LSP 120: Quantitative Reasoning and Technological Literacy Section 202 Özlem Elgün

Review from previous class: Solving Exponential Equations Solving for the rate (percent change) Solving for the initial value (a.k.a. old value, reference value) Solving for time (using Excel) TODAY: Solving for time (using log function)

Exponential function equation As with linear, there is a general equation for exponential functions. The equation for an exponential relationship is: y = P*(1+r)x P = initial value (value of y when x = 0), r is the percent change (written as a decimal), x is the input variable (usually units of time), Y is the output variable (i.e. population) The equation for the above example would be y = 192 * (1- 0.5)x Or y = 192 * 0.5x. We can use this equation to find values for y if given an x value. Write the exponential equation for other examples.

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 Bengal tigers was 2000 in 1973. In the year 2010 there were 4700 Bengal tigers in the world. If we assume that the percent change (rate of population growth) was constant, at what yearly rate did the population grew? reminder our formula is  y = P*(1+r)x In this case: y= 4700 P= 2000 x= number of time periods from the initial value to the new value=2010-1973=37 years Solve for r! 4700 = 2000 *(1+r)37 4700/2000 = (1+r)37 2.35 = (1+r)37 2.351/37 = (1+r) 1.023361 =1+r 1.023361 -1 = r 0.023361 = r 0.023 = r  2.3 % The population of Bengal tigers grew 2.3 percent every year.

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 Bengal tigers was 4700 in 2010. If we knew that the percent change (rate of population growth) was constant, and grew at a rate of 2.3 percent every year, what was its population in year 1990? reminder our formula is  y = P*(1+r)x In this case: y= 4700 x= number of time periods from the initial value to the new value=2010-1990=20 years r=2.3% (0.023 in decimals) Solve for P! 4700 = P*(1+0.023)20 4700 = P*(1.023)20 4700= P*1.575842 4700/1.575842 = P 2982.533 =P In year 1990, there were 2982 Bengal tigers in the world

Solving for time (x) using Excel 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?

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 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= 10000 P= 4700 R=0.023 Solve for x! The equation for this situation is: 10000 = 4700 * (1+0.023)X

Solving time (x) with logarithms: Continued… 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 front log(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.

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.

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.

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=x 20.546=x Since x is in units of 100 years Dead 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.

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.