# 5.6: Application Problems with Exponential and Logarithmic Functions - Mathematics

Learning Objectives

In this section, you will:

1. review strategies for solving equations arising from exponential formulas
2. solve application problems involving exponential functions and logarithmic functions

## STRATEGIES FOR SOLVING EQUATIONS THAT CONTAIN EXPONENTS

When solving application problems that involve exponential and logarithmic functions, we need to pay close attention to the position of the variable in the equation to determine the proper way solve the equation we investigate solving equations that contain exponents.

Suppose we have an equation in the form : value = coefficient(base) exponent

We consider four strategies for solving the equation:

STRATEGY A: If the coefficient, base, and exponent are all known, we only need to evaluate the expression for coefficient(base) exponent to evaluate its value.

STRATEGY B: If the variable is the coefficient, evaluate the expression for (base) exponent. Then it becomes a linear equation which we solve by dividing to isolate the variable.

STRATEGY C: If the variable is in the exponent, use logarithms to solve the equation.

STRATEGY D: If the variable is not in the exponent, but is in the base, use roots to solve the equation.

Below we examine each strategy with one or two examples of its use.

STRATEGY A: If the coefficient, base, and exponent are all known, we only need to evaluate the expression for coefficient(base)exponent to evaluate its value.

Example (PageIndex{1})

Suppose that a stock’s price is rising at the rate of 7% per year, and that it continues to increase at this rate. If the value of one share of this stock is $43 now, find the value of one share of this stock three years from now. Solution Let (y) = the value of the stock after (t) years: (y = ab^t) The problem tells us that (a) = 43 and (r) = 0.07, so (b = 1+ r = 1+ 0.07 = 1.07) Therefore, function is (y = 43(1.07)^t). In this case we know that (t) = 3 years, and we need to evaluate (y) when (t) = 3. At the end of 3 years, the value of this one share of this stock will be [y=43(1.07)^{3}=$ 52.68 onumber]

STRATEGY B: If the variable is the coefficient, evaluate the expression for (base) exponent. Then it becomes a linear equation which we solve by dividing to isolate the variable.

Example (PageIndex{2})

