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5.6: Application Problems with Exponential and Logarithmic Functions


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[24]{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[5]{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 invest $10,000 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.


Applications of Exponential and Log Functions



A series of free, online Intermediate Algebra Lessons or Algebra II lessons.
Videos, worksheets, and activities to help Algebra students.

In this lesson, we will learn

  • how to calculate compound interest (finite)
  • how to calculate compound interest (continuous)
  • how to solve exponential growth or decay word problems

Compound Interest (Finite Number of Calculations)

Compound Interest (Continuously)

Problems that involve continuous compound interest use a different equation from problems that have finitely compounded interest, but the continuous compound interest equation is also an exponential equation. We use many of the same methods for calculating continuous compound interest as we do finitely compounded interest. To calculate compound interest, we can use logarithms and methods for solving exponential equations.

Exponential Growth and Decay

Exponential decay refers to an amount of substance decreasing exponentially. Exponential decay is a type of exponential function where instead of having a variable in the base of the function, it is in the exponent. Exponential decay and exponential growth are used in carbon dating and other real-life applications.

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.


5.6: Application Problems with Exponential and Logarithmic Functions

The purpose of this lab is to use Maple to study applications of exponential and logarithmic functions. These are used to model many types of growth and decay, for example bacterial growth and radiaoctive decay. This lab also describes applications of exponential and logarithmic functions for heating and cooling and to medicine dosage

Separating the variables and integrating (see section 4.4 of the text), we have

In the case of exponential growth, we can drop the absolute value signs around , because will always be a positive quantity. Solving for , we obtain

which we may write in the form , where is an arbitrary positive constant.

where is a constant. This is the same equation as in exponential growth, except that replaces . The solution is

where is a positive constant. Physically, is the amount of material present at .

Radioactivity is often expressed in terms of an element's half-life. For example, the half-life of carbon-14 is 5730 years. This statement means that for any given sample of , after 5730 years, half of it will have undergone decay. So, if the half-life is of an element Z is years, it must be that , so that and .

where is the constant of proportionality and is the temperature of the environment. Using a technique called separation of variables it isn't hard to derive the solution

where is the temperature of the object at .

A problem facing physicians is the fact that for most drugs, there is a concentration, , below which the drug is ineffective and a concentration, , above which the drug is dangerous. Thus the physician would like the have the concentration satisfy

This means that the initial dose must not produce a concentration larger than and that another dose will have to be administered before the concentration reaches .

Sometimes you need to use experimental data to determine the value of constants in models. For example, suppose that for a particular drug, the following data were obtained. Just after the drug is injected, the concentration is 1.5 mg/ml (milligrams per milliliter). After four hours the concentration has dropped to 0.25 mg/ml. From this data we can determine values of and as follows. The value of is the initial concentration, so we have

To find the value of we need to solve the equation

which we get by plugging in and using the data . Maple commands for solving for and defining and plotting the function are shown below.

  1. In 1935 Charles F. Richter of Cal Tech developed a scale for measuring the magnitude of earthquakes. The Richter Scale formula is given by

where is the magnitude of the earthquake, is the amplitude of the largest seismic wave as measured on a standard seismograph 100 kilometers from the epicenter and is the amplitude of a reference earthquake of amplitude 1 micron on a standard seismograph at the same distance from the epicenter. A When the amplitude of an earthquake is tripled, by how much does the magnitude increase?

B In 1989, the San Francisco Bay area suffered severe damage from an earthquake of magnitude 7.1. However, the damage was not nearly as extensive as that caused by the great quake of 1906, which has been estimated to have had magnitude 8.3. What is the ratio of the amplitude of the 1906 quake to the 1989 quake?

C The largest earthquake magnitude ever measured was 8.9 for an earthquake in Japan in 1933. Determine the ratio of the amplitude of this earthquake to that of the 1906 San Francisco earthquake.


Properties of Logarithms

Use properties of logarithms to condense each logarithmic expression. Write the expression as a single logarithm whose coefficient is $1 .$ Where possible, evaluate logarithmic expressions.
$log x+3 log y$

Make Sense? Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning.
I expanded $log _ <4>sqrt>$ by writing the radical using a rational exponent and then applying the quotient rule, obtaining $frac<1> <2>log _ <4>x-log _ <4>y.$

Disprove each statement in Exercises $106-110$ by
a. letting y equal a positive constant of your choice, and
b. using a graphing utility to graph the function on each side of the equal sign. The two functions should have different graphs, showing that the equation is not true in general
$ln (x y)=(ln x)(ln y)$


Exponential and Logarithmic Functions and Examples

1. Give an example of an exponential function. Convert this exponential function to a logarithmic function, then plot the graph of both functions.

