LSP 120: Quantitative Reasoning and Technological Literacy

Section 118

Özlem Elgün

• Solving for the rate (percent change)

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

• Solving for time (using logarithms)

Review from previous class:Solving Exponential Equations

Solving Exponential EquationsRemember that exponential equations are in the form:

y = P(1+r)x

P is the initial (reference, old) valuer 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

-or can be thought of as-y = ba x

b is the initial (reference, old) value a is the growth/decay factor and is equal to either 1+r or 1-r, (r is the rate)x is timeY is the new value

-and another way-N=Pa t

P is the initial value (reference, old) valuea is the growth/decay factor and is equal to either 1+r or 1-r, (r is the rate)t is timeN is the new value

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.

Solving with logarithms: A Petri dish contains 100 bacteria cells. The number of cells increases 5% every minute. How long will it take for the number of cells in the dish to

reach 3000? Start with : Y= P * (1 + r)X. Fill the variables that you know. To use logarithms, x (time) must be your “unknown” quantity. y= 3000P= 100R=0.05Solve for x!

The equation for this situation is: 3000 = 100 * (1+.05)X

We need to solve for x: Step 1: divide both sides by 100 30 = (1+.05)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 30 = log (1+.05)X

Step 3: bring the x down in front log 30 = x * log(1+.05) Step 4: divide by log (1+.05) Enter the following into a cell in excel: =log(30)/log(1+.05) to get 69.71 (Of course you may use a calculator.)

This tells us that at 69.71 minutes, there are 3000 cells. Which also means 69 minutes and .71 seconds of a minute (which is 60 seconds) 60*.71= 42.6 seconds.

This tells us that at 69 minutes and 42.6 seconds there will be 3000 cells in the Petri dish.

Application of Exponential Models:Radioisotope 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.

• 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%

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)

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%

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.