The value of a new car depreciates (decreases) after it is purchased. Suppose that the value of the car depreciates according to an exponential decay model. Suppose that the value of the car is $12000 at the end of 5 years and that its value has been decreasing at the rate of 9% per year. Find the value of the car when it was new. Solution Let (y) be the value of the car after (t) years: (y = ab^t), (r) = -0.09 and (b = 1+r = 1+(-0.09) = 0.91) The function is (y = a(0.91)^t) In this case we know that when (t) = 5, then (y) = 12000; substituting these values gives [12000 = a(0.91)^5 onumber] We need to solve for the initial value a, the purchase price of the car when new. First evaluate (0.91)5 ; then solve the resulting linear equation to find (a). [ 1200 = a(0.624) onumber ] (a=frac{12000}{0.624} =$ 19,230.77); The car's value was 19,230.77 when it was new. STRATEGY C: If the variable is in the exponent, use logarithms to solve the equation. Example (PageIndex{3}) A national park has a population of 5000 deer in the year 2016. Conservationists are concerned because the deer population is decreasing at the rate of 7% per year. If the population continues to decrease at this rate, how long will it take until the population is only 3000 deer? Solution Let (y) be the number of deer in the national park (t) years after the year 2016: (y = ab^t) (r) = -0.07 and (b = 1+r = 1+(-0.07) = 0.93) and the initial population is (a) = 5000 The exponential decay function is (y = 5000(0.93)^t) To find when the population will be 3000, substitute (y) = 3000 [ 3000 = 5000(0.93)^t onumber] Next, divide both sides by 5000 to isolate the exponential expression [egin{array}{l} frac{3000}{5000}=frac{5000}{5000}(0.93)^{2} 0.6=0.93^{t} end{array} onumber] Rewrite the equation in logarithmic form; then use the change of base formula to evaluate. [t=log _{0.93}(0.6) onumber] (t = frac{ln(0.6)}{ln(0.93)}=7.039) years; After 7.039 years, there are 3000 deer. Note: In Example (PageIndex{3}), we needed to state the answer to several decimal places of precision to remain accurate. Evaluating the original function using a rounded value of (t) = 7 years gives a value that is close to 3000, but not exactly 3000. [y=5000(0.93)^{7}=3008.5 ext { deer } onumber ] However using (t) = 7.039 years produces a value of 3000 for the population of deer [ y=5000(0.93)^{7.039}=3000.0016 approx 3000 ext { deer } onumber] Example (PageIndex{4}) A video posted on YouTube initially had 80 views as soon as it was posted. The total number of views to date has been increasing exponentially according to the exponential growth function (y = 80e^{0.2t}), where (t) represents time measured in days since the video was posted. How many days does it take until 2500 people have viewed this video? Solution Let (y) be the total number of views (t) days after the video is initially posted. We are given that the exponential growth function is (y = 80e^{0.2t}) and we want to find the value of (t) for which (y) = 2500. Substitute (y) = 2500 into the equation and use natural log to solve for (t). [2500 = 80e^{0.12t} onumber] Divide both sides by the coefficient, 80, to isolate the exponential expression. [egin{array}{c} frac{2500}{80}=frac{80}{80} e^{0.12 t} 31.25=e^{0.12 t} end{array} onumber] Rewrite the equation in logarithmic form [ 0.12t = ln(31.25) onumber ] Divide both sides by 0.04 to isolate (t); then use your calculator and its natural log function to evaluate the expression and solve for (t). [egin{array}{l} mathrm{t}=frac{ln (31.25)}{0.12} mathrm{t}=frac{3.442}{0.12} mathrm{t} approx 28.7 ext { days } end{array} onumber] This video will have 2500 total views approximately 28.7 days after it was posted. STRATEGY D: If the variable is not in the exponent, but is in the base, we use roots to solve the equation. It is important to remember that we only use logarithms when the variable is in the exponent. Example (PageIndex{5}) A statistician creates a website to analyze sports statistics. His business plan states that his goal is to accumulate 50,000 followers by the end of 2 years (24 months from now). He hopes that if he achieves this goal his site will be purchased by a sports news outlet. The initial user base of people signed up as a result of pre-launch advertising is 400 people. Find the monthly growth rate needed if the user base is to accumulate to 50,000 users at the end of 24 months. Solution Let (y) be the total user base (t) months after the site is launched. The growth function for this site is (y = 400(1+r)^t); We don’t know the growth rate (r). We do know that when (t) = 24 months, then (y) = 50000. Substitute the values of (y) and (t); then we need to solve for (r). [5000 = 400(1+r)^{24} onumber] Divide both sides by 400 to isolate (1+r)24 on one side of the equation [egin{array}{l} frac{50000}{400}=frac{400}{400}(1+r)^{24} 125=(1+r)^{24} end{array} onumber] Because the variable in this equation is in the base, we use roots: [egin{array}{l} sqrt{125}=1+r 125^{1 / 24}=1+r 1.2228 approx 1+r 0.2228 approx r end{array} onumber] The website’s user base needs to increase at the rate of 22.28% per month in order to accumulate 50,000 users by the end of 24 months. Example (PageIndex{6}) A fact sheet on caffeine dependence from Johns Hopkins Medical Center states that the half life of caffeine in the body is between 4 and 6 hours. Assuming that the typical half life of caffeine in the body is 5 hours for the average person and that a typical cup of coffee has 120 mg of caffeine. 1. Write the decay function. 2. Find the hourly rate at which caffeine leaves the body. 3. How long does it take until only 20 mg of caffiene is still in the body? www.hopkinsmedicine.org/psyc...fact_sheet.pdf Solution a. Let (y) be the total amount of caffeine in the body (t) hours after drinking the coffee. Exponential decay function (y = ab^t) models this situation. The initial amount of caffeine is (a) = 120. We don’t know (b) or (r), but we know that the half- life of caffeine in the body is 5 hours. This tells us that when (t) = 5, then there is half the initial amount of caffeine remaining in the body. [egin{array}{l} y=120 b^{t} frac{1}{2}(120)=120 b^{5} 60=120 b^{5} end{array} onumber] Divide both sides by 120 to isolate the expression (b^5) that contains the variable. [egin{array}{l} frac{60}{120}=frac{120}{120} mathrm{b}^{5} 0.5=mathrm{b}^{5} end{array} onumber] The variable is in the base and the exponent is a number. Use roots to solve for (b): [egin{array}{l} sqrt{0.5}=mathrm{b} 0.5^{1 / 5}=mathrm{b} 0.87=mathrm{b} end{array} onumber] We can now write the decay function for the amount of caffeine (in mg.) remaining in the body (t) hours after drinking a cup of coffee with 120 mg of caffeine [y=f(t)=120(0.87)^{t} onumber ] b. Use (b = 1 + r) to find the decay rate (r). Because (b = 0.87 < 1) and the amount of caffeine in the body is decreasing over time, the value of (r) will be negative. [egin{array}{l} 0.87=1+r r=-0.13 end{array} onumber] The decay rate is 13%; the amount of caffeine in the body decreases by 13% per hour. c. To find the time at which only 20 mg of caffeine remains in the body, substitute (y) = 20 and solve for the corresponding value of (t). [egin{array}{l} y=120(.87)^{t} 20=120(.87)^{t} end{array} onumber] Divide both sides by 120 to isolate the exponential expression. [egin{array}{l} frac{20}{120}=frac{120}{120}left(0.87^{t} ight) 0.1667=0.87^{t} end{array} onumber] Rewrite the expression in logarithmic form and use the change of base formula [egin{array}{l} t=log _{0.87}(0.1667) t=frac{ln (0.1667)}{ln (0.87)} approx 12.9 ext { hours } end{array} onumber] After 12.9 hours, 20 mg of caffeine remains in the body. ## EXPRESSING EXPONENTIAL FUNCTIONS IN THE FORMS y = abt and y = aekt Now that we’ve developed our equation solving skills, we revisit the question of expressing exponential functions equivalently in the forms (y = ab^t) and (y = ae^{kt}) We’ve already determined that if given the form (y = ae^{kt}), it is straightforward to find (b). Example (PageIndex{7}) For the following examples, assume (t) is measured in years. 1. Express (y = 3500e^{0.25t}) in form (y = ab^t) and find the annual percentage growth rate. 2. Express (y = 28000e^{-0.32t}) in form (y = ab^t) and find the annual percentage decay rate. Solution a. Express (y = 3500e^{0.25t}) in the form (y = ab^t) [egin{array}{l} y=a e^{k t}=a b^{t} aleft(e^{k} ight)^{t}=a b^{t} end{array} onumber] Thus (e^k=b) In this example (b=e^{0.25} approx 1.284) We rewrite the growth function as y = 3500(1.284t) To find (r), recall that (b = 1+r) [egin{aligned} &1.284=1+r &0.284=mathrm{r} end{aligned} onumber] The continuous growth rate is (k) = 0.25 and the annual percentage growth rate is 28.4% per year. b. Express (y = 28000e^{-0.32t}) in the form (y = ab^t) [egin{array}{l} y=a e^{k t}=a b^{t} aleft(e^{k} ight)^{t}=a b^{t} end{array} onumber] Thus (e^k=b) In this example (mathrm{b}=e^{-0.32} approx 0.7261) We rewrite the growth function as y = 28000(0.7261t) To find (r), recall that (b = 1+r) [egin{array}{l} 0.7261=1+r 0.2739=r end{array} onumber] The continuous decay rate is (k) = -0.32 and the annual percentage decay rate is 27.39% per year. In the sentence, we omit the negative sign when stating the annual percentage decay rate because we have used the word “decay” to indicate that r is negative. Example (PageIndex{8}) 1. Express (y = 4200 (1.078)^t) in the form (y =ae^{kt}) 2. Express (y = 150 (0.73)^t) in the form (y =ae^{kt}) Solution a. Express (y = 4200 (1.078)^t) in the form (y =ae^{kt}) [egin{array}{l} mathrm{y}=mathrm{a} e^{mathrm{k} t}=mathrm{ab}^{mathrm{t}} mathrm{a}left(e^{mathrm{k}} ight)^{mathrm{t}}=mathrm{ab}^{mathrm{t}} e^{mathrm{k}}=mathrm{b} e^{k}=1.078 end{array} onumber] Therefore (mathrm{k}=ln 1.078 approx 0.0751) We rewrite the growth function as (y = 3500e^{0.0751t}) b. Express (y =150 (0.73)^t) in the form (y = ae^{kt}) [egin{array}{l} y=a e^{k t}=a b^{t} aleft(e^{k} ight)^{t}=a b^{t} e^{k}=b e^{k}=0.73 end{array} onumber] Therefore (mathrm{k}=ln 0.73 approx-0.3147) We rewrite the growth function as (y = 150e^{-0.3147t}) ## AN APPLICATION OF A LOGARITHMIC FUNCTON Suppose we invest10,000 today and want to know how long it will take to accumulate to a specified amount, such as $15,000. The time (t) needed to reach a future value (y) is a logarithmic function of the future value: (t = g(y)) Example (PageIndex{9}) Suppose that Vinh invests$10000 in an investment earning 5% per year. He wants to know how long it would take his investment to accumulate to $12000, and how long it would take to accumulate to$15000.

