Chabot Mathematics
§4.2 Log
Functions
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Review §
4.1
 Any QUESTIONS About
• §4.1 → Exponential Functions
 Any
QUESTIONS
About
HomeWork
• §4.1 → HW-18
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
§4.2 Learning Goals
 Define and explore
logarithmic functions
and their properties
 Use logarithms to
solve exponential
equations
 Examine applications
involving logarithms
John Napier (1550-1617) • Logarithm Pioneer
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BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Logarithm → What is it?
 Concept: If b > 0 and b ≠ 1, then
y = logbx is equivalent to x = by
 Symbolically
The exponent is the logarithm.
x = by
y = logbx
The base is the base of the logarithm.
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Logarithm Illustrated
 Consider the exponential function
y = f(x) = 3x. Like all exponential
functions, f is one-to-one. Can a
formula for the inverse Function,
x = g(y) be found?
x = 3y
y = 3x Need
y ≡ the exponent to which we
must raise 3 to get x.
f −1(x) ≡ the exponent to which we must
raise 3 to get x.
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5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Logarithm Illustrated
 Now define a new symbol to replace the
words “the exponent to which we must
raise 3 to get x”:
log3x, read “the logarithm, base 3, of x,” or
“log, base 3, of x,” means “the exponent to
which we raise 3 to get x.”
 Thus if f(x) = 3x, then f−1(x) = log3x.
Note that f−1(9) = log39 = 2, as 2 is the
exponent to which we raise 3 to get 9
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Evaluate Logarithms

Evaluate:
a) log381

b) log31
c) log3(1/9)
Solution:
a) Think of log381 as the exponent to which we
raise 3 to get 81. The exponent is 4. Thus,
since 34 = 81, log381 = 4.
b) ask: “To what exponent do we raise 3 in order
to get 1?” That exponent is 0. So, log31 = 0
c) To what exponent do we raise 3 in order to get
1/9? Since 3−2 = 1/9 we have log3(1/9) = −2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
The Meaning of logax
 For x > 0 and a a positive constant
other than 1, logax is the exponent to
which a must be raised in order to get x.
Thus,
logax = m means am = x
 or equivalently, logax is that unique
exponent for which
Chabot College Mathematics
8
a
loga x
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Exponential to Log
 Write each exponential equation in
logarithmic form.
4
2
1
1

3
c. a  7
a. 4  64
b.   
 2
16
3
 Soln
a. 4  64  log 64  3
4
4
1
1
 1
b.   
 log1 2
4
 2
16
16
c. a 2  7  log a 7  2
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Log to Exponential
 Write each logarithmic equation in
exponential form
a. log 3 243  5 b. log 2 5  x c. log a N  x
 Soln
a. log 3 243  5  243  35
b. log 2 5  x  5  2
x
c. log a N  x  N  a
Chabot College Mathematics
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x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Evaluate Logarithms
 Find the value of each of the following
logarithms
c. log1 3 9
a. log 5 25
b. log 2 16
1
d. log 7 7
e. log 6 1
f. log 4
2
 Solution
a. log 5 25  y  25  5 or 5  5
y
y2
b. log 2 16  y  16  2 or 2  2
y
y4
y
y
Chabot College Mathematics
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2
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Evaluate Logarithms
 Solution (cont.)
y
 1
2
y
c. log1 3 9  y  9    or 3  3
 3
d. log 7 7  y  7  7 y or 71  7 y
e. log 6 1  y  1  6 or 6  6
y
0
y
y 1
y0
1
1
y
1
2y
f. log 4  y   4 or 2  2
2
2
Chabot College Mathematics
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y  2
1
y
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Use Log Definition
 Solve each equation for x, y or z
1
a. log 5 x  3
b. log 3
y
27
2
d. log 2 x  6x  10  1
c. log z 1000  3

