Applications of Exponential Functions and Logarithmic Functions: All Other Applications

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For JAMB and NZQA Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
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All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

(1.) ACT The number of decibels, $d$ produced by an audio source can be modeled by the equation $d = 10\log{\left(\dfrac{I}{K}\right)}$, where I is the sound intensity of the audio source and K is a constant.
How many decibels are produced by an audio source whose sound intensity is 1,000 times the value of K?

$ F.\;\; 4 \\[3ex] G.\;\; 30 \\[3ex] H.\;\; 40 \\[3ex] J.\;\; 100 \\[3ex] K.\;\; 10,000 \\[3ex] $

Application: Intensity of sound/sound waves
Discipline: Physics

$ d = 10\log{\left(\dfrac{I}{K}\right)} \\[5ex] I = 1000 * K = 1000K \\[3ex] d = 10\log{\left(\dfrac{1000K}{K}\right)} \\[5ex] = 10 * \log{1000} \\[3ex] = 10 * 3 \\[3ex] = 30 \:\:decibels $

(2.) Atmospheric pressure P in pounds per square inch is represented by the formula $P = 14.7e^{-0.21x}$, where x is the number of miles above sea level.
To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.781 pounds per square inch?
(Hint: there are 5,280 feet in a mile)

$ P = 14.7e^{-0.21x} \\[3ex] 14.7e^{-0.21x} = P \\[3ex] P = 8.781 \\[3ex] 14.7e^{-0.21x} = 8.781 \\[3ex] e^{-0.21x} = \dfrac{8.781}{14.7} \\[5ex] e^{-0.21x} = 0.5973469388 \\[3ex] Take\;\;the\;\;natural\;\;logarithm\;\;of\;\;both\;\;sides \\[3ex] \ln e^{-0.21x} = \ln 0.5973469388 \\[3ex] -0.21x = -0.5152571974 \\[3ex] x = \dfrac{-0.5152571974}{-0.21} \\[5ex] x = 2.453605702\;\;miles \\[3ex] $ Let $k$ be the height of the mountain in feet.

Proportional Reasoning
mile	feet
$1$	$5280$
$2.453605702$	$k$

$ \dfrac{5280}{1} = \dfrac{k}{2.453605702} \\[5ex] k * 1 = 5280(2.453605702) \\[3ex] k = 12955.03811 \\[3ex] k \approx 12955\;\;feet $

(3.) The relative humidity is the ratio (expressed as a percent) of the amount of water vapor in the air to the maximum amount that it can hold at a specific temperature.
The relative humidity, R, is found using the formula: $ R = 10^{\left(\dfrac{4221}{T + 459.4} - \dfrac{4221}{D + 459.4} + 2\right)} $ where T is the air temperature (in °F) and D is the dew point temperature (in °F)

(a.) Calculate the relative humidity if the air temperature is 50°F and the dew point temperature is 46°F
(b.) Calculate the relative humidity if the air temperature is 77°F and the dew point temperature is 54°F
(c.) Calculate the relative humidity is the air temperature is the same as the dew point temperature.

$ R = 10^{\left(\dfrac{4221}{T + 459.4} - \dfrac{4221}{D + 459.4} + 2\right)} \\[7ex] (a.) \\[3ex] T = 50^\circ F \\[3ex] D = 46^\circ F \\[3ex] R = 10^{\left(\dfrac{4221}{50 + 459.4} - \dfrac{4221}{46 + 459.4} + 2\right)} \\[7ex] R = 85.98417472\% \\[5ex] (b.) \\[3ex] T = 77^\circ F \\[3ex] D = 54^\circ F \\[3ex] R = 10^{\left(\dfrac{4221}{77 + 459.4} - \dfrac{4221}{54 + 459.4} + 2\right)} \\[7ex] R = 44.40869305\% \\[3ex] $ If you observe the exponent in the formula carefully, you will notice that the first term involving T and the second term involving D is the same.
The only difference in those terms are T and D
So, if T is equal to D, the subtraction of those terms will give 0.

$ (c.) \\[3ex] T = D = 70^\circ F...assume \\[3ex] R = 10^{\left(\dfrac{4221}{70 + 459.4} - \dfrac{4221}{70 + 459.4} + 2\right)} \\[7ex] R = 10^2 \\[3ex] R = 100\% $

(4.) ACT Hikers' World Foods sells raisin-nut mix in bulk to stores.
The dollar amount per pound, P(x), for a store to purchase x pounds of raisin-nut mix from Hikers' World is given by the function below.
$P(x) = 3.50 + 0.90^x$
To the nearest $\$0.01$, which of the following dollar values is equal to the total price for a store to purchase 100 pounds of raison-nut mix in a single order from Hikers' World?