Solution

We start by writing the exponential growth function that models the value of this investment as a function of the time since the $10000 is initially invested [y=10000(1.05)^{t} onumber] We divide both sides by 10000 to isolate the exponential expression on one side. [frac{y}{10000}=1.05^{t} onumber] Next we rewrite this in logarithmic form to express time as a function of the accumulated future value. We’ll use function notation and call this function (g(y)). [mathrm{t}=mathrm{g}(mathrm{y})=log _{1.05}left(frac{mathrm{y}}{10000} ight) onumber ] Use the change of base formula to express (t) as a function of (y) using natural logarithm: [mathrm{t}=mathrm{g}(mathrm{y})=frac{ln left(frac{mathrm{y}}{10000} ight)}{ln (1.05)} onumber] We can now use this function to answer Vinh’s questions. To find the number of years until the value of this investment is$12,000, we substitute (y) = $12,000 into function (g) and evaluate (t): [mathrm{t}=mathrm{g}(12000)=frac{ln left(frac{12000}{10000} ight)}{ln (1.05)}=frac{ln (1.2)}{ln (1.05)}=3.74 ext { years } onumber] To find the number of years until the value of this investment is$15,000, we substitute (y) = $15,000 into function (g) and evaluate (t): [mathrm{t}=mathrm{g}(15000)=frac{ln left(frac{15000}{10000} ight)}{ln (1.05)}=frac{ln (1.5)}{ln (1.05)}=8.31 ext { years } onumber] Before ending this section, we investigate the graph of the function (mathrm{t}=mathrm{g}(mathrm{y})=frac{ln left(frac{mathrm{y}}{10000} ight)}{ln (1.05)}). We see that the function has the general shape of logarithmic functions that we examined in section 5.5. From the points plotted on the graph, we see that function (g) is an increasing function but it increases very slowly. If we consider just the function (mathrm{t}=mathrm{g}(mathrm{y})=frac{ln left(frac{mathrm{y}}{10000} ight)}{ln (1.05)}), then the domain of function would be (y > 0), all positive real numbers, and the range for (t) would be all real numbers. In the context of this investment problem, the initial investment at time (t) = 0 is (y) =$10,000. Negative values for time do not make sense. Values of the investment that are lower than the initial amount of $10,000 also do not make sense for an investment that is increasing in value. Therefore the function and graph as it pertains to this problem concerning investments has domain (y ≥ 10,000) and range (t ≥ 0). The graph below is restricted to the domain and range that make practical sense for the investment in this problem. ## Lesson Bacteria growth problems Regarding standard bacteria growth problems, there three major models in use. One model is a doubling period model the other is an exponential ekt-model with the base "e" of natural logarithms, and the third model is exponential model with an arbitrary base. In this lesson, I present all these models and show how to use them. ### Problem 1 A bacteria culture initially contains 1500 bacteria and doubles every half an hour. Find the size of the bacterial population after 2 hours. ### Problem 2 A bacteria culture initially contains 1500 bacteria and doubles every half an hour. Find the size of the bacterial population after 100 minutes. ### Problem 3 The doubling period of a bacterial population is 30 minutes. At time t= 120 minutes, the bacterial population was 60000. (a) What was the initial population at time t=0 ? (b) Find the size of the bacterial population after 5 hours. ### Problem 4 Bacteria in a dish double the area they cover everyday. If the dish is fully covered after 25 days, on what day was 1/2 of it covered? ### Problem 5 A bacteria culture starts with 800 bacteria. Two hours later there are 1280 bacteria. (a) Find an exponential model for the size of the culture as a function of time t in hours. (b) Use the model to predict how many bacteria there will be after 2 days. ### Problem 6 Assume that the number of bacteria follows an exponential growth model P(t) = . The count in the bacteria culture was 900 after 20 minutes and 1600 after 30 minutes. (a) What was the initial size of the culture? (b) Find the population after 80 minutes. (c) How many minutes after the start of the experiment will the population reach 10000? ### Problem 7 The count in a bacteria culture was 600 after 15 minutes and 2000 after 30 minutes. Assuming the count grows exponentially, 1) What was the initial size of the culture? 2) Find the doubling period. 3) Find the population after 120 minutes. 4) When will the population reach 14000 ? Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I. ## APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS Problem 3: If you invested$1,000 in an account paying an annual percentage rate (quoted rate) compound daily (based on a bank year of 360 days) and you wanted to have $2,500 in your account at the end of your investment time, what interest rate would you need if the investment time were 1 year, 10 years, 20 years, 100 years? Answer: 1 year = 91.75%, 10 years = 9.16%, 20 years = 4.58%, and 100 years 0.92% Solution and Explanations: Use the formula where$2,500 is the balance at the end of a certain time period, $1,000 is the beginning investment, t is the number of years, and r is the annual percentage rate. The annual rate of r% is converted to a daily interest rate since the compounding is daily (360 times per year). Take the annual interest rate of 4% and divide by 360 to obtain the daily interest rate. The exponent is 360t because there are 360 compounding periods in every year. Therefore, 360t represents the number of compounding periods during t years. To find the rate for 1 year: Step 1: Substitute 1 for t in the equation Step 2: Divide both sides of the above equation by$1,000:

Step 3: Take the natural logarithm of both sides of the above equation:

Step 4: Simplify the right side of the above equation using the third rule of logarithms:

Step 5: Divide both sides of the above equation by 360:

Step 6: Simplify the left side of the above equation:

Step 7: Convert the above equation to an exponential equation with base e and exponent 0.00254525203298:

Step 8: Simplify the left side of the above equation:

Step 9: Subtract 1 from both sides of the above equation:

Step 10: Multiply both sides of the above equation by 360:

This means the interest rate would have to be 91.75% (rounded) for the year.

This is a close enough check. Remember it will not check perfectly before we rounded the interest rate.

To find the rate for 10 year:

Step 1: Substitute 10 for t in the equation

Step 2: Divide both sides of the above equation by $1,000: Step 3: Take the natural logarithm of both sides of the above equation: Step 4: Simplify the right side of the above equation using the third rule of logarithms: Step 5: Divide both sides of the above equation by 3600: Step 6: Simplify the left side of the above equation: Step 7: Convert the above equation to an exponential equation with base e and exponent 0.000254525203298: Step 8: Simplify the left side of the above equation: Step 9: Subtract 1 from both sides of the above equation: Step 10: Multiply both sides of the above equation by 360: This means the interest rate would have to be 9.16% (rounded) per year for 10 years. This is a close enough check. Remember it will not check perfectly before we rounded the interest rate. To find the rate for 20 year: Step 1: Substitute 20 for t in the equation Step 2: Divide both sides of the above equation by$1,000:

Step 3: Take the natural logarithm of both sides of the above equation:

Step 4: Simplify the right side of the above equation using the third rule of logarithms:

Step 5: Divide both sides of the above equation by 7200:

Step 6: Simplify the left side of the above equation:

Step 7: Convert the above equation to an exponential equation with base e and exponent 0.000127262601649:

Step 8: Simplify the left side of the above equation:

Step 9: Subtract 1 from both sides of the above equation:

Step 10: Multiply both sides of the above equation by 360:

This means the interest rate would have to be 4.58% (rounded) per year for 20 years.

This is a close enough check. Remember it will not check perfectly before we rounded the interest rate.

To find the rate for 100 year:

Step 1: Substitute 100 for t in the equation

Step 2: Divide both sides of the above equation by $1,000: Step 3: Take the natural logarithm of both sides of the above equation: Step 4: Simplify the right side of the above equation using the third rule of logarithms: Step 5: Divide both sides of the above equation by 36000: Step 6: Simplify the left side of the above equation: Step 7: Convert the above equation to an exponential equation with base e and exponent 0.0000254525203298: Step 8: Simplify the left side of the above equation: Step 9: Subtract 1 from both sides of the above equation: Step 10: Multiply both sides of the above equation by 360: This means the interest rate would have to be 0.92% (rounded) per year for 100 years. This is a close enough check. Remember it will not check perfectly before we rounded the interest rate. If you would like to work another problem and check the answer and solution, click on Problem. ## 5.6: Application Problems with Exponential and Logarithmic Functions - Mathematics In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm. One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. The table below lists the half-life for several of the more common radioactive substances. Substance Use Half-life gallium-67 nuclear medicine 80 hours cobalt-60 manufacturing 5.3 years technetium-99m nuclear medicine 6 hours americium-241 construction 432 years carbon-14 archeological dating 5,715 years uranium-235 atomic power 703,800,000 years We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay: ### Example 13: Using the Formula for Radioactive Decay to Find the Quantity of a Substance How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay? ## Pros and cons of logarithm applications • The rate of change can be represented graphically, using the logarithmic functions as an operator. • The data value can be elevated for small inputs. • Logarithmic scale can be used as a statistical auxiliary (Wenger & Wolfger, 2016). • Resolving exponential functions into linear functions, logarithms can improve the understanding for people. The key advantage of the application of logarithm can be considered as transformation of exponential function into the linear function. Since exponential functions are hard to understand, for the general people, thus, using the logarithmic application, simplification can be conducted. In a similar vein, plotting large value in graph paper is not possible using ordinary mathematical approaches. In this context, logarithm considers 10 as base value of the function and thereby large values can be plotted in the graph papers quite easily. Disadvantages • Error plotting may be easy using logarithm, however, proper evaluation of error may be denied. • The process of analysis is difficult in nature (Bernstein et al. 2016). • Plotting null value is not possible using logarithm scale. • Using the same graph, both positive, as well as negative value, cannot be represented. In spite of vast field of application, logarithm applications are solely dependent on the base value of the function. Hence, according to the change of base value, output of the logarithm differs. Likewise, using the logarithm function, zero cannot be represented. Hence, null value is denied by the function and viability of information reduces thereby. Moreover, reciprocal nature of logarithm, in context to an exponential function, deprives the nature of calculation. Thus, exponential graphs cannot be prepared using logarithm graph papers. ## 5.6: Application Problems with Exponential and Logarithmic Functions - Mathematics The purpose of this lab is to familiarize you with some applications from real life involving exponential and logarithmic functions. The diversity of the processes which are described by the natural exponential function appears amazing. In this lab, you will encounter to three very different applications and try to use Maple to simplify computations and visualize behaviors of the processes in time. The lab consists of Background including both the relevant theoretical notes and description of the use of appropriate Maple commands. There are four problems, each of which has separate preliminary remarks (discussion of equations, introduction of terms, etc.) and an exercise to do. Theoretical notes. In many natural processes the rate of change of physical quantity (temperature, velocity amount of money, electric current, whatever) is proportional to the current amount of the quantity. If we also know the amount present at time t =0, call it y 0 , we can find y as a function of t by solving the following initial value problem: Differential equation: is a constant. (1) Initial conditions: y = y 0 when t = 0 . If y is positive and increasing, then k is positive, and we use Eq. (1) to say that the rate of growth is proportional to what has already been accumulated. If y is positive and decreasing, then k is negative, and Eq. (1) is used to say that the rate of decay is proportional to the amount still left. It is seen that the constant function y = 0 is a solution of Eq. (1). To find the non-zero solutions, the equation is solved in accordance with the known technique of separating variables and integrating. The solution involves the natural exponential function (whose derivatives is the function itself) and is expressed as The constant k is called the growth rate in exponential growth ( k >0) and the decay rate in exponential decay ( k < 0). In a process that can be modeled by exponential functions, the rate constant k depends only on the process and the conditions under which it is carried out. Some processes are described by differential equations similar to (1) but containing two or more constants characterizing some other circumstances in which these processes are carried out. Also, the initial conditions might be specified by more complicated expressions. The corresponding initial value problems lead to the solution having slightly different form, e.g., like (2) including a combination of additive and multiplicative constants. You will meet illustrations of that in the problems below. In some applications, a physical quantity varies on a huge range. To make this quantity more convenient to handle, special scales involving logarithms are used. This allows one to deal with the corresponding powers instead of actual values of y . Relevant Maple Means. In order to enter the exponential and natural logarithmic functions, use the exp and ln command. The syntax of these commands is similar to that of sin and cos . For example, below it is shown how to enter the function f ( x ) = e x and then evaluate it at x = 0.8. We suggest the commands involving expressions instead of functions here because they might be useful when doing exercises below. In the same manner, the function is entered and calculated. Try the following commands: To simplify expression involving logarithms, use command simplify it works as follows: The common logarithm is defined by function log10 = log , but log10 must be defined with the command readlib(log10) before use: Maple manipulates with common logarithms the same way but may return expressions including natural logarithms. The solve command is usually sufficient for solving most problems encountered in your Calculus courses. This command comes in a couple of varieties, as shown below. i. Solve linear equation: ii. Solve displaying result as an equation: Plotting the two functions f ( x ) = e x and on the same coordinate system illustrates an idea of the symmetry around the line y = x . All the three graphs can be plotted by the use of the following command: Responding this command, Maple returns the figure as shown in Figure 1. In order to get a graph of a left-hand/right-hand side of an equation obtained after symbolic transformations and/or computations, first use commands lhs and rhs respectively. The example: When analyzing graphs obtained as the result of your computation, you may need to plot supplementary or auxiliary straight lines in the same figure. The following format of the plot command illustrates how the line y = 60 from x = 0 to x = 20 can be displayed: Radioactive decay is a typical example to which the exponential decay model can be applied. In Eq. (2), y represents the mass (in grams) of an isotope, y 0 and k are constants determining from initial conditions: y 0 is the mass present originally, and k is the decay constant. k is often specified in terms of an empirical parameter, the half-life of the isotope. The half-life of a sample of a radioactive isotope is the time required for half of the atoms of that sample to decay. The half-lives of some common radioactive isotopes are as follows:  Uranium (U-238) 4,510,000,000 years Plutonium (Pu-239) 24,360 years Carbon (C-14) 5,730 years Einsteinium (Es-254) 270 days Nobelium (No-257) 23 sec The relationship between k and is set up from the condition saying that the sample of y 0 grams will contain only grams after the time , so that, referring to equation (2): and therefore: The worst nuclear accident in history happened on April 26, 1986 at 01:23 a.m. (GMT +03:00) at the Chernobyl nuclear plant 60 miles north of Kiev in the Ukraine. An explosion destroyed one of the plant's four reactors, releasing large amount of radioactive isotopes into the atmosphere. For almost two weeks the radiation continued to escape and clouds settled onto villages, towns, and forests across wide stretches of Ukraine, Belarus, and neighboring parts of Russia. About 300,000 people were evacuated from the fallout areas. The estimate of the accident's ultimate death toll varies from several thousand to half a million deaths. (For more information, see `The Chernobyl Nuclear Accident'* site at Geocities.com) Consider 20 grams of the plutonium isotope Pu-239 released in the Chernobyl nuclear accident. 1. How long will it take for the 20 grams to decay to 5 grams? 2. Plot the graph showing the decay of the mass of the plutonium isotope took place up to date (1986-1998). Does it look like a typical curve of exponential decay? Why? Discuss the graph in the context of the radioactive safety. Base 10 logarithms, often called common logarithms, appear in many scientific and applied formulas. For example, earthquake intensity is often reported on the logarithmic Richter scale. Here the formula is where a is the amplitude of the ground motion in microns at the receiving station, T is the period of the seismic wave in seconds, and B is an empirical factor that allows for the weakening of the seismic wave with increasing distance from the epicenter of the earthquake. For an earthquake 10,000 km from the receiving station, B = 6.8. Thus if the recorded vertical ground motion is a = 10 microns and the period is T = 1 sec, the earthquake's magnitude, following (7), is R = 7.8. An earthquake of this magnitude does great damage near its epicenter. Another example of the use of common logarithms is the decibel scale using, particularly, for measuring loudness. (The decibel unit is named in honor of Alexander G. Bell (1847-1922), inventor of the telephone.) If I is the intensity of sound in watts per square meter, the decibel level of the sound is where I 0 is an intensity of 10 -12 watts per square meter corresponding roughly to the faintest sound that can be heard. When tuning the rock band's equipment before the concert in a big concert hall, an audio engineer finds that in order to maintain appropriate loudness in this hall, he needs to increase the power of the amplifiers compared with the level used for the previous concert in a smaller hall. Use formula (8) (a) to find what tripling the power adds to the level of loudness in decibels (b) to determine by what factor k the engineer has to multiply the intensity of I of the sound to add 15 dB to the sound level S for the next concert of the band on the stadium. An aluminum beam brought from the outside cold into a machine shop where the regular normal temperature is maintained warms up to the temperature of the surrounding air. A hot silver ingot immersed in water cools to the temperature of the surrounding water. In situations like these, the rate at which an object's temperature is changing at any given time is approximately proportional to the difference between its temperature and the temperature of the surrounding medium. This observation is sometimes called the Newton's Law of Cooling, although, as in the case with the aluminum beam, it applies to warming as well. An equation representing this law is derived from a corresponding differential equation and can be written as where T is the temperature of the object at time is the surrounding temperature, T 0 is the value of T when t = 0. 