 Solution
a. log 5 x  3
3
x5
1
1
x 3 
5
125
Chabot College Mathematics
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
 1 
The solution set is 
.
125 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Use Log Definition
 Solution (cont.)
1
b. log 3
y
27
1
 3y
27
33  3y
3  y
c. log z 1000  3
1000  z
3
10  z
10  z
The solution set is 10.
3
3
The solution set is 3.
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Inverse Property of Logarithms
 Recall Def: For x > 0, a > 0, and a ≠ 1,
y  log a x if and only if
xa .
y
 In other words, The logarithmic function
is the inverse function of the exponential
function; e.g.
log a a  x
x
Chabot College Mathematics
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a
log a x
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Derive Change of Base Rule
 Any number >1 can be used for b, but since
most calculators have ln and log functions we
usually change between base-e and base-10
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Inverse Property
 Evaluate: log 5 23
5
.
 Solution
Remember that log523 is the exponent
to which 5 is raised to get 23. Raising 5
to that exponent, obtain
log 5 23
5
Chabot College Mathematics
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 23.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Basic Properties of Logarithms
 For any base a > 0, with a ≠ 1,
Discern from the Log Definition
1. Logaa = 1
• As 1 is the exponent to which a
must be raised to obtain a (a1 = a)
2. Loga1 = 0
• As 0 is the exponent to which a
must be raised to obtain 1 (a0 = 1)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Graph Logarithmic Function
 Sketch the graph of y = log3x
x
y = log3x
(x, y)
–3
(1/27, –3)
–2
(1/9, –2)
–1
(1/3, –1)
30 = 1
0
(1, 0)
31 = 3
1
(3, 1)
32 = 9
2
(9, 2)
 Soln:
3–3 = 1/27
Make
3–2 = 1/9
T-Table
–3 = 1/3
3
→
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Graph Logarithmic Function
 Plot the
ordered pairs
and connect
the dots with a
smooth curve
to obtain the
graph of
y = log3x
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Graph by Inverse
 Graph
y = f(x) = 3x
 Solution:
Use Inverse Relation
for Logs &
Exponentials
 Reflect the graph of
y = 3x across the line
y = x to obtain the
graph of y = f−1(x)
= log3x
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Properties of Exponential and
Logarithmic Functions
Exponential Function
f (x) = ax
Logarithmic Function
f (x) = loga x
Domain (–∞, ∞)
Range (0, ∞)
Domain (0, ∞)
Range (–∞, ∞)
y-intercept is 1
No x-intercept
x-intercept is 1
No y-intercept
x-axis (y = 0) is the
horizontal asymptote
y-axis (x = 0) is the
vertical asymptote
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Properties of Exponential and
Logarithmic Functions
Exponential Function
f (x) = ax
Logarithmic Function
f (x) = loga x
Is one-to-one , that is,
au = av
if and only if u = v
Is one-to-one, that is,
logau = logav
if and only if u = v
Increasing if a > 1
Decreasing if 0 < a < 1
Increasing if a > 1
Decreasing if 0 < a < 1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Graphs of Logarithmic Fcns
f (x) = loga x (a > 1)
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f (x) = loga x (0 < a < 1)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Common Logarithms


The logarithm with base 10 is called
the common logarithm and is denoted
by omitting the base: logx = log10x. So
y = logx if and only if x = 10y
Applying the basic properties of logs
1. log(10) = 1
2. log(1) = 0
3. log(10x) = x
4. 10logx = x
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Common Log Convention
 By this Mathematics CONVENTION the
abbreviation log, with no base written, is
understood to mean logarithm base 10,
or a common logarithm. Thus,
log21 = log1021
 On most calculators, the key for
LOG
common logarithms is marked
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Natural Logarithms
 Logarithms to the base “e” are called
natural logarithms, or Napierian
logarithms, in honor of John Napier,
who first “discovered” logarithms.
 The abbreviation “ln” is generally used
with natural logarithms. Thus,
ln 21 = loge 21.
 On most calculators, the key for natural
logarithms is marked LN
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Natural Logarithms