2. Given the following values 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, x, and y, form:
A linear equation in one variable
A linear equation in two variables
A quadratic equation
A polynomial of three terms
An exponential function
A logarithmic function
Compare all the functions written and post a note on their differences.

3. Plot a graph from the equations from above.

4. How do you think the knowledge of graphical representation of these functions will help you in your career or life?

© BrainMass Inc. brainmass.com March 4, 2021, 6:59 pm ad1c9bdddf
https://brainmass.com/math/basic-algebra/exponential-logarithmic-functions-examples-74815

Solution Summary

This solution shows the graphical representation of various types of functions and explains real-life applications.


5.6: Application Problems with Exponential and Logarithmic Functions

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?


College Algebra Problems With Answers sample 3 : Exponential and Logarithmic Functions

College algebra problems on logarithmic and exponential function with answers, are presented along with solutions are at the bottom of the page.

  1. Let the logarithmic function f be defined by f(x) = 2ln(2x - 1).
    a) Find the domain of f.
    b) Find vertical asymptote of the graph of f.
  2. Let the exponential function h be defined by h(x) = 2 + e x
    a) Find the range of h.
    b) Find the horizontal asymptote of the graph of h.
  3. The population of city A changes according to the exponential function
    A(t) = 2.9 (2) 0.11 t (millions)
    and the population of city B changes according to the exponential function
    B(t) = 1.7 (2) 0.17 t (millions)
    where t = 0 correspond to 2009.
    a) Which city had larger population in 2009?
    b) When will the sizes of the populations of the two cities be equal?
  4. Find the inverse of the logarithmic function f defined by f(x) = 2 Log5 (2x - 8) + 3
  5. Find the inverse of the exponential function h defined by h(x) = - 2*3 -3x + 9 - 4
  6. Solve the logarithmic equation defined by
    ln(2x - 2) + ln(4x - 3) = 2 ln(2x)
  7. A, B and k in the exponential function f given by
    f(x) = A e k x + B
    are constants. Find A, B and k if f(0) = 1 and f(1) = 2 and the graph of f has a horizontal asymptote y = -4.

Answers to the Above Questions

    1. solve 2x - 1 > 0 to find domain: x > 1/2
    2. solve 2x - 1 v = 0 to find vertical asymptote: x = 1/2
    1. range of h: (2, +infinity)
    2. horizontal asymptote: y = 2
    1. A(0) = 2.9 millions , B(0) = 1.7 millions, city A had larger population.
    2. solve 2.9 (2) 0.11 t = 1.7 (2) 0.17 t , to find t.
      take ln of both sides of the equation
      ln[ 2.9 (2) 0.11 t ] = ln[ 1.7 (2) 0.17 t ]
      ln(2.9) + 0.11t ln(2) = ln(1.7) + 0.17t ln(2)
      solve for t:
      t = (ln1.7 - ln2.9) / (0.11ln2 - 0.17ln2) = 13 (approximated to the nearest unit)
      The size of the two populations will be the same in 2009 + 13 = 2022.
      solve the equation: x = 2 Log5 (2y - 8) + 3 for y to obtain the inverse of the function.
      f -1 (x) = (1/2) 5 (x-3)/2 + 4
      solve the equation: x = - 2*3 -3y + 9 - 4 for y to obtain the inverse of the function.
      h -1 (x) = (-1/3)Log3 [(x+4)/-2] + 3
      Rewrite the given equation as follows
      ln(2x - 2)(4x - 3) = ln(2x) 2
      The above gives the algebraic equation
      (2x - 2)(4x - 3) = (2x) 2
      Solve the above quadratic equation for x
      x = 3 and x = 1/2
      Check the two values of x and only x = 3 is a solution to the given equation.
      Horizontal asymptote y = - 4 gives B = - 4.
      f(0) = A + B = 1
      which gives A = 5 since B = - 4
      f(1) = 5e k - 4 = 2
      solve for k to obtain: k = ln(6 / 5)