$ A.\;\; \$350.00 \\[3ex] B.\;\; \$359.00 \\[3ex] C.\;\; \$440.00 \\[3ex] D.\;\; \$616.00 \\[3ex] E.\;\; \$903.50 \\[3ex] $

$ P(x) = 3.50 + 0.90^x \\[3ex] x = 100 \\[3ex] P(100) = 3.50 + 0.90^{100} \\[3ex] P(100) = 3.5 + 0.00002656139889 \\[3ex] P(100) = 3.500026561 \;\;per\;\;pound \\[3ex] For\;\;100\;\;pounds: \\[3ex] P(100) = 3.500026561(100) \\[3ex] P(100) = 350.0026561 \\[3ex] P(100) \approx \$350.00 $

(5.) The pH of a chemical solution is given by the formula: $pH = -\log_{10}[H^+]$, where $[H^+]$ is the concentration of hydrogen ions in moles per liter.
Values of pH range from 0 (acidic) to 14 (alkaline).

(a.) What is the pH of a solution for which $[H^+]$ is 0.01?
(b.) What is the pH of a solution for which $[H^+]$ is 0.001?
(c.) What is the pH of a solution for which $[H^+]$ is 0.0001?
(d.) What happens to pH as the hydrogen ion concentration decreases?
(e.) Determine the hydrogen ion concentration of a solution with pH = 3.6
(f.) Determine the hydrogen ion concentration of a solution with pH = 7.8

$ pH = -\log_{10}[H^+] \\[4ex] (a.) \\[3ex] [H^+] = 0.01 \\[3ex] pH = -\log (0.01) \\[3ex] = - \log 10^{-2} \\[4ex] = - (-2) \log 10 ...Law\;5...Log \\[3ex] = 2 * 1 ...Law\;4...Log \\[3ex] = 2 \\[5ex] (b.) \\[3ex] [H^+] = 0.001 \\[3ex] pH = -\log (0.001) \\[3ex] = - \log 10^{-3} \\[4ex] = - (-3) \log 10 ...Law\;5...Log \\[3ex] = 3 * 1 ...Law\;4...Log \\[3ex] = 3 \\[5ex] (c.) \\[3ex] [H^+] = 0.0001 \\[3ex] pH = -\log (0.0001) \\[3ex] = - \log 10^{-4} \\[4ex] = - (-4) \log 10 ...Law\;5...Log \\[3ex] = 4 * 1 ...Law\;4...Log \\[3ex] = 4 \\[3ex] $ (d.) Based on the three examples we did, we can see that as the hydrogen ion concentration decreases (from 0.01 to 0.0001), the pH value increases (2 to 4)

$ pH = -\log_{10}[H^+] \\[4ex] -\log_{10}[H^+] = pH \\[4ex] \log_{10}[H^+] = -pH \\[4ex] [H^+] = 10^{-pH}...Relationship\;\;Between\;\;Exp\;\;and\;\;Log \\[4ex] (e.) \\[3ex] pH = 3.6 \\[3ex] [H^+] = 10^{-3.6} \\[4ex] = 0.0002511886432 \\[3ex] = 2.511886432 * 10^{-4} \\[5ex] (f.) \\[3ex] pH = 7.8 \\[3ex] [H^+] = 10^{-3.6} \\[4ex] = 0.00000001584893192 \\[3ex] = 1.584893192 * 10^{-8} $

(6.) If a single pane of glass obliterates 1% of the light passing through it, the percent p of light that passes through n successive panes is given approximately by the function: $p(n) = 100(0.99)^n$

(a.) What percent of light will pass through 5 panes?
(b.) What percent of light will pass through 10 panes?
(c.) Explain the meaning of the base 0.99 in the function.

$ p(n) = 100(0.99)^n \\[4ex] (a.) \\[3ex] n = 5\;\;panes \\[3ex] p(5) = 100(0.99)^5 \\[4ex] p(5) = 95.09900499\% \\[5ex] (b.) \\[3ex] n = 10\;\;panes \\[3ex] p(10) = 100(0.99)^{10} \\[4ex] p(10) = 90.4382075\% \\[3ex] $ (c.) The base: 0.99 (99%) implies that each pane allows only 99% to pass through, while obliterating 0.01 (1%).

(7.) ACT The total amount of a certain substance present in a laboratory experiment is given by the formula $A = A_o\left(2^{\dfrac{h}{5}}\right)$, where A is the total amount of the substance h hours after an initial amount ($A_o$) of the substance began accumulating.
Which of the following expressions gives the number of hours it will take an initial amount of 10 grams of this substance to accumulate to 100 grams?