The logarithm with base e is called the
natural logarithm and is denoted by ln
x. That is, ln x = loge x. So
y = lnx if and only if x = ey
Applying the basic properties of logs
1. ln(e) = 1
2. ln(1) = 0
3. ln(ex) = x
4. elnx = x
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Sound Intensity
 This function is sometimes
 I 
used to calculate
d  10log  
sound intensity
 I0 
 Where
• d ≡ the intensity in decibels,
• I ≡ the intensity watts per unit of area
• I0 ≡ the faintest audible sound to the
average human ear, which is 10−12 watts
per square meter (1x10−12 W/m2).
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Sound Intensity
 Use the Sound Intensity Equation
(a.k.a. the “dBA” Eqn) to find the
intensity level of sounds at a decibel
level of 75 dB?
 Solution: We need
to isolate the
intensity, I, in
the dBA eqn
Chabot College Mathematics
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 I 
d  10log   ,
 I0 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Sound Intensity
 Solution (cont.) in the dBA eqn
substitute 75 for d and 10−12 for I0 and
then solve for I
 I 
75  10  log 
12 
 10

 I  1012  I  1012  107.5
7.5  log 
12
12 
10
 10

I
4.5
7.5
10
I
 10
10
12
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Sound Intensity
 Thus the Sound
Intensity at 75 dB is
10−4.5 W/m2 =
10−9/2 W/m2
 Using a Scientific
calculator and find
that I = 3.162x10−5
W/m2
= 31.6 µW/m2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Sound Intensity
 Check
If the sound intensity is 10−4.5 W/m2 ,
verify that the decibel reading is 75.
 104.5 
d  10log 

 1012 


7.5
d  10log10
d  10  7.5
Chabot College Mathematics
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d  75

Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Summary of Log Rules
 For any positive numbers M, N, and a
with a ≠ 1, and whole number p
log a ( MN )  log a M  log a N ;
log a M
p
 p log a M ;
Power Rule
M
log a
 log a M  log a N ;
N
k
log a a  k .
Chabot College Mathematics
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Product Rule
Quotient Rule
Base-to-Power Rule
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Typical Log-Confusion
 Beware that Logs do NOT behave
Algebraically. In General:
log a ( MN )  (log a M )(log a N ),
M log a M
log a

,
N log a N
log a ( M  N )  log a M  log a N ,
log a ( M  N )  log a M  log a N .
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Change of Base Rule
 Let a, b, and c be positive real
numbers with a ≠ 1 and b ≠ 1.
Then logbx can be converted to a
different base as follows:
log a x
log x
ln x
log b x 


log a b
log b
ln b
(base a) (base 10) (base e)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Derive Change of Base Rule
 Any number >1 can be used for b, but since
most calculators have ln and log functions we
usually change between base-e and base-10
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Evaluate Logs
 Compute log513 by changing to
(a) common logarithms
(b) natural logarithms
 Soln
Chabot College Mathematics
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log13
a. log 5 13 
log 5
 1.59369
ln13
b. log 5 13 
ln 5
 1.59369
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Evaluate Logs
 Use the change-of-base formula to
calculate log512.
• Round the answer to four decimal places
 Solution
 Check
Chabot College Mathematics
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log12
log 5 12 
log5
 1.5440
51.5440  12.0009  12

Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Evaluate Logs
 Find log37 using the change-of-base
formula
log10 7
 Solution log3 7 
log10 3
0.84509804

0.47712125
Substituting into
log a M
logb M 
.
log a b
1.7712
3
 7.000
 1.7712
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Use The Rules
 Express as an equivalent expression
using individual logarithms of x, y, & z
x3
xy
a) log 4
b) logb 3
yz
z7
3
x
3 – log yz
a)
log
=
log
x
4
 Soln
4
4
yz
a)
= 3log4x – log4 yz
= 3log4x – (log4 y + log4z)
= 3log4x – log4 y – log4z
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Use The Rules
1/ 3
 Soln
xy
 xy 
3
b)
log
 logb  
b
b)
7
7
z 
z
1
xy
  logb
7
3
z

