Transcendental Functions:

Working with Exponential and Logarithmic Models

Lesson Overview

Instructor Presentation

Introduction

The rules for solving equations are the same no matter which type of equation needs to be solved.  The goal: ISOLATE THE VARIABLE.  The method: INVERSE OPERATIONS

Goals

Students will be able to solve complicated exponential and logarithmic equations.

Students will create exponential and logarithmic models depicting real-life applications and use the solutions and graphs to analyze and predict situations, parameters, or values in the future..

Notes

Exponential models include:

Population growth (bacteria, insects, etc.)

Financial applications (savings accounts, loan interest, etc.)

Logarithmic models include:

Earthquake study (Richter scale)

Loudness of sound (decibel scale)

Studies in the Physical Sciences (Newton's Law of Cooling, Acidity tests, etc.)

Logistic growth models include:

Population growth (cities)

Concept Exploration

Bases of Exponential functions

Solving Equations

Take a look at the following examples shown step by step.  The non-obvious steps in solving the equations are highlighted in blue.

Example A

({{({{(1.06^x)=4.1}, {ln((1.06^x))=ln(4.1)}, {x ln (1.06)=ln(4.1)}, {x = ((ln(4.1))/(ln (1.06)))}, {x ≈ 24.215092}})
}})

To remove the variable from inside the exponent, we must perform the inverse operation, which is this case is to apply a logarithm.  It doesn't matter which base is used, so I chose the natural logarithm (base e).

Example B

({{({{3 - log(x + 2) = 5}, {-log(x + 2)  = 2}, {log(x + 2)  = -2}, {(10^(log(x + 2)))  = (10^(-2))}, {x + 2  = 0.01}, {x  = -1.99}})}})

To remove the variable from inside the logarithm, we must perform the inverse operation after all other algebra is performed.  In this case, it is to apply an exponential function of the same base to both sides.

Example C

Now, use Mathematica to check your answers.  Use the Solve command, and remember to include a double equals sign.

Solve[(1.06^x)==4.1,x]

We already know that Mathematica is using "inverse functions," so this answer is fine.

Solve[3-Log[10,x+2]==5,x]

Examples

The Value of an Account

If interest is compounded k times per year, the following formula can be used to determine the amount (A) in the account after t years is...

A = P(1 + ((r/k))^kt)

A = amount in account after t years
P = initial amount
t = time in years
r = interest rate
k = number of compoundings per year

Example 1 - Finance

An account, with an initial balance of $2,500 and an annual interest rate of 1.5% compounded monthly needs to reach $3,000.  How long will it take for this account to reach the desired amount?

Solve[2500(((1+(.015/12)))^(12t))==3000,t]

What initial amount would allow the account to mature in half the time?  In a quarter of the time?  Move the slider in the demonstration below.  This slider represents the initial value of the account.  The red line indicates the goal ($3,000).

Manipulate[Plot[{P*(((1+.015/12))^(12*t)),3000},{t,0,12},PlotRange->{0,3000},PlotStyle->{{Thick,Blue},{Thick,Red}},ImageSize->{500,375}],{P,1,3000}]

Initial amount for maturation in half the time:

Initial amount for maturation in one-quarter the time:

Example 2 - Population growth

Use the data in the Table to determine whether a linear, logarithmic, exponential, power, or logistic regression equation is the best model for the data.

years=Range[1900,2000,10]

popValues={63.6, 64.4, 55.0, 59.2, 72.5, 128.6, 226.2, 302.6, 401.9, 550.0, 626.9}

Grid[Join[{{Style["Year",FontFamily->"Arial",Bold],Style["Population of Alaska in thousands",FontFamily->"Arial",Bold]}},
Partition[Riffle[years,popValues],2]],Dividers->{2->True,2->True}]

Partition[Riffle[years,popValues],2]

ListPlot[%,Joined->True, Filling->Axis,ColorFunction->"BlueGreenYellow",AxesLabel->{"year","Alaska Pop. in 1000's"},LabelStyle->Directive[Bold,FontFamily->"Helvetica"]]

From "PreCalculus: Graphical, Numerical, Algebraic - 6th edition" by Demana, Waits, Foley, and Kennedy

For a look at some interactive Mathematica demonstrations involving population growth, check out http://demonstrations.wolfram.com/search.html?query=population+growth&submit.x=0&submit.y=0

Example 3 - Exponential Decay

Radioactive Decay of Elements

Radioactive iodine is used to determine the health of the thyroid gland.  It decays according to the equation y = (ae^(-0.0856t)), where t is in days.  Find the half-life of this substance.  Work this problem graphically and algebraically.

***The "Half-life" of a substance is the time it takes for the substance to decay to half of its original mass.

({{y = ((ae)^(-0.0856t))}, {0.5a = ((ae)^(-0.0856t))}, {0.5 = (e^(-0.0856t))}, {ln (0.5) = ln ((e^(-0.0856t)))}, {ln (0.5) = -0.0856t}, {((ln (0.5))/(-0.0856)) = ((-0.0856t)/(-0.0856))}, {8.09751 =t }})

From Glencoe McGraw-Hill Mathematics "Algebra 2", 2003.

Now, move the slider on the half-life model below.  The a represents the initial amount of radioactive material.  What can you deduce about the half-life of the radioactive isotope as the initial amount changes?

Manipulate[Plot[{a*Exp[-0.0856t],.5a},{t,0.001,20},PlotRange->{0,5},PlotStyle->{{Thick,Blue},{Thick,Red}},Axes->True,AxesLabel->{"time (t) in years","Initial amount"}],{a,0.001,5}]

For a look at some interactive Mathematica demonstrations involving radioactive decay, check out http://demonstrations.wolfram.com/search.html?query=radioactive+decay&submit.x=0&submit.y=0

Examples Answer Key

Example 1 - Finance

Approximately 12 years to reach maturity.

~$2750 to halve the time and ~$2, 890 to one - third the time.

Example 2 - Population Growth

Exponential or possibly logistic.  To determine whether or not the regression should be logistic might require a few values beyond the year 2000.

Example 3 - Exponential Decay

Approximately 8 years.

The half-life of the radioactive isotope does not depend on the initial amount present.