$ F.\;\; 5 \\[3ex] G.\;\; 25 \\[3ex] H.\;\; \log_2{(50)} \\[3ex] J.\;\; 5\log_2{(10)} \\[3ex] K.\;\; 5\log_{20}{(100)} \\[3ex] $

$ A = A_o\left(2^{\dfrac{h}{5}}\right) \\[5ex] A = 100\;grams \\[3ex] A_o = 10\;grams \\[3ex] \implies \\[3ex] 100 = 10\left(2^{\dfrac{h}{5}}\right) \\[5ex] 10\left(2^{\dfrac{h}{5}}\right) = 100 \\[5ex] 2^{\dfrac{h}{5}} = \dfrac{100}{10} \\[5ex] 2^{\dfrac{h}{5}} = 10 \\[5ex] Introduce\;\;\log\;\;to\;\;both\;\;sides \\[3ex] \log 2^{\dfrac{h}{5}} = \log 10 \\[5ex] \dfrac{h}{5} \log 2 = \log 10 \\[5ex] \dfrac{h}{5} = \dfrac{\log 10}{\log 2} \\[5ex] h = 5 \dfrac{\log 10}{\log 2} \\[5ex] h = 5\log_{2}{10}...Law\;6...Log \\[3ex] h = 5\log_2{(10)} $

(8.) The atmospheric pressure p on a balloon or an aircraft decreases with increasing height.
This pressure, measured in millimeters of mercury, is related to the height h (in kilometers) above sea level by the formula: $p = 760e^{-0.145h}$

(a.) Determine the height of an aircraft if the atmospheric pressure is 319 millimeters of mercury.
(b.) Determine the height of a mountain if the atmospheric pressure is 658 millimeters of mercury.

$ p = 760e^{-0.145h} \\[4ex] 760e^{-0.145h} = p \\[4ex] e^{-0.145h} = \dfrac{p}{760} \\[5ex] Introduce\;\;\ln\;\;to\;\;both\;\;sides \\[3ex] \ln \left(e^{-0.145h}\right) = \ln \left(\dfrac{p}{760}\right) \\[5ex] -0.145h = \ln \left(\dfrac{p}{760}\right)...Law\;4...Log \\[5ex] h = \dfrac{\ln \left(\dfrac{p}{760}\right)}{-0.145}...in\;\;km \\[7ex] (a.) \\[3ex] p = 319\;mm \\[3ex] h = \dfrac{\ln \left(\dfrac{319}{760}\right)}{-0.145}...in\;\;km \\[7ex] h = 5.987085038\;km \\[5ex] (b.) \\[3ex] p = 658\;mm \\[3ex] h = \dfrac{\ln \left(\dfrac{658}{760}\right)}{-0.145}...in\;\;km \\[7ex] h = 0.9938862204\;km $

(9.) Between 12:00 PM and 1:00 PM, cars arrive at a local bank's drive-thru at the rate of 3 cars per hour (0.05 car per minute).
The probability that a car will arrive within t minutes of 12:00 PM is modeled by the function: $F(t) = 1 - e^{-0.05t}$

(a.) Determine the probability that a car will arrive within 35 minutes of 12:00 PM (that is, before 12:35 PM)
(b.) Determine the probability that a car will arrive within 45 minutes of 12:00 Pm (that is, before 12:45 PM)
(c.) What value does F approach as t becomes unbounded in the positive direction?
(d.) Graph F using a graphing utility.
(e.) Using the TRACE function on the graphing utility or an acceptable algebraic method, how many minutes are needed for the probability to reach 50%?

$ F(t) = 1 - e^{-0.05t} \\[4ex] (a.) \\[3ex] t = 35\;minutes \\[3ex] F(35) = 1 - e^{-0.05 * 35} \\[4ex] F(35) = 0.8262260565 \\[5ex] (b.) \\[3ex] t = 45\;minutes \\[3ex] F(45) = 1 - e^{-0.05 * 45} \\[4ex] F(45) = 0.8946007754 \\[5ex] (c.) \\[3ex] As\;\;t \rightarrow \infty, \;\; e^{-0.05t} \rightarrow 0 \\[3ex] \implies \\[3ex] 1 - e^{-0.05t} \rightarrow 1 - 0 = 1 \\[3ex] \therefore, \;\;As t \rightarrow \infty,\;\; F(t) = 1 \\[3ex] $ (d.) The graph of F(t) versus t is:

Number 9

$ (e.) \\[3ex] F(t) = 1 - e^{-0.05t} \\[4ex] F(t) = 50\% = 0.5 \\[3ex] \implies \\[3ex] 0.5 = 1 - e^{-0.05t} \\[4ex] e^{-0.05t} = 1 - 0.5 \\[4ex] e^{-0.05t} = 0.5 \\[3ex] Introduce\;\;\ln\;\;to\;\;both\;\;sides \\[3ex] \ln e^{-0.05t} = \ln 0.5 \\[4ex] -0.05t = -0.6931471806 \\[3ex] t = \dfrac{-0.6931471806}{-0.05} \\[5ex] t = 13.86294361\;minutes $

(10.) Between 12:00 PM and 1:00 PM, cars arrive at a local bank's drive-thru at the rate of 18 cars per hour (0.3 car per minute).
The following formula from statistics can be used to determine the probability that a car will arrive within t minutes of 12:00 PM: $F(t) = 1 - e^{-0.3t}$

(a.) How many minutes are needed for the probability to reach 40%?
(b.) How many minutes are needed for the probability to reach 90%?
(c.) Is it possible for the probability to equal 100%? Explain.

$ F(t) = 1 - e^{-0.3t} \\[4ex] e^{-0.3t} = 1 - F \\[4ex] Introduce\;\;\ln\;\;to\;\;both\;\;sides \\[3ex] \ln e^{-0.3t} = \ln (1 - F) \\[4ex] -0.3t = \ln (1 - F) \\[3ex] t = \dfrac{\ln (1 - F)}{-0.3} \\[5ex] (a.) \\[3ex] F = 40\% = 0.4 \\[3ex] t = \dfrac{\ln (1 - 0.4)}{-0.3} \\[5ex] t = 1.702752079\;minutes \\[5ex] (b.) \\[3ex] F = 90\% = 0.9 \\[3ex] t = \dfrac{\ln (1 - 0.9)}{-0.3} \\[5ex] t = 7.675283643\;minutes \\[3ex]' (c.) \\[3ex] Assume\;\; F = 100\% = 1 \\[3ex] t = \dfrac{\ln (1 - 1)}{-0.3} \\[5ex] \ln 0 \;\;is\;\;NA \\[3ex] $ The domain of a logarithmic function is a positive number.
Hence, it is not possible for the probability to be equal to 100%.

(11.) ACT A 500-square-mile national park in Kenya has large and small protected animals.
The number of large protected animals at the beginning of 2014 is given in the table below.

Large animal	Number
Elephant Rhinoceros Lion Leopard Zebra Giraffe	600 100 200 300 400 800
Total	2,400

At the beginning of 2014, the number of all protected animals in the park was 10,000.
Zoologists predict that for each year from 2015 to 2019, the total number of protected animals in the park at the beginning of the year will be 2% more than the number of protected animals in the park at the beginning of the previous year.

Let t be a positive integer less than 6.
Based on the zoologists' prediction, which of the following expressions represents the number of protected animals in the park t years after the beginning of 2014?

$ F.\;\; 10,000 + 0.02t \\[3ex] G.\;\; 10,000 + 0.2t \\[3ex] H.\;\; 10,000(1 + 0.02^t) \\[3ex] J.\;\; 10,000(1 + 0.02)^t \\[3ex] K.\;\; 10,000(1 + 0.2)^t \\[3ex] $

Reviewing the way the question was worded: total number of protected animals in the park at the beginning of the year will be 2% more than the number of protected animals in the park at the beginning of the previous year. (present = 2% more than previous); the function is not linear.
So, Options F. and G. are eliminated.

Student: I do not understand.
How do you know the function is not linear?
Teacher: If it was worded this way: "present is 2 more than the previous" ...
rather than 2% more than the previous, then we would consider the linear function
Saying 2% more than the previous year means that first, we have to find 2% of the previous year, then add it the previous year in order to get the present year
Hence, it is exponential because each present year involves multiplication of the previous year and addition of that product to the previous year.
If we simplify it for several years, exponents begin to build up for later years... leading to an exponential function.
Student: An example would be very helpful.
Teacher: Sure, let's do it.

