Exponential and Logarithmic Functions 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
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
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.

For each problem:
specify the application
specify the discipline/subject
show all work.

(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 $
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(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


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(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)} $
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(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.
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(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 = Nt

$ 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 $
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