Logarithmic Models
Raja Almukahhal
Larame Spence
Mara Landers
Nick Fiori
Art Fortgang
Melissa Vigil
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Printed: March 2, 2017
AUTHORS
Raja Almukahhal
Larame Spence
Mara Landers
Nick Fiori
Art Fortgang
Melissa Vigil
www.ck12.org
C HAPTER
Chapter 1. Logarithmic Models
1
Logarithmic Models
Learning Objectives
Here you will explore more complex log functions and real-world applications of logarithmic functions.
In prior lessons, you used an exponential model to predict the population of a town based on a constant growth rate
such as 6% per year.
In the real world however, populations often do not just grow continuously and without limit. A town originally
founded near a convenient water source may grow very quickly at first, but the expansion will slow dramatically
as houses and businesses run out of room near the water source, and need to begin transporting water further and
further away.
How can a situation like this be modeled with an equation?
Logarithmic Models
In a prior lesson, we considered the solutions of simple log equations. Now we return to that topic and explore some
more complex examples. Solving more complicated log equations can be less difficult than you might think, by
using our knowledge of log properties.
For example, consider the equation log2 (x) + log2 (x - 2) = 3. We can solve this equation using a log property.
TABLE 1.1:
log2 (x) + log2 (x - 2) = 3
log2 (x(x - 2)) = 3
log2 (x2 - 2x) = 3 ⇒
23 = x2 - 2x
x2 - 2x - 8 = 0
(x - 4) (x + 2) = 0
x = -2, 4
logb x + logb y = logb (xy)
write the equation in exponential form.
Solve the resulting quadratic
The resulting quadratic has two solutions. However, only x = 4 is a solution to our original equation, as log2 (-2) is
undefined. We refer to x = -2 as an extraneous solution.
MEDIA
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Examples
Example 1
Earlier, you were asked how a situation could be modeled with an equation.
A population that increases continuously at a constant rate may be modeled with an exponential function.
A population that increases rapidly and then levels off may be modeled with a logarithmic function.
Example 2
Solve each equation.
a. log (x + 2) + log 3 = 2
TABLE 1.2:
log (3(x + 2)) = 2
log (3x + 6) = 2
102 = 3x + 6
100 = 3x + 6
3x = 94
x = 94/3
logb x + logb y = logb (xy)
Simplify the expression 3(x+2)
Write the log expression in exponential form
Solve the linear equation
b. ln (x + 2) - ln (x) = 1
TABLE 1.3:
x+2
=
x
x+2
1
e = x
ln
1
ex = x + 2
ex − x = 2
x(e − 1) = 2
2
x = e−1
logb x − logb y = logb
x
y
Write the log expression in exponential form.
Multiply both sides by x.
Factor out x.
Isolate x.
The solution above is an exact solution. If we want a decimal approximation, we can use a calculator to find that x
≈ 1.16. We can also use a graphing calculator to find an approximate solution. Consider again the equation ln (x +
2) - ln (x) = 1. We can solve this equation by solving a system:
(
y = ln(x + 2) − ln(x)
y=1
If you graph the system on your graphing calculator, you should see that the curve and the horizontal line intersection
at one point. Using the INTERSECT function on the CALC menu (press <2nd>[CALC]), you should find that the x
coordinate of the intersection point is approximately 1.16. This method will allow you to find approximate solutions
for more complicated log equations.
Example 3
Use a graphing calculator to solve each equation:
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Chapter 1. Logarithmic Models
a. log(5 - x) + 1 = log x
The graphs of y = log (5 - x) + 1 and y = log x intersect at x ≈ 4.5454545.
Therefore the solution of the equation is x ≈ 4.54.
b. log2 (3x + 8) + 1 = log3 (10 - x)
First, in order to graph the equations, you must rewrite them in terms of a common log or a natural log. The resulting
+ 1 and y = log(10−x)
equations are: y = log(3x+8)
log2
log3 .
The graphs of these functions intersect at x ≈ -1.87. This value is the approximate solution to the equation.
Example 4
Consider population growth:
TABLE 1.4:
Year
1
5
10
20
30
40
Population
2000
4200
6500
8800
10500
12500
If we plot this data, we see that the growth is not quite linear, and it is not exponential either.
We can find a logarithmic function to model this data. First enter the data in the table in L1 and L2. Then press STAT
to get to the CALC menu. This time choose option 9. You should get the function y = 930.4954615 + 2780.218173
ln x. If you view the graph and the data points together, as described in the Technology Note above, you will see
that the graph of the function does not touch the data points, but models the general trend of the data.
Note about technology: you can also do this using an Excel spreadsheet. Enter the data in a worksheet, and create
a scatterplot by inserting a chart. After you create the chart, from the chart menu, choose “add trendline.” You will
then be able to choose the type of function. Note that if you want to use a logarithmic function, the domain of your
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data set must be positive numbers. The chart menu will actually not allow you to choose a logarithmic trendline if
your data include zero or negative x values. See below:
Example 5
Solve for x : log2 x − log2 (x − 4) = 12.
To solve log2 x − log2 (x − 4) = 12:
x
= 12 : Using logx y − logx z = logx yz
log2 x−4
212 =
x
x−4
4, 096 =
: Write in exponential form
x
x−4
: With a calculator
4, 096x − 16384 = x : Multiply both sides by x − 4
4, 095x = 16, 384 : Simplify
x = 4 : Divide
Example 6
Biologists use the formula n = k · logA to estimate the number of species n that live in a given area A by multiplying
by a constant k which changes by location. If a particular rain forest has a constant k of 943 how many species would
be estimated to live in an area of 950km2 ?
To find the number of species in an area of 950km2 :
n = 943 · log950 : Substitute the given k and A values
n = 943 · 2.977 : With a calculator
n = 2, 807
Therefore 2,807 species would likely live in the area.
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Chapter 1. Logarithmic Models
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Review
Express 1-7 in exponential form:
1
1. log12 1728
= −3
2. log216 6 = 13
3. log 1 19 = 3
3
4.
5.
6.
7.
1
log 1 16
=2
4
log5 125 = 3
log15 225 = 2
log25 5 = 21
For questions 8-13, solve for x.
8.
9.
10.
11.
12.
13.
logx 64 = 2
log3 6561 = x
log5 x = 4
logx 27 = 3
log2 x = 6
log4 64 = x
For questions 14-19, solve for x.
14. 4log( 5x ) + log( 625
4 ) = 2logx
log5 125
15. log5 z + log5 x = 72
16. logp = 2−logp
logp
17. 2logx − 2log(x + 1) = 0
18. log(25 − z3 ) − 3log(4 − z) = 0
3)
19. log(35−y
log(5−y) = 3
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.10.
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