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LSP 120: Quantitative Reasoning and Technological Literacy Section 118

LSP 120: Quantitative Reasoning and Technological Literacy Section 118. Özlem Elgün. EXPONENTIAL FUNCTIONS.

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LSP 120: Quantitative Reasoning and Technological Literacy Section 118

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  1. LSP 120: Quantitative Reasoning and Technological Literacy Section 118 ÖzlemElgün

  2. EXPONENTIAL FUNCTIONS We can use Excel to determine if a relationship is exponential by filling the neighboring column with the percent change from one Y to the next. As with linear, do not put this formula in the cell next to the first pair of numbers but in the cell next to the second.

  3. If doing this calculation on a calculator, you will need to multiply by 100 to convert to a percent. In Excel, just click on the % icon on the toolbar. • If the column is constant, then the relationship is exponential. So this function is exponential.

  4. 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 • where P is the starting value (value of y when x = 0), r is the percent change (written as a decimal), and x is the input variable (usually time). • The equation for the above example would be y = 192 * (1-.5)x or y = 192 * .5x. • We can use this equation to find values for y if given an x value. Try the other examples.

  5. Why are exponential relationships important? Where do we encounter them? • Populations tend to growth exponentially not linearly • When an object cools (e.g., a pot of soup on the dinner table), the temperature decreases exponentially toward the ambient temperature (the surrounding temperature) • Radioactive substances decay exponentially • Bacteria populations grow exponentially • Money in a savings account with at a fixed rate of interest increases exponentially • Viruses and even rumors tend to spread exponentially through a population (at first) • Anything that doubles, triples, halves over a certain amount of time • Anything that increases or decreases by a percent

  6. If a quantity changes by a fixed percentage, it grows or decays exponentially.

  7. How to increase/decrease a number by a percent • There are 2 “ways” to do this: N = P + P * r or N= P * (1 + r) • According to the distributive property, these two formulas are the same. For the work we will be doing later in the quarter, the second version is preferred. • Similarly to the formula above, N is the ending value, P is the starting value and r is the percent (written in decimal form). To write a percent in decimal form, move the decimal 2 places to the left. Remember that if there is a percent decrease, you will be subtracting instead of adding.

  8. Exponential growth or decay is increasing or decreasing by same percent over and over • If a quantity P is growing by r % each year, after one year there will be P*(1 + r) • So, P has been multiplied by the quantity 1 + r.   • If  P*(1 + r) is in turn increased by r percent, it will be multiplied by (1 + r) again.  • So after two years, P has become: P*(1+r)*(1+r) = P*(1 + r)2 • So after 3 years, you have P*(1+r)3, and so on. Each year, the exponent increases by one since you are multiplying what you had previously by another (1+r). • If a quantity P is decreasing by r%, then by the same logic, the formula is P*(1 r).

  9. Example: A bacteria population is at 100 and is growing by 5% per minute. 2 questions:1. How many bacteria cells are present after one hour?2. How many minutes will it take for there to be 1000 cells. • If the population is growing at 5% a minutes, this means it is being multiplied by 1.05 each minute. Using Excel, we can set up a table to calculate the population at each minute. Note that the time column begins with 0. • The formula in the second cell of the column is the cell above it multiplied by 1 + the percent written as a decimal. (Note that you do not enter the exponent in the formula) Filling the column will give us the population at each minute. Reminder formula for calculating the new population after each period: y=P*(1 + r) where P= starting (reference, old) value r= rate (percent change in decimals) y= the new value

  10. If you wanted to calculate the population after one hour (60 minutes), you could drag the columns of the excel table down to 60 or you could use the following formula which I refer to as the “by hand” formula. Note that here because you are doing the calculation in one step you do use the exponent. Y = P *(1+r)x • Here x is the number of minutes. • Therefore, the population after 60 minutes = y = 100*(1+.05)60 = 1868. (To enter an exponent in Excel use the ^ key which is above the number 6.) Remember to round your answer appropriately. For this example, we should round to the nearest whole number. • If you wanted to know how long it would take for the population to reach 1000, you could set up the table in Excel and drag the columns down until the population (the Y value) reaches 1000. Doing so, you should find that the population will reach 1000 after 48 minutes. We will discuss a more mathematical and exact way to solve this problem in the near future.

