LSP 120: Quantitative Reasoning and Technological Literacy Section 903

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LSP 120 Quantitative Reasoning and Technological
Literacy Section 903

• Özlem Elgün

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Review from previous session 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)
• Now Solving for time (using log function)

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Exponential function equation

• As with linear, there is a general equation for
  exponential functions. The equation for an
  exponential relationship is
• y  P(1r)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.

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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 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(1r)x
• In this case
• y 4700
• P 2000
• x number of time periods from the initial value
  to the new value2010-197337 years
• Solve for r!
• 4700 2000 (1r)37
• 4700/2000 (1r)37
• 2.35 (1r)37
• 2.351/37 (1r)
• 1.023361 1r
• 1.023361 -1 r
• 0.023361 r
• 0.023 r ? 2.3
• The population of Bengal tigers grew 2.3
  percent every year.

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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(1r)x
• In this case
• y 4700
• x number of time periods from the initial value
  to the new value2010-199020 years
• r2.3 (0.023 in decimals)
• Solve for P!
• 4700 P(10.023)20
• 4700 P(1.023)20
• 4700 P1.575842
• 4700/1.575842 P
• 2982.533 P
• In year 1990, there were 2982 Bengal tigers in
  the world

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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?

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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 agt0
• This property is used to solve for the variable x
  (usually time), where x is the exponent.

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• 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
• R0.023
• Solve for x!
• The equation for this situation is 10000 4700
  (10.023)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 (10.023)X
• 2.12766 (10.023)X
• USE THE LOG property you learned earlier ?
  log ax x log a for agt0
• Step 2 take the log of both sides
• log(2.12766) log (10.023)X
•  
• Step 3 bring the x down in front
• log(2.12766) x log (10.023)
•  
• 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

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Application of Exponential ModelsCarbon 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.

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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) xlog (0.98798)
• -0.107905397 x(-0.005251847)
• -0.107905397/-0.005251847x
• 20.546x
• 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)

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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
• Y60.51
• 60.61 Beryllium-11 remains after 10 seconds.
•  

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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) xlog (0. 951)
• -0.301029996 x(-0.021819483)
• -0.301029996 /-0.021819483x
• 13.796x
• It would take 13.796 seconds for Beryllium-11 to
  reach its half life.