3. From the graphs, make an estimate when the soup should be eaten if it is supposed to be best at C? Thomas Robert Malthus (1766-1834), English economist and sociologist, was a pioneer in modern world population study. In An Essay on the Principle of Population (1798 rev. ed. 1803) he suggested that populations tend to increase exponentially, whereas resources which support populations (e.g., food) tend to grow only linearly. So poverty, distress, wars, famine are unavoidable because population increases faster than the means of subsistence. The differential equation (1) may be regarded as a mathematical model of the changing population according to Malthus. Thus population P ( t ) with constant birth rate and constant death rate is given by where P 0 = P (0) and is called the Malthusian parameter. If t is measured in years, it represents the annual growth rate, which, depending on the relationship between and , can be positive, negative, or zero. However, equation (3) describes only the simplest ideal situation, in which the inner impulse of the population to expand is given a completely free rein it does not take into account any of the inhibiting factors that put a ceiling on the possible size of a real population. It is obvious, for example, that the human population of the earth can never expand to the stage where there will be only a small fraction of an acre of usable land per person. Long before the point is reached at which the whole surface of the earth becomes teeming slum, the rate of population growth will be forced down social, psycological, and economic effects will depress the birth rate and increase the death rate . In about 1840 the Belgian mathematician Pierre-Francoise Verhulst proposed a population growth model which takes into account the possible limitation of population size due to limited resources. It is called logistic model. When the population P ( t ) is small, it tends to grow at a rate proportional to itself but when it becomes larger, P ( t ) grows at a smaller rate. If L is the largest population beyond which the population cannot grow, the rate of growth of the population is assumed to be proportional to P and to ( L - P ), so that The solution of equation (4) subject to the condition that P = P 0 when t = 0 is obtained by separation of variables and eventually yields Since and depend on time, parameter k in (5) is interpreted as not their difference but rather as the maximum possible rate of population growth which is the net effect of birth and death rates. In accordance with some environmental and demographic studies, the earth will not support a population of more than 16 billion. It is also known that there were 2 billion people in 1925 and 4 billion people in 1975. 1. Find the exact population in 2015 using both models. Plot graphs of both models. From the graphs, estimate when, following the Malthus exponential growth model, the population will reach 16 billion, and then estimate the size of the population in this year in accordance with the Verhulst logistic model. 2. Compute when in accordance with the logistic model, the birth rate will be equal to the death rate, in other words, find the inflection point of the logistic graph. Next: About this document . Up: Labs and Projects for Previous: Labs and Projects for ## Major Assignment Here are some papers of mine that I would like to share with you. These papers should give you a good idea of what research looks like, and how it is sourced at the academic level. I don’t expect you to produce a paper of this scope, it is just there for you to use as a reference on how to write a report. Enjoy! Day 13 (Wed March 22): UNIT #4 Trigonometric Functions Homework (please refer to unit outline) Day 14 (Wed April 8): Addition and Subtraction Formulas & Double Angle Formulas Day 15 (Mon April 13): Trig Proofs pt.2 / Review / Test First Half- Trig Proofs and Review Second Half- Test: Trigonometry Pt. 1 Day 16 (Wed April 15): UNIT#5 Trigonometry Pt 2 Graphing basic trig functions (sin, cos, tan) and inverses (csc, sec, cot) Homework (please refer to the course outline for problems) Day 17 (Mon April 20): Trig Applications and Equations NOTES WILL BE PROVIDED IN CLASS Homework (please refer to unit outline) Day 18 (Wed April 22): Trig Pt 2 Review and Test Average/Instantaneous Rate of Change of Change Review (graphing, application problems, solving equations) Day 19 (Mon April 27): UNIT #6: Logairthms Review Exponential Function Logarithmic and Exponential Form Transformations of Exponential Functions Application Problems using Exponential and Logarithmic Forms Day 20 (Wed April 29): Logarithms Applications and Solving Logarithmic Equations Transformations of Logarithmic Functions Solving Exponential and Logarithmic Equations Day 21 (Mon May 4): Review and Logarithmic Functions Test (FINAL) Day 22 (Wed May 6): Composition of Functions and Exam Review ## The Exponential Function and Its Applications in Science Consider the pairs of values for x and y shown in the table. Note that each time x increases by 1, the value of y is increased by a factor of 2. Any such function in which the dependent variable increases by a constant factor is called a “geometric progression”. In this case, the function can be generated by the expression: If we take the natural log of both sides of this equation, we get: If we then take the exponential of both sides, we find: From this we can see that any geometric progression can be represented by an exponential function with base e. Consider a geometric progression where y increases by a factor of 5 for each increase by 1 of x. (a) Assuming the first value of y is 5, calculate the first five values of y. Plot a graph of the function. (b) Find the value of λ such that y = e λ x > represents this progression. #### Derivative of the Exponential Function A key property (in fact, the defining property for the base e) of the exponential function is that it is it’s own derivative. d d x e x = e x >e^=e^> (See Stewart, section 3.1) This leads to a very important result that has many applications in the sciences. Whenever a quantity is changing in such a way that the time rate of change is proportional to the quantity itself, the quantity can be represented by an exponential function of time. (Note that since the independent variable now represents time, the symbol x is replaced by t.) For example, suppose y is a quantity that has the property: Dividing by y and multiplying by dt, we get: Integrating both sides of the equation gives: ln y = λ t + k where k is an arbitrary constant. Taking the exponential of each side of the equation gives: #### Exponential Decay Now consider a geometric progression where the factor is ½ (ie the dependent variable changes by a factor of ½ for each increment of 1 of the independent variable). The expression is: Again taking the natural log of both sides we get: Finally taking the exponential of both sides: and is characterized by the fact that the dependent variable decreases by a constant factor each time the independent variable increases by a certain amount. As a consequence, the exponent is negative. (a) Calculate y for x = 1, 2, 3, 4, 5. (b) In the corresponding geometric progression, what is the constant factor by which y changes each time x increases by 1? Now consider the case where the dependent variable decreases with time in proportion to its own value. As seen in section 1.2, the solution to this “differential equation” is an exponential, but this time with a negative exponent. The equation for an exponential decay is often written using a different parameter called the “time constant” which is defined as: τ = 1 λ >,!> The exponential decay equation becomes: Finally consider what we get when we take the natural log of each side of the equation: y = y o e − t / τ → l n y = l n y o − 1 τ t e^<-t/ au > ightarrow lny=lny_-< au >>t> Comparing this to the generic equation for a straight line, y = b + m x, (note that the variable y in this equation is different than the y in all other equations) we see that a graph of ln y vs t will yield a straight line with slope − 1 / τ ( or − λ ). This is a powerful experimental tool for demonstrating whether or not a certain measured quantity follows an exponential decay. The table shows the values of some quantity z as a function of time. ### Applications in Physics There are many physical processes that follow an exponential function. As discussed in section 1, this occurs whenever the rate of change of some quantity is proportional to the quantity itself. #### Radioactive Decay Reference: Benson, section 43.4 The atomic nucleus is made up of neutrons and protons. Only nuclei with certain ratios of the number of neutrons to the number of protons (called isotopes) are stable, all others will change spontaneously into another combination by giving off a quantum of radiation. These radiations were called alpha, beta and gamma rays by the early researchers. We now know that the alpha ray is a helium nucleus, the beta ray is an electron or positron and the gamma ray is a high energy photon. Because this process is governed by quantum mechanics, it happens at random. In a sample of radioactive nuclei, there is no way to know which individual nucleus will decay next or when the next decay will occur. All that is known is that there is a certain probability that a given nucleus will decay within a certain period of time. The probability of decay per unit time is λ and the probability in a period of time Δ t is λ Δ t. Since the probability is exactly the same for all nuclei of the same species, then the number of nuclei that will decay during the next unit of time is N λ Δ t where N is the number of nuclei present. Since the number of nuclei present in the sample decreases by this same amount, then: In a real experiment, one usually measures the number of particles or “rays” emitted by the sample per unit time. Assuming that each particle emitted represents the decay of one nucleus, this is the same as the number of nuclei that decay per second and is known as the decay rate R. It is given by: R = λ N ## 5.6: Application Problems with Exponential and Logarithmic Functions - Mathematics Objectives: - Students should be able to graph exponential functions and evaluate exponential expressions. - Students should be able to use exponential functions to model real life situations such as the depreciation of a cars value. On the board when the students enter the room are these two graphs: Questions to be asked: Which one of these graphs represents exponential decay? Why? Any other suggestions? What is the domain? Range? How did you get that? These graphs are representations of exponential functions. The equation looks like f(x)=A^x where A is any real number greater than 1. Students will have graphing calculators Questions to be asked: Why do you think A can not be 1? 0? Why do you think A can not be a negative number? The above graph shows -A^x but the graph of (-A)^x is only points on the graph. Everyone put these equations into the y= screen: y1= 2^x y2=-2^x y3=(-2)^x ( you will need to use the carrot button).Now go to the table of values- What is different about y1 and y2 ? How is y3 different from y1 or y2? If you have y=-2^x and let x=2 then what is y? y= (-1)2*2= -4 What about y=(-2)^x and let x=2 then what is y? y=(-2)(-2)=4 ***In the homework tonight when it says to sketch the graphs of f(x)= -6^x then you will sketch the graph of y=-1*(6^x). Now students will explore different types of exponential functions. They will be allowed to use any type of technology that is available. I will suggest that the students clear everything in your y= screen and graph y= 3^x ( you will need to use the carrot button). The questions that they will need to answer and hand in for each equation below is What is the Domain? Range?. Now what would happen to this graph if: y= 3^(x)-3 y= 3^(x+2)-5 y= 3^x+2 y= 3^(x-3)+5 y= 3^x-3 y= -3^x remember that this is -1*(3^x) y= 3^x+2 Anyone have any idea why we use these types of functions? One reason why we learn how to use these exponential functions is one day you may want to sell your car and you will need to know how to calculate the present day value ( so no one will rip you off) ? We will us a formula that is almost the same as the one you used to find compound interest on yesterdays test. A=P(1+(r/n))^nt this formula represents growth we will use A=P(1-(r/n))nt but sense n will always be 1 b/c the value only decreases one time a yr. A=P(1-r)^t where A is the value of the car P is the price that you paid for the car r is the rate in which it decreases t is the age of the car Transparency of the following: Suppose that you purchased a new car for 8000 in 1991. If the value of the car decreases by 10% each year to 90% of its previous value what is the car worth today? On the board A= P(1-r)^t What is P? A= 8000(1-r)^t What is r? r needs to be in decimal form A= 8000(1-.10)^t What is t? (1991-1997) A= 8000(1-.10)^6 A= 4251 ?? How can we make a graph of this function? Remember what an exponential function looks like What about y= 8000(.90)^x graph this on your graphing calculator What do you see? What should we set the window to be? xmin=0 xmax=20 ymin= 0 ymax= 8000 yscl=1000 This graph will show us the value of the car at any age. ?? Use your trace key and find the value of the car if it is 15 yr. old? =1647 What is the value to 3 decimal places? =1647.129 (go to table) Students will be able to explore with different values and years. They should get a good understanding how the independent and dependent varaible works. Go over transparency #2 In 1990, you bought a television for$600. Each year, for t years, the value, v, of the television decreases by 8%. Write an exponential model that describes this situation.