1
 logb xy  logb z 7
3

1
  logb x  logb y  7logb z 
3
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Caveat on Log Rules
 Because the product and quotient
rules replace one term with two, it
is often best to use the rules within
parentheses, as in the previous
example
1
xy
  logb
3
z7
1
  logb x  logb y  7logb z 
3
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Cesium-137 ½-Life
 A sample of radioactive Cesium-137
has been Stored, unused, for cancer
treatment for 2.2 years. In that time, 5%
of the original sample has decayed.
 What is the half-life
(time required to reduce
the radioactive substance
to half of its starting
quantity) of Cesium-137?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Cesium-137 ½-Life
 SOLUTION:
 Start with the math model
kt
A t  A0 e
for exponential Decay
 Recall the Given information: after 2.2
years, 95% of the sample remains
 Use the Model and given data to find k
 Use data in Model: 0.95 A0  A0e  k 2.2 
 Divide both sides by A0: 0.95  e 2.2 k

Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Cesium-137 ½-Life
 Now take the ln of both Sides

ln 0.95  e 2.2 k


 ln 0.95  ln e 2.2 k

 Using the Base-to-Power Rule


ln 0.95  ln e 2.2 k  log e e 2.2 k
 ln 0.95  2.2k
 Find by
Algebra
 Now set the amount, A, to ½ of A0
A0
At HL  
 A0 e 0.0233t HL
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Cesium-137 ½-Life
 After dividing both
0.5  e 0.0233t
sides by A0
 Taking the ln of Both Sides

ln 0.5  ln e 0.0233t HL

HL
  0.6931  0.0233t HL
 Solving for
 0.6931
t HL 
 29.73
the HalfLife
 0.0233
 State: The HalfLife of Cesion-137 is
approximately 29.7 years
Chabot College Mathematics
47
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Compound Interest
 In a Bank Account that Compounds
CONTINUOUSLY the relationship
between the $-Principal, P, deposited,
the Interest rate, r, the Compounding
time-period, t, and the $-Amount, A, in
the Account:
1 A
t  ln
r P
Chabot College Mathematics
48
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Compound Interest
 If an account pays 8% annual interest,
compounded continuously, how long will
it take a deposit of $25,000 to produce
an account balance of $100,000?
 Familiarize
In the Compounding Eqn replace P with
25,000, r with 0.08, A with $100,000,
and then simplify.
Chabot College Mathematics
49
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Compound Interest
 Solution
1
100, 000
t
ln
0.08 25, 000
1
t
ln 4
0.08
t  17.33
Substitute.
Divide.
Approximate using a calculator.
 State Answer
The account balance will reach
$100,000 in about 17.33 years.
Chabot College Mathematics
50
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Example  Compound Interest
 Check:
1
A
17.33 
ln
0.08 25, 000
A
1.3864  ln
25, 000
1.3864  ln A  ln 25, 000
1.3864  ln 25, 000  ln A
11.513  ln A
e11.513  A
100, 007.5  A
 Because 17.33 was not the exact time,
$100,007.45 is reasonable for the Chk
Chabot College Mathematics
51
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
WhiteBoard Work
 Problems From §4.2
• P72 → Atmospheric
Pressure at Altitude
– See also: B. Mayer,
“Small Signal Analysis
of Source Vapor Control
Requirements for
APCVD”, IEEE
Transactions on
Semiconductor
Manufacturing, vol. 9,
no. 3, 1996, pg 355
Chabot College Mathematics
52
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
All Done for Today
Napier’s
MasterWork
Year 1619
Chabot College Mathematics
53
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
54
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
ConCavity Sign Chart
ConCavity
Form
d2f/dx2 Sign
++++++
Critical (Break)
Points
Chabot College Mathematics
55
−−−−−−
a
Inflection
−−−−−−
b
NO
Inflection
++++++
c
Inflection
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
x
Chabot College Mathematics
56
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Chabot College Mathematics
57
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx
Chabot College Mathematics
58
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx