LSP 120: Quantitative Reasoning and Technological Literacy

Topic 2: Exponential ModelsLecture notes 2.2

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Solving Exponential Equations

Remember that exponential equations are in the form:

y = P(1+r)x

• P is the initial (reference, old) value• r is the rate, a.k.a. percent change (and it can be

either positive or negative)• x is time (years, minutes, hours, seconds decades

etc…)• Y is the new value

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Important reminder

• You use y = P*(1+r)x formula if r is increasing yearly/monthly etc. and you are asked to calculate exponential change over a number of time periods.

– Example: the population is growing at 0.01% yearly. The population is 1 million now; what will it be in 10 years?

• You use y = P*(1+r) formula (without x) to calculate exponential change, if you are given an r value that does not grow annually/monthly etc., but represents overall percent change across time.

– Example: The population is was 1 million in 2000. It grew 15% in 10 years. What was the population in 2010?

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Solving for rate(percent change) and the initial value

• Solving for the rate(percent change):– We already know that we solve for percent change

between two values that are one time period apart:

old

oldnew

original

differencegeercentChan

P

year population (in millions) percent change (in decimal format) percent change (%)1995 1251996 127.625 (127.625-125)/125= .021 2.1%

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Solving for the rate(percent change)

What if we had to solve for the rate (percent change) but the new value is not one time period but several time periods ahead? How do we solve for the rate then?

Example: The population of country A was 125 million in 1995. In the year 2010 its population was 170.7246 million people. If we assume that the percent change (rate of population growth) was constant, at what rate did the population grew?

reminder our formula is y = P*(1+r)x

In this case:y= 170.7246 millionP= 125 million x= number of time periods from the initial value to the new value=2010-1995=15 yearsSolve for r!170.7246 = 125 *(1+r)15

170.7246/ 125 = (1+r)15

1.3657968 = (1+r)15

1.36579681/15 = (1+r)1.020999994 =1+r1.020999994 -1 = r0.020999994 = r0.021 = r 2.1%

The population grew 2.1 percent every year in country A from 1995 to 2010. Prepared by Ozlem Elgun 5

Solving for the initial value (a.k.a. old value, reference value)

What if we had to solve for the initial value?

Example: The population of a country was 125 million in 1995. If we knew that the percent change (rate of population growth) was constant, and grew at a rate of 2.1 percent every year, what was its population in year 1990?

reminder our formula is y = P*(1+r)x

In this case:y= 125 million x= number of time periods from the initial value to the new value=1995-1990=5 yearsr=2.1% (0.021 in decimals)Solve for P!

125 = P*(1+0.021)5

125 = P*(1.021)5

125= P*1.109503586125/1.109503586= P112.6629977=P

In year 1990, there were 112.66 million people in country A.Prepared by Ozlem Elgun 6

Solving for time (x) using ExcelExample: The population of Bengal tigers was 4700 in 2010. If we assume that the rate of population

growth will remain constant, and will be at a rate of 2.3 percent every year, how many years have to pass for Bengal tiger population to reach 10,000?

yearbengal tiger population

percent change

2010 47000.023

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Solving for time (using logarithms)

• To solve for time, you can get an approximation by using Excel. To solve an exponential equation algebraically for time, you must use logarithms.

• There are many properties associated with logarithms. We will focus on the following property:

log ax = x * log a for a>0

• This property is used to solve for the variable x (usually time), where x is the exponent. Prepared by Ozlem Elgun 9

Solving time (x) with logarithms:Example: The population of Bengal tigers was 4700 in 2010. If we assume that the rate of population growth will remain constant, and will be at a rate of 2.3 percent every year, how many years have to pass for Bengal tiger population to reach 10,000?

Start with : Y= P * (1 + r)X. Fill the variables that you know. To use logarithms, x (time) must be

your “unknown” quantity. y= 10000P= 4700R=0.023Solve for x!

The equation for this situation is: 10000 = 4700 * (1+0.023)X

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Solving time (x) with logarithms: Continued….We need to solve for x: Step 1: divide both sides by 4700 10000/4700 = (1+0.023)X

2.12766 = (1+0.023)X

USE THE LOG property you learned earlier log ax = x * log a for a>0

Step 2: take the log of both sides log(2.12766) = log (1+0.023)X

Step 3: bring the x down in frontlog(2.12766) = x * log (1+0.023) Step 4: divide both sides by log (1+.023) log(2.12766) /log(1+.023) = x to get 33.20316

Step 5: Write out your answer in words:

If we assume that the rate of population growth will remain constant, and will be at a rate of 2.3 percent every year 33.2 years have to pass for Bengal tiger population to reach 10,000.