V= 600(1-.08)^t
Next I am going to do a quick review. (Students to the board)
Positive Integer Exponent What does 4^3 mean? = 4*4*4 or an= a*a*a*. *a (n times)
Zero Exponent What is 5^0 ? = 1 or a0= 1
Rational Exponent 3^(1/4)= or a^(1/n)= 3^(5/4)= or a^(m/n)=
Negative Exponent 3^(-1/4)= or a(-1/n) =
Homework: p. 402: 1-4,7-11 odd, 13-20, 27-39 odd, 45-52, 53-60, 65,66
Assessment: Having students go to the board and answer the review questions above will give me a chance to see how the students learn or recall information. I allowed time at the end of class to start homework, and this gave me the opportunity to walk around the room to observe their style of learning.

Objectives: Evaluate logarithmic expressions algebraically and with calculators
Convert between exponential and logarithmic form

I. Warm-up:
If a^n=a^x, then x=n. Ex. If 3^2=3^x, then x = 2.
Solve for x:
2^x=16
3^x=27
5^x=1/25
-2^x=-8
Evaluate: 3^-3, 5^-3, 10^0, 597^0, 10^-2, 10^3
NO CALCULATORS!

II. Questions from last night's homework and completion of unfinished work from yesterday

III. Introduction of logarithms and converting between logarithmic and exponential form
A. Conduct whole-class discussion about evaluating 2^x=6. May need to note 2^2=4 and 2^3 = 9, so 2<x<3.
B. Briefly speak of John Napier, the inventor of logs.
C. When are we EVER going to use this?
Briefly explain about real-life applications of logarithms: decibels, Richter Scale, Carbon Dating.
D. Logarithms are just exponents!
E. Go through example 1 in textbook while writing some of the following on the board:
Logarithmic form Exponential form
log(2 )16=4 24=16
log 10=1 10^1=10
log(3 )1=0 3^0=1
log 0.1=-1 10^-1=0.1 or 1/10
log(4 )64=x _______
log(5) 125=y _______
log(7 )49=z _______

IV. Evaluating logarithmic expressions
log(4 )16
log(5 )1
log(9 )3
log(3 )-1
In whole class discussion, go through the answers to these problems. Ask students how they got their answers and call some to the board to write their methods down if necessary. Use exponential form, if necessary. Go back to examples already written on the board (log4 64=x, log5 125=y, and log7 49=z) and have students evaluate them for x, y, and z, respectively.

V. Special cases and quick things to point out
A. On the board or overhead, write:
log(a) 1=0 because a0=1
log(a) a=1 because a1=a
log(a) a^x=x because _____
Conduct a brief discussion as to what is to be written in the blank. (ax=ax)
B. Conduct a discussion as to why we cannot have base 1 nor take the log of anything less than or equal to zero. Include in the discussion the following examples:
Logarithmic form Exponential form
log(1) 500=x 1^x=500
log(3 )0 =x 3^x=0
log7 (-5)=x 7x=-5
C. Explain that log x can be written as log x. This is the common logarithm.
If time, show students on their graphing calculators that the "log" button means log10. Have them evaluate on the calculator: log 1000000 and log 106. Does 106 = 1000000? Explain that base 10 is the base used most often in our lives.

VI. Change of base formula and wrap up.
A. Go back to the problem 2^x=6 and explain/conduct discussion on how to solve. Confirm why we take log2 of both sides. log(2) 2^x=log2 6. So, x= log2 6.

B. Observe with students that there is no way to enter base on many calculators (specifically, the TI-82). Write the formula on the board for students to use to find x:

So, log(2) 6 = log 6/log2 ª 2.585. Go through the formula with students.

Homework Assignment:
In textbook, p. 409: 1-5, 11-14, 15-33 odd, 37-39, 41-55 odd. Due next class day (Monday, Feb. 17). See attached.
On homework, students MUST copy the problem down and show some work. Homework will be graded A, B, C, F, or 0 based on effort.

