decimal places.

13. To verify D) is shown, recall known graph of log (x) with vertical asymptote at x = 0. Eliminate A), B), C), and E), choices by evaluating for a test point, say x = 1, with log on calculator & find a value that is not shown. Observe that the domain of f(x) in figure is x > 3. Problem 12 (39) Select the matching logarithmic function to the graph. f(x) = log (x - 3) It is now shifted to ( h, k ) = (3, 0) with a vertical asymptote at x = 3. A) log(3 – 1) = 0.3B) 3 – log(1) = 3C) log(1) – 3 = – 3E) log(1/3) = –.48

14. Evaluate A) & B) for a test point, say x = -3, with y x on calculator. A): 0.32 -3 = 30.5 is not on graph. B): 0.65 -3 = 3.6 is on graph. Eliminate C), D), and E), choices since they are of form a x, a >1, which increases & is not shown. [y = 0.65 -x = (1/.65) x = (1.538) x.] Problem 13 (44) Select the matching exponential function to the graph. Thus, the graph is of function f(x) = a x = 0.65 x. f(x) = a x = 0.65 x

15. Problem 14 (33) Solve the problem. log 5 (x) = - 3 One computes the exponential function 5 x of both sides (which must be equal because of the given equation) to get: 5 log 5 (x) = 5 -3 = 1/5 3 = 1/125. Then, one recalls that a log a (x) = x. Hence, x = 1/125.

16. Problem 15 (33) Solve the problem. log 5 (25) = x Without a calculator one notes, x = log 5 (25) = log 5 (5 2 ) = 2 log 5 (5) = 2. Or, evaluate one, or other, change-of-base formula: log 5 (25) = log(25)/log(5), log 5 (25) = ln(25)/ln(5).

17. Problem 16 (72) The formula D = 6e -0.04h can be used to find the number of milligrams D of a certain drug in a patient's bloodstream h hours after the drug has been given. When the number of milligrams reaches 2, the drug is to be given again. What is the time between injections? Rearrange 2 = 6e -0.04h ___________ 1/e -0.04h = 6/2 or e 0.04h = 3 _____ ln(e 0.04h ) = 0.04h ln(e) = 0.04h = ln(3) h = ln(3)/0.04 = 27.47 hours

18. Problem 17 (39) Solve the problem. pH = -log 10 [H+] Find the pH, if the [H+] = 5.2 x 10 -3. Use your calculator. ___________ pH = -log 10 [5.2 x 10 -3 ] ___________ = -(-2.284) _ __________ = 2.28

19. Problem 18 (56) Solve the equation. log 5x = log 4 + log (x - 3) Compute the exponential function 10 x of both sides (which must be equal because of the given equation) and use a log a (x) = x to get: 10 log 5x = 10 [log 4 +log(x –3)] = 10 log 4 10 log(x –3)] 5x = 4(x - 3) __________ 5x - 4x = - 12________ x = -12 Note: testing => log arguments, 5x= -60 & x-3= -15 are < 0; this is not allowed. Thus, there is no solution. _ ________

20. Problem 19 (56) Solve the equation. log y 8 = 5 Compute the exponential function y x of both sides (which must be equal because of the given equation) and use a log a (x) = x to get: y log y 8 = 8 = y 5 y 5 = 8 __________ y = 8 1/5 ________ {8 1/5 }

21. Problem 20 (100) Find the value of the expression. Let log b A = 3 and log b B = -4. Find log b [(AB) 1/2 ]. log b [(AB) 1/2 ] = (1/2)log b (AB) = _ (1/2)[log b (A)(+ log b (B)] = (1/2)[ (3)(+ (-4) ] = - 1/2