$ 2\% = \dfrac{2}{100} = 0.02 \\[5ex] \underline{Beginning\;\;of\;\;2014} \\[3ex] Number = 10,000 \\[3ex] \underline{Beginning\;\;of\;\;2015} \\[3ex] 1\;\;year\;\;after\;\;2014 \\[3ex] Number = 10000 + 2\%\;\;of\;\;10000 \\[3ex] = 10000 + 0.02(10000) \\[3ex] = 10000(1 + 0.02)^1 \\[3ex] = 10000(1 + 0.02) \\[3ex] \underline{Beginning\;\;of\;\;2016} \\[3ex] 2\;\;years\;\;after\;\;2014 \\[3ex] Number = 10000(1 + 0.02) + 2\%\;\;of\;\;10000(1 + 0.02) \\[3ex] = 10000(1 + 0.02) + 0.02[10000(1 + 0.02)] \\[3ex] = 10000(1 + 0.02)[1 + 0.02] \\[3ex] = 10000(1 + 0.02)(1 + 0.02) \\[3ex] = 10000(1 + 0.02)^2 \\[3ex] Similarly: \\[3ex] t\;\;years\;\;after\;\;2014 \\[3ex] Number = 10000(1 + 0.02)^t \\[3ex] $ Student: You did not determine the exact numbers for each year?
Teacher: Yes, because I wanted to find a result that would resemble one of the options.
Student: But...
May I find the exact answers for those two years (2015 and 2016) and then compare my answers with each option?
Teacher: Yes..., that is one of the ways of solving the question.
I would suggest, however that you mark the question and come back to it later after you have completed all other questions because testing each option for 2 years will likely take more than a minute.

(12.) The concentration of alcohol in a person's blood is measurable.
Suppose that the relative risk, R (given as a percent) of having an accident while driving a car can be modeled by the equation $R = 3e^{kx}$ where x is the variable concentration of alcohol in the blood and k is a constant.

(a.) Suppose that a concentration of alcohol in the blood of 0.04 results in a 10% relative risk (R = 10) of an accident.
Determine the constant, k in the equation.

(b.) Using the value of k in part (a.), what is the relative risk if the concentration is 0.10?

(c.) Using the same value of k found in part (a.), what concentration of alcohol corresponds to a relative risk of 100%?

(d.) Using the value of k found in part (a.), if the law asserts that anyone with a relative risk of having an accident of 15% or more should not have driving privileges, at what concentration of alcohol in the blood should a driver be arrested and charged with a DUI?

$ R = 3e^{kx} \\[4ex] 3e^{kx} = R \\[4ex] e^{kx} = \dfrac{R}{3} \\[5ex] Introduce\;\;\ln\;\;\;to\;\;both\;\;sides \\[3ex] \ln (e^{kx}) = \ln \left(\dfrac{R}{3}\right) \\[5ex] kx = \ln \left(\dfrac{R}{3}\right) \\[5ex] To\;\;find\;\;k: \\[3ex] k = \dfrac{\ln \left(\dfrac{R}{3}\right)}{x} \\[7ex] To\;\;find\;\;x: \\[3ex] x = \dfrac{\ln \left(\dfrac{R}{3}\right)}{k} \\[7ex] (a.) \\[3ex] x = 0.04 \\[3ex] R = 10 \\[3ex] k = \dfrac{\ln \left(\dfrac{10}{3}\right)}{0.04} \\[7ex] k = 30.09932011 \\[5ex] (b.) \\[3ex] k = 30.09932011 \\[3ex] x = 0.1 \\[3ex] R = 3e^{kx} \\[4ex] R = 3e^{30.09932011 * 0.1} \\[4ex] R = 60.85806196...in\;\% \\[3ex] R = 60.85806196\% \\[5ex] (c.) \\[3ex] k = 30.09932011 \\[3ex] R = 100 \\[3ex] x = \dfrac{\ln \left(\dfrac{100}{3}\right)}{30.09932011} \\[7ex] x = 0.1164995716 \\[5ex] (d.) \\[3ex] k = 30.09932011 \\[3ex] R = 15 \\[3ex] x = \dfrac{\ln \left(\dfrac{15}{3}\right)}{30.09932011} \\[7ex] x = 0.0534709059 $

(13.) The equation governing the amount the current I (in amperes) after time, t (in seconds) in a single RL circuit consisting of a resistance R (in ohms), an inductance L (in henrys), and an electromotive force, E (in volts) is:

$ I = \dfrac{E}{R}\left[1 - e^{-\left(\dfrac{R}{L}\right)t}\right] \\[5ex] $ (a.) If E = 19 volts, R = 18 ohms, and L = 1 henrys, how long does it take to obtain a current of 0.5 ampere?
(b.) How long does it take to obtain a current of 1.0 ampere?
(c.) Using a graphing utility, display the graph of I versus t