    More References and links


    5.6: Application Problems with Exponential and Logarithmic Functions

    The purpose of this lab is to use Maple to study applications of exponential and logarithmic functions. These are used to model many types of growth and decay, for example bacterial growth and radiaoctive decay. This lab also describes applications of exponential and logarithmic functions for heating and cooling and to medicine dosage

    Separating the variables and integrating (see section 4.4 of the text), we have

    In the case of exponential growth, we can drop the absolute value signs around , because will always be a positive quantity. Solving for , we obtain

    which we may write in the form , where is an arbitrary positive constant.

    where is a constant. This is the same equation as in exponential growth, except that replaces . The solution is

    where is a positive constant. Physically, is the amount of material present at .

    Radioactivity is often expressed in terms of an element's half-life. For example, the half-life of carbon-14 is 5730 years. This statement means that for any given sample of , after 5730 years, half of it will have undergone decay. So, if the half-life is of an element Z is years, it must be that , so that and .

    where is the constant of proportionality and is the temperature of the environment. Using a technique called separation of variables it isn't hard to derive the solution

    where is the temperature of the object at .

    A problem facing physicians is the fact that for most drugs, there is a concentration, , below which the drug is ineffective and a concentration, , above which the drug is dangerous. Thus the physician would like the have the concentration satisfy

    This means that the initial dose must not produce a concentration larger than and that another dose will have to be administered before the concentration reaches .

    Sometimes you need to use experimental data to determine the value of constants in models. For example, suppose that for a particular drug, the following data were obtained. Just after the drug is injected, the concentration is 1.5 mg/ml (milligrams per milliliter). After four hours the concentration has dropped to 0.25 mg/ml. From this data we can determine values of and as follows. The value of is the initial concentration, so we have

    To find the value of we need to solve the equation

    which we get by plugging in and using the data . Maple commands for solving for and defining and plotting the function are shown below.


    SOLUTION: Need help with solving this problem, thank you for helping! Find the values of the constants C and b so that the curve y = Cb^(x) contains the points (4,3) and (5,6). Express yo

    go back to the first original equation (either one will be ok), and replace b with 2 to get:
    first original equation of 3 = cb^4 becomes:
    3 = c*2^4 which becomes:
    3 = c*16
    divide both sides of this equation by 16 to get:
    c = 3/16

    your solution should be that b = 2 and c = 3/16

    replace c with 3/16 and b with 2 and x with 4 in your first original equation to get:
    y = cb^x becomes:
    y = 3/16 *2^4 which becomes:
    y = 3/16 * 16 which becomes:
    y = 3
    solution is confirmed to be good in the first equation because y = cb^x becomes y = 3 when x = 4.

    replace c with 3/16 and b with 2 and x with 5 in your second original equation to get:
    y = cb^x becomes:
    y = 3/16 * 2^5 which becomes:
    y = 3/16 * 32 which becomes:
    y = 6
    solution is confirmed to be good in the second equation because y = cb^x becomes y = 6 when x = 5.


    Solving Exponential and Logarithmic Equations

    Evaluate and simplify algebraic expressions, for example: products/quotients of polynomials, logarithmic expressions and complex fractions and solve and graph linear, quadratic, exponential and logarithmic equations and inequalities, and solve and graph systems of equations and inequalities.

    Applications of Functions

    Non-Linear Equations

    Determine, use and/or interpret minimum and maximum values over a specified interval of a graph of a polynomial, exponential or logarithmic function.

    • Families of functions exhibit properties and behaviors that can be recognized across representations. Functions can be transformed, combined, and composed to create new functions in mathematical and real world situations.
    • Mathematical functions are relationships that assign each member of one set (domain) to a unique member of another set (range), and the relationship is recognizable across representations.
    • Numbers, measures, expressions, equations, and inequalities can represent mathematical situations and structures in many equivalent forms.
    • Patterns exhibit relationships that can be extended, described, and generalized.
    • Relations and functions are mathematical relationships that can be represented and analyzed using words, tables, graphs, and equations.
    • There are some mathematical relationships that are always true and these relationships are used as the rules of arithmetic and algebra and are useful for writing equivalent forms of expressions and solving equations and inequalities.
    • Algebraic properties, processes and representations
    • Exponential functions and equations
    • Polynomial functions and equations
    • Quadratic functions and equations
    • Represent a polynomial function in multiple ways, including tab les , graphs, equations, and contextual situations, and make connections among representations relate the solution of the associated polynomial equation to each representation.
    • Represent a quadratic function in multiple ways, including tab les , graphs, equations, and contextual situations, and make connections among representations relate the solution of the associated quadratic equation to each representation.
    • Represent exponential functions in multiple ways, including tab les , graphs, equations, and contextual situations, and make connections among representations relate the growth/decay rate of the associated exponential equation to each representation.