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Application of Exponential Models:Carbon Dating

• A radioisotope is an atom with an unstable nucleus, which is a nucleus characterized by excess energy which is available to be imparted either to a newly-created radiation particle within the nucleus, or else to an atomic electron. The radioisotope, in this process, undergoes radioactive decay, and emits a gamma ray(s) and/or subatomic particles. These particles constitute ionizing radiation. Radioisotopes may occur naturally, but can also be artificially produced.

• Radiocarbon dating, or carbon dating, is a radiometric dating method that uses the naturally occurring radioisotope carbon-14 (14C) to determine the age of carbonaceous materials up to about 58,000 to 62,000 years

• One of the most frequent uses of radiocarbon dating is to estimate the age of organic remains from archaeological sites.

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• The Dead Sea Scrolls are a collection of 972 documents, including texts from the Hebrew Bible, discovered between 1946 and 1956 in eleven caves in and around the ruins of the ancient settlement of Khirbet Qumran on the northwest shore of the Dead Sea in the West Bank.

• We date the Dead Sea Scrolls which have about 78% of the normally occurring amount of Carbon 14 in them. Carbon 14 decays at a rate of about 1.202% per 100 years. I make a table of the form.

• Using excel and extending the table we find that the Dead Sea Scrolls would date from between 2100 to 2000 years ago.

Years (after death) % Carbon remaining percent change percent change (%)0 100

100 98.798 -0.01202 -1.202%200

.

.

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Years (after death) % Carbon remaining percent change percent change (%)0 100

100 98.798 -0.01202 -1.202%200 97.61044804 -0.01202 -1.202%300 96.43717045 -0.01202 -1.202%400 95.27799567 -0.01202 -1.202%500 94.13275416 -0.01202 -1.202%600 93.00127845 -0.01202 -1.202%700 91.88340309 -0.01202 -1.202%800 90.77896458 -0.01202 -1.202%900 89.68780143 -0.01202 -1.202%

1000 88.60975405 -0.01202 -1.202%1100 87.54466481 -0.01202 -1.202%1200 86.49237794 -0.01202 -1.202%1300 85.45273956 -0.01202 -1.202%1400 84.42559763 -0.01202 -1.202%1500 83.41080194 -0.01202 -1.202%1600 82.4082041 -0.01202 -1.202%1700 81.41765749 -0.01202 -1.202%1800 80.43901725 -0.01202 -1.202%1900 79.47214026 -0.01202 -1.202%2000 78.51688513 -0.01202 -1.202%2100 77.57311217 -0.01202 -1.202%2200 76.64068337 -0.01202 -1.202%

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Use logarithms to solve for time

• When were the dead sea scrolls created?78 = 100*(1-0.01202)X

.78=(0.98798) X

Log (.78)= x*log (0.98798)-0.107905397= x*(-0.005251847)

-0.107905397/-0.005251847=x20.546=x

Since x is in units of 100 yearsDead Sea Scrolls date back 2054.6 years

(Current estimates are that a 95% confidence interval for their date is 150 BC to 5 BC)

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Activity 3, Question 1: Beryllium-11 is a radioactive isotope of the alkaline metal Beryllium. Beryllium-11 decays at a rate of 4.9% every second.

a) Assuming you started with 100%, what percent of the beryllium-11 would be remaining after 10 seconds? Either copy and paste the table or show the equation used to answer the question.

y = 100*(1-0.049)10

=100*(0.951) 10

Y=60.51

60.61 % Beryllium-11 remains after 10 seconds.

seconds % Beryllium 11 remainig percent change percent change (%)0 1001 95.1 -0.049 -4.9%2 90.4401 -0.049 -4.9%3 86.0085351 -0.049 -4.9%4 81.79411688 -0.049 -4.9%5 77.78620515 -0.049 -4.9%6 73.9746811 -0.049 -4.9%7 70.34992173 -0.049 -4.9%8 66.90277556 -0.049 -4.9%9 63.62453956 -0.049 -4.9%

10 60.50693712 -0.049 -4.9%

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Activity 3, Question 1 Beryllium-11 is a radioactive isotope of the alkaline metal Beryllium. Beryllium-11 decays at a rate of 4.9% every second.

b) How long would it take for half of the beryllium-11 to decay? This time is called the half life. (Use the "solve using logs" process to answer the question) Show your work.

50 =100*(1-0.049) X

.50 =(0.951) X

Log (.50) = x*log (0. 951)-0.301029996= x*(-0.021819483)-0.301029996 /-0.021819483=x

13.796=x

It would take 13.796 seconds for Beryllium-11 to reach its half life.

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