$ I = \dfrac{E}{R}\left[1 - e^{-\left(\dfrac{R}{L}\right)t}\right] \\[7ex] I * \dfrac{R}{E} = 1 - e^{-\left(\dfrac{R}{L}\right)t} \\[5ex] e^{-\left(\dfrac{R}{L}\right)t} = 1 - \dfrac{IR}{E} \\[5ex] e^{-\left(\dfrac{R}{L}\right)t} = \dfrac{E}{E} - \dfrac{IR}{E} \\[5ex] e^{-\left(\dfrac{R}{L}\right)t} = \dfrac{E - IR}{E} \\[5ex] Introduce\;\;\ln\;\;to\;\;both\;\;sides \\[3ex] \ln \left[e^{-\left(\dfrac{R}{L}\right)t}\right] = \ln \left(\dfrac{E - IR}{E}\right) \\[7ex] -\left(\dfrac{R}{L}\right)t = \ln \left(\dfrac{E - IR}{E}\right) \\[5ex] t = \dfrac{\ln \left(\dfrac{E - IR}{E}\right)}{-\left(\dfrac{R}{L}\right)} \\[7ex] (a.) \\[3ex] E = 19\;V \\[3ex] R = 18\;\Omega \\[3ex] L = 1\;H \\[3ex] I = 0.5\;A \\[3ex] t = \dfrac{\ln \left(\dfrac{19 - 0.5(18)}{19}\right)}{-\left(\dfrac{18}{1}\right)} \\[7ex] t = 0.0356585492\;seconds \\[5ex] (b.) \\[3ex] E = 19\;V \\[3ex] R = 18\;\Omega \\[3ex] L = 1\;H \\[3ex] I = 1\;A \\[3ex] t = \dfrac{\ln \left(\dfrac{19 - 1(18)}{19}\right)}{-\left(\dfrac{18}{1}\right)} \\[7ex] t = 0.1635799433\;seconds \\[3ex] $ (c.) The graph of I versus t is:

Number 13

(14.) ACT The isotope iodine-131 has a half-life of 8 days, which means that the amount of iodine-131 remaining after t days is $N\left(\dfrac{1}{2}\right)^{\dfrac{t}{8}}$, where N is the number of grams of iodine-131 at t = 0.
How many grams of iodine-131 will remain after 16 days if there were 32 grams of iodine-131 at t = 0?

$ A.\;\; 0 \\[3ex] B.\;\; 2 \\[3ex] C.\;\; 8 \\[3ex] D.\;\; 16 \\[3ex] E.\;\; 128 \\[3ex] $

Let the amount of iodine-131 remaining after t days = N_t

$ t = 16\;days \\[3ex] N = 32\;g \\[3ex] N_t = N\left(\dfrac{1}{2}\right)^{\dfrac{t}{8}} \\[5ex] N_t = 32 * \left(\dfrac{1}{2}\right)^{\dfrac{16}{8}} \\[5ex] = 32 * \left(\dfrac{1}{2}\right)^2 \\[5ex] = 32 * \dfrac{1}{2} * \dfrac{1}{2} \\[5ex] = 8\;g $

(16.) The loudness L(x), measured in decibels, of a sound of intensity x, measured in Watts per square meter, is defined as $L(x) = 10\log\left(\dfrac{x}{I_0}\right)$, where $I_0 = 10^{-12}$ Watt per square meter is the least intense sound that a human ear can detect.
Determine the loudness, in decibels, of a sound with intensity of $x = 10^{-2.3}$ Watt per square meter.

$ L(x) = 10\log\left(\dfrac{x}{I_0}\right) \\[5ex] x = 10^{-2.3}\;W/m^2 \\[4ex] I_0 = 10^{-12}\;W/m^2 \\[4ex] L(x) = 10\log\left(\dfrac{10^{-2.3}}{10^{-12}}\right) \\[7ex] L(x) = 97\;decibels $

(18.) The price p, in dollars, of a specific car that is x years old is modeled by the function: $p(x) = 22295(0.89)^x$

(a.) Calculate the cost of a 2-year old car.
(b.) Calculate the cost of a 7-year old car.
(c.) What is the meaning of the base, 0.89 in the function?

$ p(x) = 22295(0.89)^x \\[3ex] (a.) \\[3ex] x = 2\;years \\[3ex] p(2) = 22295(0.89)^2 \\[4ex] p(2) = 17659.8695 \\[3ex] p(2) \approx \$17659.87 \\[5ex] (b.) \\[3ex] x = 7\;years \\[3ex] p(2) = 22295(0.89)^7 \\[4ex] p(2) = 9861.376115 \\[3ex] p(2) \approx \$9861.38 \\[3ex] $ (c.) The base, 0.89 in the exponential function implies that after each year, the car is worth 89% of its value of the previous year.

(20.) When using the Richter scale, all earthquakes are compared to a zero-level earthquake whose seismographic reading measures 0.001 millimeter at a distance of 100 kilometers from the epicenter.
An earthquake whose seismographic reading measures x millimeters has magnitude M(x), given by $M(x) = \log \left(\dfrac{x}{x_0}\right)$, where $x_0 = 10^{-3}$ is the reading of a zero-level earthquake the same distance from its epicenter.
Determine the magnitude of an earthquake with a seismographic reading of 20,979 millimeters 100 kilometers from the center.