    Objectives

    In this lesson, students will write and solve exponential and logarithmic equations. Students will: [IS.2 - Struggling Learners]

    1. convert to and from exponential and logarithmic form.
    2. use the change of base formulas with the common logarithm and natural logarithm.
    3. solve real-world application problems using exponential and logarithmic equations.
    4. identify the domain and range of exponential and logarithmic functions.
    5. identify characteristics of the graphs of exponential and logarithmic functions.
    6. translate from one representation of an exponential or logarithmic function to another representation.
    7. identify what happens to the graph of an exponential or logarithmic function when the parameters change.

    Essential Questions

    1. How can we determine if a real-world situation should be represented by a quadratic, polynomial, or exponential function?
    2. How do you explain the benefits of multiple methods of representing exponential functions (tables, graphs, equations, and contextual situations)?

    Vocabulary

    1. Asymptote: A line such that a point, tracing a given curve and simultaneously receding to an infinite distance from the origin, approaches indefinitely near to the line a line such that the perpendicular distance from a moving point on a curve to the line approaches zero as the point moves off an infinite distance from the origin. [IS.1 - Struggling Learners]
    2. Exponential Equation: An equation in the form of y=ax an equation in which the unknown occurs in an exponent, for example, 9 (x + 1) = 243.
    3. Logarithmic Equation: An equation in the form of y=logax , where x=ay the inverse of an exponential equation.
    4. Domain: The set of all x-values or input values for an equation.
    5. Range: The set of all y-values or output values for an equation.
    6. Common Logarithm: Logarithm with base 10 if a = 10 x , then log a = x.
    7. Natural Logarithm: Logarithm with base e also ln, Napierian logarithm, Euler logarithm. The base, e, is approximately 2.71828.

    Duration

    120&ndash180 minutes/2&ndash3 class periods [IS.3 - All Students]

    Prerequisite Skills

    Materials

    1. Solving Exponential and Logarithmic Applications Worksheet (M-A2-4-2_Solving Exponential and Logarithmic Applications Worksheet.docx)
    2. Lesson 2 Exit Ticket (M-A2-4-2_ Lesson 2 Exit Ticket.docx)
    3. Graphing Exponential and Logarithmic Function Notes (M-A2-4-2_Graphing Exponential and Logarithmic Function Notes and KEY.docx)
    4. Graphing Practice Worksheet (M-A2-4-2_Graphing Practice Worksheet.docx)
    5. graph paper

    Related Unit and Lesson Plans

    Related Materials & Resources

    The possible inclusion of commercial websites below is not an implied endorsement of their products, which are not free, and are not required for this lesson plan.

    1. Solving Exponential and Logarithmic Applications Worksheet (M-A2-4-2_Solving Exponential and Logarithmic Applications Worksheet.docx)
    2. Lesson 2 Exit Ticket (M-A2-4-2_ Lesson 2 Exit Ticket.docx)
    3. Graphing Exponential and Logarithmic Function Notes (M-A2-4-2_Graphing Exponential and Logarithmic Function Notes and KEY.docx)
    4. Graphing Practice Worksheet (M-A2-4-2_Graphing Practice Worksheet.docx)
    5. graph paper

    Formative Assessment

    1. The Think-Pair-Share activity (Part 2) uses the Graphing Practice Worksheet. Students can evaluate their own and their partners&rsquo understanding of ways to accurately and appropriately represent logarithmic functions graphically. Remind students to apply the principles they already know about functions to make sure their graphs have one and only one y-value for each x. [IS.10 - Struggling Learners]
    2. The Lesson 2 Exit Ticket includes a growth/decay model of a real-world application of logarithms and requires student understanding of how to use logarithms as tools to represent a practical problem. Ask students to consider the reasonableness of their answers before completing the work.