Graphs and applications of logarithms

Goals: Reinforce evaluation of logarithmic expressions algebraically and with calculators
Reinforce conversion between exponential and logarithmic form
Graph simple logarithmic functions
Use logarithms in a real-life situation

log(6) 36=x
log(3) 0=x
log(2) (-4)=z
log(2) 8=x
log(5) 125=x
log(3) x=4
log(2) (x+4)=5

II. Questions from last night's homework and completion of unfinished work from yesterday

III. Graphs of basic logarithmic functions
With your graphing calculators graph y=log(2) x

B. Conduct whole class discussion of how to graph this and other logarithmic functions such as log x, log(5 )x. Use T-tables, graphing calculator and dry-erase graph board. Have student volunteer show how to come up with the graph without graphing it on the calculator. Be sure to note why we cannot start the T-table with x=0 (When will 2^y=0?). Students should notice what happens to the graph as base increases. Change of base formula may need to be noted: . To make the T-table, exponential form will likely be used.
C. On graphing calculator, graph alog (x+b) +c, varying a, b, and c. Conduct a class discussion about how varying these values affects the graphs.

On the overhead you will see that the top curve is when you add c, and the bottom curve is when you subtract c.

IV. Noticing inverse relations
A. With students, graph y=log x on graphing calculators. (Make sure students know this is logarithm base 10)
B. y=10^x. Notice anything?
C. y=x. Explain that these two functions are inverses of each other.

Slope of a beach, s, is related to the diameter, d, of the sand particles on it by this equation:
s=0.159+0.118log d.

Have students work in groups of 2 to find the slope when d=0.25. Have student volunteer work it out on board. (Answer is roughly 9/100). Teacher should sketch a rough graph of what the slope of the beach looks like. The students can also use the trace key to find a solution. Tell the students to get as close to .25 as possible.

B. Have students evaluate d=0.125 (fine sand) in same groups. Have student volunteer graph what the slope of the beach looks like.

Homework Assignment:
1. In textbook, p. 409-411:6, 61-66, 69-77 odd, 79-83, 88-90. Due tomorrow. (Tuesday, Feb. 18).
2. For extra credit, Mixed Review in textbook p. 412. Due in four class days. (Friday, Feb. 21)
On homework, students MUST copy the problem down and show some work. Homework will be graded A, B, C, F, or 0 based on effort. Evaluation scheme for the Mixed Review has yet to be determined.

Goals/ Objectives:
-Students should be able to use properties of logarithms.
-Students should be able to expand and condense logarithms.

When I introduced log properties I wanted the students to make a connection with something that they already know therefore, I had the following warm up problems on the board:

1- 4^ 3 * 4^ 2 = 4*4*4*4*4= 45 2- 2^ 4 * 2^ 5 = 2*2*2*2*2*2*2*2*2= 512
3- 3^ 6 /3^ 4 = (3*3*3*3*3*3)/(3*3*3*3)= (6-4) 32= 9
4- 7^ 2 /7^ 5 = (2-5) = 1/73
5- (7^ 3 )^ 2= (7*7*7)^2 = (7*7*7)(7*7*7)= 76
My goal for the students to come up with the following on their own.
Therefore a^na^m = a^(n+m) (a^n)^m = a^nm a^n/a^m = a^(n-m)

Log properties are formed the same way. This is still on the board:
a^na^m = a^(n+m) (a^n)^m = a^nm a^n/a^m = a^(n-m)
?? What do you think loga^(nm)? ***loga means log base a

Let's prove that loga (nm) = loga n + loga m
Proof: Let loga m= x and loga n= y.
a^x= m and a^y= n Change to exponential form.
nm= a^xa^y Find the product of m & n.
nm= a^(x+y) Multiplying same base means adding exponets
loga nm= loga a^(x+y) Change to logarithmic form.
loga nm= (x+y) b/c loga a^x= x
loga nm= loga n + loga m Substitute for x & y.
Therefore loga nm = loga n + loga m
--Everyone get your calculators out and lets see if this is true.
The first 3 rows see what log(8*32) , and the last 3 rows see what log8+ log32?
Has anyone got an answer? 2.408 2.408 Are they the same?
?? What about loga m/n? ***loga means log base a

Let's prove that loga m/n = loga m- loga n together
Proof: Let ax= m and ay= n
loga m =x and loga n =y Change to exponential form
a^x/a^y = m/n Find the quotient of m/n.
a^(x-y)= m/n Property of exponents
loga a^(x-y) = loga (m/n) Property of equality for log functions.
x-y = loga (m/n) Definition of inverse function.
loga m - loga n = loga (m/n) Substitute for x and y.
Therefore loga m/n = loga m- loga n
-- Lets see if this property holds.
The first 3 rows see what log(27/3), and the last 3 rows see what log27- log3?
Has anyone got an answer? .954 .954 Are they the same?
?? Can any one guess what the loga n^m ?
loganm = m logan
(logan)^m (loga n) is an exponent to a power m, so you multiply them together.
(an)^m = n is the exponent to a power of m, so you multiply them together.
example: log10^2=2 or 2 log 10=2
Let's try this property out. The first 3 rows see what log8^4, and the other 3 row see what 4 log 8? 3.612 3.612 Are they the same?
Without your calculators lets do some more examples:
Let log2= .301 and log3= .477
?? What is log2/3? log 2/3= log2- log3
.301- .477 = -.176
What is log6? log6= log (2*3)
= log2+ log3
=.301 + .477 = .778
What is log9? log9= log3^2
= 3 log2 = 3(.301)= .954
Now, how can I rewrite log7x3?
= log7+ logx^3 Is there anything else that can be done?
= log7+ 3 logx This is called Expanding the expression.

= logx/3 This is called Condensing the expression .

-- log 3xy^2?
= log3 +logx + 2logy
Is this condensing or expanding? How do you know?

-- log2- 2 logx?
= log2 - logx2= log(2/x2)
Being able to expand & condense helps to solve log functions
Examples:
logx- log3=2 log7x^3= 2.345
log(x/3)= 2 log7 + 3 logx= 2.345
(x/3) =10^2 .8 + 3 logx= 2.345
x/3= 100 3 logx= 1.5
x= 300 logx= .5
x= 10^.5= 3.162

Homework: p. 416: 1-4,5,7,16-36 even
Assessment: I will allow time at the end of class to assess the students as they work on their homework.

8.3 Properties of Logarithms

Goals/ Objectives:
-Students should be able to solve real- life problems using logarithms.
-Students will learn how to evaluate the intensity of earthquakes.
-Students will examine the patterns of decibels and intensity as they relate to sound.

Lesson Plans:
On the overhead when the students enter the room:
Expand the following problems: Condense the following problems:
log16x log6- log3/2
log6/5 10 logx+ (2/3)log64

**Students will answer the questions at the board.

?? Any questions?
Today we are going to us logarithms to solve problems

?? Has anyone ever felt an earthquake?
-- Transparency of the Richter Scale
This is a Richter scale. It is used to measure the strength of an earthquake. Each increase corresponds to a ten- times increase in intensity. For example, if an earthquake registers 8 on the Richter scale is ten times as intense as the one registering 7.

The formula that is used for the Richter scale is R= logI where R is the magnitude, I is the intensity ( intensity is a measure of the wave energy of an earthquake per unit of area).

How would we find the intensity of the Great San Francisco Earthquake in 1906, R= 8.3
R= logI 8.3 = logI How can we rewrite this into Exponential form?
I=10^8.3 = 199,526,231

Let's try another one. How would we find the intensity of the San Francisco Bay Area Earthquake in 1989, R= 7.1?
R= logI 7.1= logI How do we rewrite this one?
I= 10^7.1 = 12,589,254

If the intensity, I, is 92,523, what is R?
R= logI R= log 92,523 = 4.966 = 5
Another application that we are going to talk about today is Sounds and Decibels.

--Transparency of the Decibel Scale
This is a scale that represents different everyday sounds and the level of that sound.
Sounds travel by waves. These waves are picked up by our ears.
?? According to the chart what is the level of this classroom?

This worksheet that I am handing out is due by the end of the class. Make sure you put your name on the worksheet. Everyone needs to split up into groups of 3 and one group will have 4.