$ M(x) = \log \left(\dfrac{x}{x_0}\right) \\[5ex] x = 20979\;mm \\[3ex] x_0 = 0.001\;mm \\[3ex] M(x) = \log \left(\dfrac{20979}{0.001}\right) \\[5ex] M(x) = 7.321784783 $

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(24.) The percentage of patients, P who survived t years after initial diagnosis of a certain disease is modeled by the function: $P(t) = 100(0.6)^t$.

(a.) According to the model, what percent of patients survive 1 year after initial diagnosis?
(b.) What percent of the patients survive 3 years after the initial diagnosis?
(c.) Explain the meaning of the base, 0.6 in the function.

$ P(t) = 100(0.6)^t \\[3ex] (a.) \\[3ex] t = 1\;year \\[3ex] P(1) = 100(0.6)^1 \\[3ex] P(t) = 60\% \\[5ex] (b.) \\[3ex] t = 3\;years \\[3ex] P(1) = 100(0.6)^3 \\[3ex] P(t) = 21.6\% \\[3ex] $ (c.) The base, 0.6 in the function implies that 60% (0.6) of the previous year's survivors survive after each year.

(25.) The equation governing the amount the current I (in amperes) after time, t (in seconds) in a single RL circuit consisting of a resistance R (in ohms), an inductance L (in henrys), and an electromotive force, E (in volts) is:

$ I = \dfrac{E}{R}\left[1 - e^{-\left(\dfrac{R}{L}\right)t}\right] \\[5ex] $ Number 25

1st:
(a.) If E = 180 volts, R = 20 ohms, and L = 5 henrys, calculate the amount of current flowing after 0.3 seconds.
(b.) How much current is flowing after 0.5 second?
(c.) How much current is flowing after 1 second?
(d.) What is the maximum current, I₁(t)?
(e.) Graph the function: I = I₁(t), measuring I along the y-axis and t along the x-axis.
2nd:
(a.) If E = 180 volts, R = 10 ohms, and L = 10 henrys, calculate the amount of current flowing after 0.3 seconds.
(b.) How much current is flowing after 0.5 second?
(c.) How much current is flowing after 1 second?
(d.) What is the maximum current, I₂(t)?
(e.) Graph the function: I = I₂(t), measuring I along the y-axis and t along the x-axis.

$ \boldsymbol{1st:} \\[3ex] (a.)\;\; \\[3ex] E = 180\;V \\[3ex] R = 20\;\Omega \\[3ex] L = 5\;H \\[3ex] t = 0.3\;s \\[3ex] I(t) = \dfrac{E}{R}\left[1 - e^{-\left(\dfrac{R}{L}\right)t}\right] \\[7ex] I(0.3) = \dfrac{180}{20}\left[1 - e^{-\left(\dfrac{20}{5}\right) * 0.3}\right] \\[7ex] I(0.3) = 6.289252093\;A \\[5ex] (b.)\;\; t = 0.5\;s \\[3ex] I(0.5) = \dfrac{180}{20}\left[1 - e^{-\left(\dfrac{20}{5}\right) * 0.5}\right] \\[7ex] I(0.5) = 7.781982451\;A \\[5ex] (c.)\;\; t = 1\;s \\[3ex] I(1) = \dfrac{180}{20}\left[1 - e^{-\left(\dfrac{20}{5}\right) * 1}\right] \\[7ex] I(1) = 8.83515925\;A \\[5ex] (d.)\;\; \\[3ex] As\;\;t \rightarrow \infty, \;\;e^{-\left(\dfrac{R}{L}\right)t} \rightarrow 0 \\[4ex] \implies \\[3ex] 1 - e^{-\left(\dfrac{R}{L}\right)t} \\[3ex] = 1 - 0 \\[3ex] = 1 \\[3ex] \implies \\[3ex] I(t) = \dfrac{E}{R}\left[1 - e^{-\left(\dfrac{R}{L}\right)t}\right] \\[7ex] = \dfrac{E}{R} * 1 \\[5ex] = \dfrac{E}{R} \\[5ex] \therefore I(t)_{max} = \dfrac{E}{R} \\[5ex] I(t)_{max} = \dfrac{180}{20} \\[5ex] I(t)_{max} = 9\;A \\[3ex] $ (e.) The graph of the function: I₁(t) versus t is:
Number 25-first