    Suggested Instructional Supports

    Students will be learning about solving exponential and logarithmic equations. Solving equations is such an important aspect in making predictions about different situations. Students will be evaluated through observation, exit tickets, and an assessment. Students will also be learning about graphing exponential and logarithmic functions. Graphs are important visuals of functions. Students will be evaluated through observation, exit tickets, and an assessment.

    Students will be interested in today&rsquos lesson because many students at this age like crime shows and mysteries. Exponential and logarithmic functions occur in many realistic situations.

    Students will work in pairs today as well as on their own. They will write notes, which they will use to complete the lesson&rsquos tasks.

    Students will be able to reflect and revisit the problems they do during the class review. Students will then take that information and revise their thought-processes on the next task. You will be walking around while students are working and give them feedback throughout this time.

    Students will be able to evaluate themselves when they check their work with a partner. Their peers might be able to give them some more insight on their understanding.

    This lesson is tailored to collaboration, in which students are grouped at similar ability levels or different ability levels. There is also an extension problem for students who need more practice or for students who work quicker than their peers.

    This lesson has several parts and each part has either individual work or partner work. We will go over each problem and discuss the problems as a class. The discussions will transition the class from activity to activity.

    Consider the following steps with regard to vocabulary for struggling learners:

    1. Use of a graphic organizer (e.g., Frayer Model, Verbal Visual Word Association, Concept Circles).
    2. Introduce new vocabulary using student friendly definitions and examples and non-examples.
    3. Review words with students.
    4. Provide opportunities for students to apply the new/reviewed terms.

    Instructional Procedures

    &ldquoToday we are going to learn how exponential and logarithmic equations are used to solve real-world applications. Who can tell me what the most basic exponential equation is and what each part of the equation means?&rdquo [IS.4 - Struggling Learners][y = ab x or y = ab x + k a &ne 0 (initial value) b is greater than 0 and &ne 1 (multiplier: describes a percentage increase or decrease) and k = asymptote (a value that the function gets close to but never touches)]

    Exponential and logarithmic functions are used in the real world. Most notably, exponential functions are used in population growth, interest, and bacterial growth. Logarithmic functions are used to measure light and sound intensity, as well as measuring magnitudes of earthquakes. Review how to convert back and forth from exponential form to logarithmic form since students will be doing this when graphing logarithmic equations.

    &ldquoToday we are going to learn how to graph exponential and logarithmic functions without the use of a calculator. We will begin with the equation, make a table of values with a few points, and sketch the graph.&rdquo

    &ldquoSuppose we have the exponential function, y=3x we can use a table of values to graph the function.&rdquo

    &ldquoLet&rsquos fill in our table.&rdquo Using a projector or interactive whiteboard, display the following chart: [IS.5 - Struggling Learners]

    3 x

    3 &minus2

    &minus2, .11 (point A)

    3 &minus1

    &minus1, .33 (point B)

    3 0

    3 1

    1, 3 (point D)

    3 2

    2, 9 (point E)

    &ldquoNow, we can create our graph.&rdquo Display the following graph:

    &ldquoNotice that the graph approaches a horizontal asymptote of y = 0&rdquo

    &ldquoNow, let&rsquos graph a logarithmic function!&rdquo

    &ldquoSince a logarithmic function is the inverse of an exponential function, we simply graph the exponential function that is the inverse, draw the line of symmetry, y = x, and plot the reverse coordinates for each point on the exponential function. An illustration will make this process easier to understand.&rdquo

    &ldquoLet&rsquos take our exponential function from before, y=3x. The inverse of this function is log3x .&rdquo

    &ldquoLet&rsquos look at our table from before and insert another column for the ordered pair containing the reversed coordinates.&rdquo [IS.6 - Struggling Learners]

    3 x

    Coordinates for log3x

    3 &minus2

    3 &minus1

    3 0

    3 1

    3 2

    &ldquoWe will now graph the points for the logarithmic function.&rdquo Display the following graph:

    &ldquoNow we simply have to connect the points of the logarithmic function. Note the vertical asymptote of x = 0.&rdquo

    &ldquoBefore we can get to the application problems, we have to learn about a few formulas. Let&rsquos say we have to solve 5x = 50. What do we do to solve for x?&rdquo (divide both sides by 5) &ldquoDivision is the inverse of multiplication. So what is the inverse of exponents?&rdquo (logarithms)

    &ldquoIf we have a problem like 2 x &ndash 1 = 8, we can simply rewrite 8 as 2 3 and then set the exponents equal to each other and solve for x.&rdquo

    &ldquoBut what if we don&rsquot have the same bases to work with? We can take the logarithm of each side of the equation.&rdquo Put the following formulas on the board and do the examples as a whole class.