REAL- WORLD APPLICATIONS: DECIBEL LEVEL

For the following questions us the formula:

B= 10 log (I/I(0))
where B is the sound level in Decibels,
I is the intensity of the sound in watts per square centimeter,
I(0) is an intensity of 10^-16 watts per square centimeter.

1- What is the sound intensity in a conversation?

2- What is the sound intensity with the decibel level of 90? Where are you when you experience this level of sound?

3- What is the sound intensity of a quiet room?

4- Choose 3 additional sounds you hear daily and estimate where you would place them in the scale shown on the transparency.
Homework: p.416-418: 40-46 even, 47-51 odd, 61,63, 68- 74.

Assessment: I will be walking around watching the students work and making sure that they are working. I will be looking for the leaders of the group so that we will be able to group them differently on the next lab. See rubric for evaluation for the homework.

8.3 Properties of Logarithms (T-I 82 Lab)

Goals/ Objectives:
-Students should be able to use their calculator to do logarithms functions.

Lesson Plans:
On the board when students come into class:
(Students to board)
1-Is the graph of y= 3(.25)^x exponential growth or exponential decay.
2-Use a calculator to evaluate log3 to three decimal places.
3-Evaluate log(5)6 following to three decimal places using the change-of-base formula.
4-Expand log(x^3/y^4)
5-Condense log3+ 2logt

Today is a lab day.
Mini- Lab 38 Logarithmic Functions 2

Solve for x: 2=log(x-7)- log (x+3)

Sketch the graph of 2= log(x-7)- log (x+3)

What is the domain and the range of log(x-7)- log (x+3)?

Solve for x: 2= log(x-7) + log (x+3)

Sketch the graph of 2= log(x-7)+og (x+3)

What is the domain and the range of log(x-7)+ log (x+3)?

This lab is due at the end of the class, and no one should have problems with finishing.
Everyone needs to stay in their desk and do the lab on their own.

Assessment: I will be walking around helping and observing the students. The lab be graded on effort like the homework.

Goals: Students should know how to use the number e as a base of an exponential function.
Students will use the natural base e in real-life applications

Identify the following functions as representing exponential growth or exponential decay: f(x)=3(0.25)^x g(x)=2(2.1)^x

III. Introduction of the natural base e
A. Briefly mention Leonhard Euler, for whom e takes its name.
B. Show/describe that e=1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + .
ª 2.718281828459
(may need to review factorial notation)
C. Review that x and y are examples of variables, but e is a number. If time, show that it is irrational.
D. Have students evaluate eb for several values of b.
E. When are we EVER going to use this? Natural base e applies to interest rates on accounts, air pressure, population, and radioactive decay.

IV. Simplifying natural base expressions
Briefly go over the rules of exponents which apply to e just as they apply to any number or variable.
(e^2)(e^3) = e^2+3 = e^5
(6e^2)/(2e^1) = 3e^2-1 = 3e
e^-4 = 1/e^4
(2e^-1)^2 = 2*2^(e-2) = 4^(e-2) = 4/e^2

V. Graph of the base e function
A. Have students graph y=2^x and y=3^x, zooming in to see the graphs more clearly.
B. Next, have students graph y=e^x. Ask them about where the graph of y=e^x lies and why they think this must be so. 2<e<3.

C. Go over some shifts of the y=ex curve. Graph y=Ce^bx +d + a varying a, d and C at first, then b last. Lastly, make b negative to yield y=Cebx . Let students discover that if b is negative, the function represents decay, and if b is positive, the function represents growth.
D. Go through some examples of exponential decay and growth:
e^(6x),-3e^(6x),0.1e^(4x),3e^-x,-5e^-4x

VI. Using base e to growth of money with a continuously compounding interest rate.
A. Recall with students the formula for compounding interest n times per year: A=P(1+r/n)^nt. Prompt students for what the variables mean: A - balance, P-initial principle, r- annual interest rate, n- number of times per year that interest is compounded.
B. What if interest were compounded continuously, that is, roughly every second?
A=Pe^(rt) where all the variables remain the same except we have no n.
C. Using the graphing calculator and previous formulas, go through second example with students:
You just deposited $200 in a savings account which has an APR of 5.0%. If interest were compounded continuously, what would the balance be after five years ($256.81)? What if interest was compounded daily? ($256.80) Our y= screen should look like this: Use your trace key to find x=5 Give other values for the students to find, so they will use the trace key. D. If time, go through more examples with students. Initial principle =$5000, interest = 3%. Compare compunding semi-annually and continuously.
Note that with smaller initial deposits, the continuous compounding does not differ much from compounding daily.

Homework Assignment:
In textbook, p. 423-425: 1-4, 9-21 odd, 25-32, 39-41, 45-52, 58-60.
On homework, students MUST copy the problem down and show some work. Homework will be graded A, B, C, F, or 0 based on effort.

Goals/Objectives: Students should be able to exhibit knowledge of the material covered in the past 8 days (and, of course, all the days before).

Conduct the following review (which Parallels the test) and then open the class to any other questions

Compare the following graphs?
y= 5x+7 y= 5(x+7)

Expand the following:
log(3/a) logx^4

Condense the following:
log3+ logx 7logb

The value of the new car you bought in 1990 for $17800 decreases by 16% each year. What will the car be worth in 1998? ?? Does anyone have any other questions from your homework? Evaluate to 3 decimal places log(2) 32 log99 Solve for x log(x) 64= 6 log2x + log25= 4 Is f(x)= 10.2e^.04t an example of exponential growth or decay?$600 is deposited in an account that pay 7% annual interest compounded continuously. Use the formula A= Pe^rt to find the balance after 10 years.

?? Does anyone have any other questions from your homework?

Assessment: I will be walking around helping the students that have any questions.

1. Evaluate e^2 to three decimal places.

2. Compare the graphs of f(x)=3^x and g(x)=3^x+1.

3. Craig and Jessica bought a new car in 1984 for $16,000. The car's value decreased by 13% each year. What was the car's value in 1992? 4. Solve for x. log 8 - (1/3)logx = log 2 5. Expand the expression log(4) 5x^2. 6. Condense the expression log(4) x - log(4 )5. 7. Solve for x. log(4) (1/16) = x 8. Which of the following is undefined? log(3) 9, log(3) 1, log(3) (-3) 9. Evaluate log(6) 30 to three decimal places. 10. Solve for x. log7 2401 = x 11. Use the grid above to sketch the graph of f(x) = log(4) (x+1). 12. Evaluate 6^e to three decimal places. 13. Is f(x) = 13.7e^-0.04t an example of exponential growth or exponential decay? 14.$1000 is deposited in an account that pays 7% annual interest compounded CONTINUOUSLY. Use the formula A = Pert to find the balance after 10 years.

The way in which our sense-organs $($eye, ear, etc.$)$ perceive the outside world $($light, sound, etc.$)$ is logarithmic e.g., if a sound becomes $a^n$ times stronger, we only perceive it as n times stronger.

You can use it in mortgage calculation. If you have a limit value to pay monthly for your house mortgage and if you wonder how many months needs to pay , you need to use logarithm. (It gives you idea about your budget and payment time)

With a fixed rate mortgage (interest is r), the borrower agrees to pay off the loan P completely at the end of the loan's term, so the amount owed at month N must be zero. For this to happen, the monthly payment c can be obtained from the previous equation to obtain: $egin c & <> = frac<1-(1+r)^<-N>>P end$

You plan to get $<$>100,000$from bank and interest rate is .15$ and you plan to pay each month \$1,000 ,

It gives you an idea about how many months you need to pay your mortgage payments with your payment budget. Please see Reference wiki page for detailed mortgage formulas