Post on 17-Dec-2015
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 1
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot Mathematics
§4.2 Log
Functions
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 2
Bruce Mayer, PE Chabot College Mathematics
Review §
Any QUESTIONS About• §4.1 → Exponential Functions
Any QUESTIONS About HomeWork• §4.1 → HW-18
4.1
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 3
Bruce Mayer, PE Chabot College Mathematics
§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
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 4
Bruce Mayer, PE Chabot College Mathematics
Logarithm → What is it?
Concept: If b > 0 and b ≠ 1, then
y = logbx is equivalent to x = by
Symbolically
x = by y = logbx
The exponent is the logarithm.
The base is the base of the logarithm.
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 5
Bruce Mayer, PE Chabot College Mathematics
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?
f −1(x) ≡ the exponent to which we must raise 3 to get x.
y = 3x x = 3y
y ≡ the exponent to which we must raise 3 to get x.
Need
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 6
Bruce Mayer, PE Chabot College Mathematics
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
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 7
Bruce Mayer, PE Chabot College Mathematics
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
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 8
Bruce Mayer, PE Chabot College Mathematics
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
xa xa log
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 9
Bruce Mayer, PE Chabot College Mathematics
Example Exponential to Log
Write each exponential equation in logarithmic form.
a. 43 64 b. 1
2
4
1
16c. a 2 7
Soln a. 43 64 log4 64 3
b. 1
2
4
1
16 log1 2
1
164
c. a 2 7 loga 7 2
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 10
Bruce Mayer, PE Chabot College Mathematics
Example Log to Exponential
Write each logarithmic equation in exponential form
a. log3 243 5 b. log2 5 x c. loga N x
Soln a. log3 243 5 243 35
b. log2 5 x 5 2x
c. loga N x N ax
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 11
Bruce Mayer, PE Chabot College Mathematics
Example Evaluate Logarithms
Find the value of each of the following logarithmsa. log5 25 b. log2 16 c. log1 3 9
d. log7 7 e. log6 1 f. log4
1
2 Solution
a. log5 25 y 25 5y or 52 5y y 2
b. log2 16 y 16 2y or 24 2y y 4
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 12
Bruce Mayer, PE Chabot College Mathematics
Example Evaluate Logarithms
Solution (cont.)
d. log7 7 y 7 7y or 71 7y y 1
e. log6 1 y 1 6y or 60 6y y 0
f. log4
1
2y
1
24 y or 2 1 22 y y
1
2
c. log1 3 9 y 9 1
3
y
or 32 3 y y 2
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 13
Bruce Mayer, PE Chabot College Mathematics
Example Use Log Definition
Solve each equation for x, y or z
a. log5 x 3 b. log3
1
27y
c. logz 1000 3 d. log2 x2 6x 10 1
a. log5 x 3
x 5 3
x 1
53 1
125
The solution set is 1
125
.
Solution
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 14
Bruce Mayer, PE Chabot College Mathematics
Example Use Log Definition
Solution (cont.)
b. log3
1
27y
1
273y
3 3 3y
3 y
c. logz 1000 3
1000 z3
103 z3
10 z
The solution set is 3 .
The solution set is 10 .
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 15
Bruce Mayer, PE Chabot College Mathematics
Inverse Property of Logarithms
Recall Def: For x > 0, a > 0, and a ≠ 1,
y loga x if and only if x ay .
In other words, The logarithmic function is the inverse function of the exponential function; e.g.
xaxa xxa
a loglog
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 16
Bruce Mayer, PE Chabot College Mathematics
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
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 17
Bruce Mayer, PE Chabot College Mathematics
Example Inverse Property
Evaluate: 5log 235 . Solution
Remember that log523 is the exponent to which 5 is raised to get 23. Raising 5 to that exponent, obtain
5log 235 23.