$ \boldsymbol{2nd:} \\[3ex] (a.)\;\; \\[3ex] E = 180\;V \\[3ex] R = 10\;\Omega \\[3ex] L = 10\;H \\[3ex] t = 0.3\;s \\[3ex] I(t) = \dfrac{E}{R}\left[1 - e^{-\left(\dfrac{R}{L}\right)t}\right] \\[7ex] I(0.3) = \dfrac{180}{10}\left[1 - e^{-\left(\dfrac{10}{10}\right) * 0.3}\right] \\[7ex] I(0.3) = 4.665272028\;A \\[5ex] (b.)\;\; t = 0.5\;s \\[3ex] I(0.5) = \dfrac{180}{10}\left[1 - e^{-\left(\dfrac{10}{10}\right) * 0.5}\right] \\[7ex] I(0.5) = 7.082448125\;A \\[5ex] (c.)\;\; t = 1\;s \\[3ex] I(1) = \dfrac{180}{10}\left[1 - e^{-\left(\dfrac{10}{10}\right) * 1}\right] \\[7ex] I(1) = 11.37817006\;A \\[5ex] (d.)\;\; \\[3ex] As\;\;t \rightarrow \infty, \;\;e^{-\left(\dfrac{R}{L}\right)t} \rightarrow 0 \\[4ex] \implies \\[3ex] 1 - e^{-\left(\dfrac{R}{L}\right)t} \\[3ex] = 1 - 0 \\[3ex] = 1 \\[3ex] \implies \\[3ex] I(t) = \dfrac{E}{R}\left[1 - e^{-\left(\dfrac{R}{L}\right)t}\right] \\[7ex] = \dfrac{E}{R} * 1 \\[5ex] = \dfrac{E}{R} \\[5ex] \therefore I(t)_{max} = \dfrac{E}{R} \\[5ex] I(t)_{max} = \dfrac{180}{10} \\[5ex] I(t)_{max} = 18\;A \\[3ex] $ (e.) The graph of the function: I₁(t) versus t is:
Number 25-second

(32.) Between 5:00 PM and 6:00 PM, cars arrive at an eatery drive-thru at the rate of 20 cars per hour.
The probability that x cars will arrive between 5:00 PM and 6:00 PM is given by the function: $P(x) = \dfrac{20^x e^{-20}}{x!}$ where $x! = x \cdot (x - 1) \cdot (x - 2)\cdot ... \cdot 3 \cdot 2 \cdot 1$

(a.) Calculate the probability that 11 cars will arrive between 5:00 PM and 6:00 PM.
(b.) Calculate the probability that 23 cars will arrive between 5:00 PM and 6:00 PM.

$ P(x) = \dfrac{20^x e^{-20}}{x!} \\[5ex] (a.) \\[3ex] x = 11\;cars \\[3ex] P(11) = \dfrac{20^{11} e^{-20}}{11!} \\[5ex] P(11) = 0.0105751028 \\[5ex] (b.) \\[3ex] x = 23\;cars \\[3ex] P(23) = \dfrac{20^{23} e^{-20}}{23!} \\[5ex] P(23) = 0.0668814737 $

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(44.) Assume the function: $D(h) = 7e^{-0.37h}$ is used to find the number of milligrams D of a certain drug that is in a patient's bloodstream h hours after the drug has been administered.
How many milligrams will be present after:
(a.) 1 hour.
(b.) 7 hours.

$ D(h) = 7e^{-0.37h} \\[4ex] (a.) \\[3ex] h = 1\;hour \\[3ex] D(1) = 7e^{-0.37 * 1} \\[4ex] D(1) = 4.835140314\;mg \\[5ex] (b.) \\[3ex] h = 7\;hours \\[3ex] D(7) = 7e^{-0.37 * 7} \\[4ex] D(7) = 0.5251402806\;mg $

(46.) Assume the function: $D(h) = 20e^{-0.5h}$ is used to find the number of milligrams D of a certain drug that is in a patient's bloodstream h hours after the drug has been administered.
When the number of milligrams reaches 3, the drug is to be administered again.
What is the time between injections?

$ D(h) = 20e^{-0.5h} \\[4ex] 20e^{-0.5h} = D \\[4ex] e^{-0.5h} = \dfrac{D}{20} \\[5ex] Introduce\;\;\ln\;\;to\;\;both\;\;sides \\[3ex] \ln \left(e^{-0.5h}\right) = \ln \left(\dfrac{D}{20}\right) \\[5ex] -0.5h = \ln \left(\dfrac{D}{20}\right) \\[5ex] h = \dfrac{\ln \left(\dfrac{D}{20}\right)}{-0.5} \\[7ex] D = 3\;mg \\[3ex] h = \dfrac{\ln \left(\dfrac{3}{20}\right)}{-0.5} \\[7ex] h = 3.79423997\;hours $