    &ldquoRemember that ln refers to the natural logarithm, not the base 10 logarithm. It&rsquos important to keep in mind that these are two different bases. The base of the natural logarithm is approximately 2.71828 and is quite useful in many fields of mathematics.&rdquo

    &ldquoWhen we write log without a base next to it, it is the Common Log, base 10.&rdquo [IS.7 - Struggling Learners]

    &ldquoLet&rsquos try some examples.&rdquo Examples should be worked out together as a class. Note that there are multiple ways to solve these equations.

    1. 2 x = 10 Answer for number 1:

    x=log10log2 2. 2 x = 10 Work for number 2: log2x=log10 xlog2=log10 x=log10log2 3. 5 x = 45 Work for number 3: log5x=log45 xlog5=log45 x=log45log5 4. 8 x -1 = 100 Work for number 4: log8x - 1=log100 (x-1)log8=log100 x-1=log100log8 x=log100log8 + 1 5. 6 2 x + 3 = 50 Work for number 5: log62x+3=log50 2x+3log6=log50 2x+3=log50log6 2x=log50log6-3 x=log50log6-32

    &ldquoLet&rsquos look at a natural logarithm example. Suppose we have the exponential equation: 4e3x + 5 = 10. We can use the natural logarithm to solve the equation, since we have e as a base. The base of the natural logarithm, e, operates in the same way as base 10. A logarithm is the inverse of an exponential function. Log (1000) = 3 because 10 3 = 1000. In the same way, e 3 &asymp 20.08553, so ln (20.08553) &asymp 3.&rdquo Work through with students: [IS.8 - Struggling Learners]

    &ldquoIf we are given a logarithm and asked to evaluate, we can use the change of base formula. We can also convert the logarithm to another base.&rdquo

    &ldquoLet&rsquos explore how to evaluate a logarithm in terms of common logarithms using the change of base formula first .&rdquo (Review this concept as a class.)

    *For all positive numbers b, c, and M, where b &ne 1 and c &ne 1.

    &ldquoFor example, take this logarithm.&rdquo

    &ldquoNow, we can also convert this logarithm to another base. Let&rsquos convert it to base 6.&rdquo Work through with students:

    &ldquoNow we are going to get into some fun problems. We will do an example as a class, then you will solve a few problems in groups.&rdquo

    Give the following problem.

    Aunt Helen likes drinking tea, but she is specific about the temperature at which she drinks it. She boiled the water (100 o C) and poured it over the tea leaves. Five minutes later she came back and the tea was 65 o . Aunt Helen keeps her house at a cool 20 o . Write an equation that represents the temperature of Aunt Helen&rsquos tea.

    &ldquoFirst we need to determine what we know. We know that at time t = 0 (the time at which the water has come to a boil), y(the temperature of the tea) is 100 and that when t = 5 (the number of minutes it has been left to sit after it came to a boil), y is 65. What else do we know from the problem?&rdquo (Room temperature is 20 o , which is the asymptote, since nothing will cool down more than room temp.) &ldquoWe will substitute the first point into an exponential equation and first solve for a. Then we will substitute the second point and solve for b. This will tell us the percentage rate at which the tea cools per minute.&rdquo Work through with students:

    &ldquoWe can use this equation to make predictions. Let&rsquos say Aunt Helen only likes to drink her tea when it is 50 o . How long will she have to wait to drink her tea?&rdquo

    50 = 80(0.8913) t + 20
    30 = 80(0.8913) t
    0.375 = 0.8913 t

    &ldquoNow we have to use the change of base formula to solve for t.&rdquo

    t = log 0.375 ÷ log 0.8913 or ln 0.375 ÷ ln 0.8913 &asymp 8.5 minutes [IS.9 - Struggling Learners]


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