  11. Another example: Country A had population of 125 million in 1995. Its population was growing 2.1% a year. Country B had a population of 200 million in 1995. Its population was decreasing 1.2% a year. What are the populations of the countries this year? Has the population of country A surpassed the population of B? If not, in what year will the population of country A surpassed the population of B? • Create your own Excel table. The formula in cell B3 is =B2*(1+0.021) and in C3 is =B2*(1-.012). Then each column is filled. Since the A’s population was not larger than B’s in 2006, the columns needed to be extended down farther.

  12. 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 -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 time Y is the new value -and another way- N=Pa t P is the initial value (reference, old) value a is the growth/decay factor and is equal to either 1+r or 1-r, (r is the rate) t is time N is the new value

  13. 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:

  14. 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 million P= 125 million x= number of time periods from the initial value to the new value=2010-1995=15 years Solve 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+r 1.020999994 -1 = r 0.020999994 = r 0.021 = r  2.1% The population grew 2.1 percent every year in country A from 1995 to 2010.

  15. 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 years r=2.1% (0.021 in decimals) Solve for P! 125 = P*(1+0.021)5 125 = P*(1.021)15 125= P*1.109503586 125/1.109503586= P 112.6629977=P In year 1990, there were 112.66 million people in country A.

  16. 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: logax = x * loga for a>0 • This property is used to solve for the variable x (usually time), where x is the exponent.

  17. Solving for time (x): example 1 1) 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? Solving without logarithms: One way to solve this problem is to use excel. If you continue the chart, you will find that after 69 minutes, there were 2898 cells and at 70 minutes, there were 3043 cells. Obviously somewhere between minute 69 and 70, the number of cells reached 3000. But we do not know exactly when. Using logarithms we can get an exact answer.

  18. 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= 3000 P= 100 R=0.05 Solve 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  logax = x * loga 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.

  19. Solving for time (x): example 2 If you $100 deposited into a savings account grows at 3.4% compounded annually, how long will it take for your balance to double? Solving without logarithms: One way to solve this problem is to use excel. If you continue the chart, you will find that after 20 years, there will be $195.17 in your account and after 21 years, there will be $201.80 in your account. Obviously somewhere between years 20 and 21 the amount in your account will double from $100 to $200. However, using logarithms we can get an exact answer.

  20. Solving with logarithms: If you $100 deposited into a savings account grows at 3.4% compounded annually, how long will it take for your balance to double? Start with : Y= P * (1 + r)X. Fill the variables that you know. To use logarithms, x (time) must be your “unknown” quantity. y= 200 P= 100 R=0.034 Solve for x! The equation for this situation is: 200 = 100 * (1+.034)X We need to solve for x: Step 1: divide both sides by 100 2 = (1+.034)X USE THE LOG property you learned earlier  logax = x * loga for a>0 Step 2: take the log of both sides log 2= log (1+.034)X Step 3: bring the x down in front log 2= x * log(1+.034) Step 4: divide by log (1+.034) Enter the following into a cell in excel: =log(32)/log(1.034) to get 20.731 (Of course you may use a calculator.) This tells us that at there will be $200 in your account in 20.731 years or it means that will take 20.731 years for the money in your account to double. Which also means 20 years and .731 days of a year (in which there are 365 days) 365*.731=  266.815 days This tells us that at there will be $200 in your account in 20 years and 267 days.

  21. Problem 3 - Practice • Let’s revisit the bacteria situation above. Assuming we allowed the bacteria population to reach 3000 then put in an antibiotic that killed the cells at a rate of 22.5% a minute, how long would it take for the population to decline to 60 cells? • Answer the problem using both Excel and Logarithms. The exact answer (using logarithms) is 15.34775 minutes.

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