SOLUTIONS Name: Section (circle one): LSA (001) Engin (002) Math 110 Quiz 5 (Sections 5.1 5.3) Available: Wednesday, November 8 Friday, November 10 1. You have 30 minutes to complete this quiz. Keeping track of time is your responsibility. If you do not return this quiz after 30 minutes, your grade on this quiz may be lowered. 2. Show an appropriate amount of work (including appropriate explanation) for each problem, so that graders can see not only your answer but how you obtained it. 3. Math Lab staff will not answer questions about the content of this quiz. 4. You may use a calculator on this quiz. 5. No other aids (notes, textbooks, computers, phones, smartwatches, etc.) are allowed. 6. Turn off or silence all phones and other electronic devices. 7. Do not discuss the contents of this quiz with anyone other than a Math 110 instructor until Friday, November 10 at 4:00 PM. Problem Points Score 1 9 2 7 3 5 4 6 Total: 27 Math Lab manager (print and sign name): Start time stamp: End time stamp:
Math 110 Quiz 5 (Fall 2017) page 2 1. [9 points] In each of the following problems, solve for the variable x. Show each step of your work, and leave all answers in exact form. Note that log denotes the base-10 logarithm. a. [2 points] ln(3x) = log(5 + π) Solution: Take the exponential of both sides: and divide by 3. e ln(3x) = e log(5+π) 3x = e log(5+π) b. [2 points] log(10 x ) = 3x e Answer: x = 1 3 elog(5+π) Solution: log(10 x ) is equal to x, so the equation is just x = 3x e. c. [2 points] ln x + ln x 2 = 60 Answer: x = e 2 Solution: Using the fact that ln(ab) = ln(a) + ln(b), we have ln(x 3 ) = 60. Take the exponential of both sides to get x 3 = e 60 and take each side to the power 1/3 to get x = (e 60 ) 1/3 = e 20. Answer: x = e 20
Math 110 Quiz 5 (Fall 2017) page 3 d. [3 points] 15e 4x+2 = 3 x Solution: Take the logarithm of each side: ln(15e 4x+2 ) = ln(3 x ). Using the properties ln(ab) = ln a + ln b and ln(a b ) = b ln a, we have ln(15) + ln(e 4x+2 ) = ln(3 x ) ln(15) + (4x + 2) = x ln 3 Rearrange and combine like terms: 4x x ln 3 = ln 15 2 (4 ln 3)x = ln 15 2 Finally, divide both sides by (4 ln 3). Answer: x = ln 15 2 4 ln 3 2. [7 points] Scientist Selena has a solution of Chemical C, which is radioactive, in her lab. Scientist Selena observes that her quantity of Chemical C is decaying exponentially. Three (3) days after she acquires her supply of Chemical C, only 80% of her original supply remains. a. [2 points] Let C 0 be the original quantity, in grams, of Scientist Selena s supply of Chemical C. In terms of C 0, give a function Q(d) that models Scientist Selena s quantity of Chemical C, in grams, d days after she acquires her original supply. Answer: Q(d) = C 0 (0.8) d/3 b. [3 points] What is the half-life of Chemical C? Include units in your answer. Solution: We need to determine when half of the original amount will remain. We started with C 0 grams of the chemical, so we should determine when 1 2C will remain. 1 2 C 0 = C 0 (0.8) d/3 1 2 = (0.8)d/3 Take the logarithm of each side: ( ) 1 ln = ln ( 0.8 d/3) 2 ( ) 1 ln = d ln 0.8 2 3 and solve for d: d = 3 ln(1/2) ln 0.8 9.319 Answer: about 9.319 days
Math 110 Quiz 5 (Fall 2017) page 4 c. [2 points] What is the continuous daily rate of decay of Scientist Selena s quantity of Chemical C? Give your final answer as a percent accurate to three (3) decimal places. Solution: We should write our exponential function Q(d) = C 0 (0.8) d/3 in the form Q(d) = ae kd. In this form, the number k is the continuous growth/decay rate. Since a = Q(0), we have a = C 0. If we write our exponential function in the form we see that Taking logarithms gives Q(d) = C 0 (0.8) d/3 or Q(d) = C 0 e kd Q(d) = C 0 ( 0.8 1/3 ) d or Q(d) = C0 (e k ) d, 0.8 1/3 = e k. k = ln ( 0.8 1/3) 0.07438 Answer: 7.438 % 3. [5 points] A family adopts three adult rabbits: Frisky, Sneezy, and Slippers. The vet tells them that Slippers is a neutered male and that Frisky and Sneezy are females. However, after 30 days, the family discovers that it has 15 rabbits. Now that the family is taking care of so many baby rabbits, it does not have time to spay or neuter the rabbits it has. Assume that the population of rabbits is modeled by an exponential function r(t), where t is the number of days since the adoption of Frisky, Sneezy, and Slippers so that t = 0 is the day of adoption, and on day t = 30 the family has 15 rabbits. a. [2 points] Find an equation for r(t). Answer: r(t) = 3 5 t/30 b. [3 points] How many days after the adoption of Frisky, Sneezy, and Slippers will the family have 100 rabbits? Show your work, then round your answer to the nearest day. Solution: We need to determine when r(t) = 100. 100 = 3 5 t/30 Divide by 3 then take the logarithm of each side: ( ) 100 ln = ln ( 5 t/30) 3 ( ) 100 ln = t 3 30 ln 5 Finally, solve for t to get t = 30 ln(100/3) ln 5 65.36 Answer: after about 65 days
Math 110 Quiz 5 (Fall 2017) page 5 4. [6 points] For each function F given below, determine the domain of F (x) and the equations of any horizontal and vertical asymptotes of the graph of y = F (x). If the graph does not have horizontal or vertical asymptotes, write none in the corresponding blank. a. [3 points] F (x) = 90e 1200x + 550 Solution: The domain of F (x) is all real numbers, since there is no value of x for which F (x) is undefined (you can raise e to any power, you can multiply any number by 90, you can add 550 to any number). The function 90e 1200x is an increasing exponential function (exponential growth) which we know looks like y y = 90e 1200x 90 with a horizontal asymptote at y = 0. The graph of our function F (x) is just this graph shifted up by 550. x domain: all real numbers horizontal asymptotes: y = 550 vertical asymptotes: none
Math 110 Quiz 5 (Fall 2017) page 6 ( ) x b. [3 points] F (x) = ln 125 50 + 30 Solution: The domain of the function ln(x) is x > 0. So for our function F (x) to be defined, we need x 125 50 > 0 which gives x > 6250. The graph of y = F (x) is some transformation of the graph of y = ln(x), which looks like y y = ln x x and has a vertical asymptote at x = 0. Note that the graph of y = ln x does not have a horizontal asymptote it is growing very slowly as x gets larger, but does get arbitrarily large. The graph of y = F (x) will have a vertical asymptote at the point where the input of ln is equal to 0, which is at x = 6250. domain: x > 6250 horizontal asymptotes: none vertical asymptotes: x = 6250