Transcript Slide 1

2
Nonlinear Functions
and Models
Copyright © Cengage Learning. All rights reserved.
2.3
Logarithmic Functions and Models
Copyright © Cengage Learning. All rights reserved.
Logarithmic Functions and Models
Today, computers and calculators have done away with
that use of logarithms, but many other uses remain.
In particular, the logarithm is used to model real-world
phenomena in numerous fields, including physics, finance,
and economics.
From the equation
23 = 8
we can see that the power to which we need to raise 2 in
order to get 8 is 3.
3
Logarithmic Functions and Models
We abbreviate the phrase “the power to which we need to
raise 2 in order to get 8” as “log2 8.”
Thus, another way of writing the equation 23 = 8 is
log2 8 = 3.
The power to which we need to raise 2 in
order to get 8 is 3.
This is read “the base 2 logarithm of 8 is 3” or “the log,
base 2, of 8 is 3.”
4
Logarithmic Functions and Models
Here is the general definition.
Base b Logarithm
The base b logarithm of x, logb x, is the power to which
we need to raise b in order to get x.
Symbolically,
logb x = y
Logarithmic form
means
by = x.
Exponential form
5
Logarithmic Functions and Models
Quick Examples
The following table lists some exponential equations and
their equivalent logarithmic forms:
6
Logarithmic Functions and Models
Common Logarithm, Natural Logarithm
The following are standard abbreviations.
Base 10: log10 x = log x
Common Logarithm
log(x)
Base e: loge x = ln x
Natural Logarithm
ln(x)
Quick Examples
Logarithmic Form
1. log 10,000 = 4
2. ln e = 1
Exponential Form
104 = 10,000
e1 = e
7
Logarithmic Functions and Models
Change-of-Base Formula
Change-of-base formula
Quick Examples
log(9)/log(11)
ln(9)/ln(11)
8
Example 1 – Solving Equations with Unknowns in the Exponent
Solve the following equations
a. 5–x = 125
b. 32x – 1 = 6
c. 100(1.005)3x = 200
Solution:
a. Write the given equation 5–x = 125 in logarithmic form:
–x = log5 125
This gives x = –log5 125
= –3.
9
Example 1 – Solution
cont’d
b. In logarithmic form, 32x – 1 = 6 becomes
2x – 1 = log3 6
2x = 1 + log3 6
giving
10
Example 1 – Solution
cont’d
c. We cannot write the given equation, 100(1.005)3x = 200,
directly in exponential form.
We must first divide both sides by 100:
11
Logarithmic Functions and Models
Logarithmic Function
A logarithmic function has the form
f(x) = logb x + C
(b and C are constants with b > 0, b  1)
or, alternatively,
f(x) = A ln x + C.
(A, C constants with A  0)
Quick Examples
1. f(x) = log x
2. g(x) = ln x – 5
12
Logarithmic Functions and Models
Logarithm Identities
The following identities hold for all positive bases a  1 and
b  1, all positive numbers x and y, and every real number
r. These identities follow from the laws of exponents.
Identity
Quick Examples
13
Logarithmic Functions and Models
Relationship with Exponential Functions
The following two identities demonstrate that the operations
of taking the base b logarithm and raising b to a power are
inverse to each other.
Identity
1. logb(bx) = x
In words: The power to which you
raise b in order to get bx is x (!)
Quick Examples
log2(27) = 7
14
Logarithmic Functions and Models
Identity
2. blogb x = x
In words: Raising b to the power to
which it must be raised to get x,
yields x. (!)
Quick Examples
5log5 8 = 8
15
Applications
16
Example 3 – Investments: How Long?
Global bonds sold by Mexico are yielding an average of
2.51% per year. At that interest rate, how long will it take a
$1,000 investment to be worth $1,200 if the interest is
compounded monthly?
Solution:
Substituting A = 1,200, P = 1,000, r = 0.0251, and m = 12 in
the compound interest equation gives
17
Example 3 – Solution
cont’d
≈ 1,000(1.004333)12t
and we must solve for t.
We first divide both sides by 1,000, getting an equation in
exponential form:
1.2 = 1.00209212t.
In logarithmic form, this becomes
12t = log1.002092(1.2).
18
Example 3 – Solution
cont’d
We can now solve for t:
log(1.2)/(log(1.002092)*12)
≈ 7.3 years
Thus, it will take approximately 7.3 years for a $1,000
investment to be worth $1,200.
19
Applications
Exponential Decay Model and Half-Life
An exponential decay function has the form
Q(t) = Q0e−kt.
Q0, k both positive
Q0 represents the value of Q at time t = 0, and k is the
decay constant.
The decay constant k and half-life th for Q are related by
thk = ln 2.
20
Applications
Quick Example
If th = 10 years, then 10k = ln 2, so k =
the decay model is
≈ 0.06931 and
Q(t) = Q0e−0.06931t.
Exponential Growth Model and Doubling Time
An exponential growth function has the form
Q(t) = Q0ekt.
Q0, k both positive
Q0 represents the value of Q at time t = 0, and k is the
growth constant.
21
Applications
The growth constant k and doubling time td for Q are
related by
tdk = ln 2.
Quick Example
P(t) = 1,000e0.05t
$1,000 invested at 5% annually with
interest compounded continuously
22
Logarithmic Regression
23
Logarithmic Regression
If we start with a set of data that suggests a logarithmic
curve we can, by repeating the methods, use technology to
find the logarithmic regression curve y = logb x + C
approximating the data.
24
Example 5 – Research & Development
The following table shows the total spent on research and
development by universities and colleges in the U.S., in
billions of dollars, for the period 1998–2008 (t is the number
of years since 1990).
Find the best-fit logarithmic model of the form
S(t) = A ln t + C
and use the model to project total spending on research by
universities and colleges in 2012, assuming the trend
continues.
25
Example 5 – Solution
We use technology to get the following regression model:
S(t) = 19.3 ln t – 12.8.
Coefficients rounded
Because 2012 is represented by t = 22, we have
S(22) = 19.3 ln(22) – 12.8
≈ 47.
Why did we round the result
to two significant digits?
So, research and development spending by universities
and colleges is projected to be around $47 billion in 2012.
26