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 18
Bruce Mayer, PE Chabot College Mathematics
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)
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 19
Bruce Mayer, PE Chabot College Mathematics
Graph Logarithmic Function
Sketch the graph of y = log3x
Soln:MakeT-Table→
x y = log3x (x, y)
3–3 = 1/27 –3 (1/27, –3)
3–2 = 1/9 –2 (1/9, –2)
3–3 = 1/3 –1 (1/3, –1)
30 = 1 0 (1, 0)
31 = 3 1 (3, 1)
32 = 9 2 (9, 2)
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 20
Bruce Mayer, PE Chabot College Mathematics
Graph Logarithmic Function
Plot the ordered pairs and connect the dots with a smooth curve to obtain the graph of y = log3x
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 21
Bruce Mayer, PE Chabot College Mathematics
Example Graph by Inverse Graph y = f(x) = 3x
Solution:Use Inverse Relationfor Logs & Exponentials
Reflect the graph of y = 3x across the line y = x to obtain the graph of y = f−1(x) = log3x
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 22
Bruce Mayer, PE Chabot College Mathematics
Properties of Exponential and Logarithmic Functions
Exponential Function f (x) = ax
Logarithmic Function f (x) = loga x
Domain (0, ∞) Range (–∞, ∞)
Domain (–∞, ∞) Range (0, ∞)
x-intercept is 1 No y-intercept
y-intercept is 1 No x-intercept
x-axis (y = 0) is the horizontal asymptote
y-axis (x = 0) is the vertical asymptote
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 23
Bruce Mayer, PE Chabot College Mathematics
Properties of Exponential and Logarithmic Functions
Exponential Function f (x) = ax
Logarithmic Function f (x) = loga x
Is one-to-one, that is, logau = logav if and only if u = v
Is one-to-one , that is, au = av if and only if u = v
Increasing if a > 1 Decreasing if 0 < a < 1
Increasing if a > 1 Decreasing if 0 < a < 1
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 24
Bruce Mayer, PE Chabot College Mathematics
Graphs of Logarithmic Fcns
f (x) = loga x (0 < a < 1)f (x) = loga x (a > 1)
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 25
Bruce Mayer, PE Chabot College Mathematics
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 logs1. log(10) = 1
2. log(1) = 0
3. log(10x) = x
4. 10logx = x
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 26
Bruce Mayer, PE Chabot College Mathematics
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 common logarithms is marked
LOG
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 27
Bruce Mayer, PE Chabot College Mathematics
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
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 28
Bruce Mayer, PE Chabot College Mathematics
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 logs1. ln(e) = 1
2. ln(1) = 0
3. ln(ex) = x
4. elnx = x
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 29
Bruce Mayer, PE Chabot College Mathematics
Example Sound Intensity
This function is sometimes used to calculate sound intensity
010log
Id
I
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).
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 30
Bruce Mayer, PE Chabot College Mathematics
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 0
10log ,I
dI
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 31
Bruce Mayer, PE Chabot College Mathematics
Example Sound Intensity
Solution (cont.) in the dBA eqn substitute 75 for d and 10−12 for I0 and then solve for I
1275 10 log
10
I
127.5 log
10
I
7.5
1210
10
I
12 12 7.512
10 10 1010
I
4.510 I
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 32
Bruce Mayer, PE Chabot College Mathematics
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
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 33
Bruce Mayer, PE Chabot College Mathematics
Example Sound Intensity
CheckIf the sound intensity is 10−4.5 W/m2 , verify that the decibel reading is 75.
4.5
1210
10log10
d
7.510log10d
10 7.5d
75d
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 34
Bruce Mayer, PE Chabot College Mathematics
Summary of Log Rules
For any positive numbers M, N, and a with a ≠ 1, and whole number p
log log log ;a a aM
M NN
log log ;pa aM p M
log .ka a k
log ( ) log log ;a a aMN M N Product Rule
Power Rule
Quotient Rule
Base-to-Power Rule
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 35
Bruce Mayer, PE Chabot College Mathematics
Typical Log-Confusion
Beware that Logs do NOT behave Algebraically. In General:
loglog ,
loga
aa
MM
N N
log ( ) (log )(log ),a a aMN M N
log ( ) log log ,a a aM N M N
log ( ) log log .a a aM N M N
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 36
Bruce Mayer, PE Chabot College Mathematics
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:
logb x loga x
loga b
log x
logb
ln x
lnb
(base a) (base 10) (base e)
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 37
Bruce Mayer, PE Chabot College Mathematics
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
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 38
Bruce Mayer, PE Chabot College Mathematics
Example Evaluate Logs
Compute log513 by changing to (a) common logarithms (b) natural logarithms
Soln
b. log5 13 ln13
ln 51.59369
a. log5 13 log13
log 5
1.59369
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 39
Bruce Mayer, PE Chabot College Mathematics
Use the change-of-base formula to calculate log512. • Round the answer to four decimal places
Solution
Example Evaluate Logs
5
log12log 12
log5
1.5440
1.54405 12.0009 12 Check
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 40
Bruce Mayer, PE Chabot College Mathematics
Find log37 using the change-of-base formula
Solution
Example Evaluate Logs
Substituting into log
log .log
ab
a
MM
b
0.84509804
0.47712125
1.7712
103
10
log 7log 7
log 3
000.73 7712.1
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 41
Bruce Mayer, PE Chabot College Mathematics
Example Use The Rules
Express as an equivalent expression using individual logarithms of x, y, & z
Solna)
334 7
a) log b) logbx xy
yz z
= log4x3 – log4 yz
= 3log4x – log4 yz
= 3log4x – (log4 y + log4z)
= 3log4x – log4 y – log4z
3
4a) log x
yz
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 42
Bruce Mayer, PE Chabot College Mathematics
Example Use The Rules
Solnb)
1/ 33
7 7 b) log logb b
xy xy
z z
71
log3 b
xy
z
71log log
3 b bxy z
1log log 7log
3 b b bx y z
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 43
Bruce Mayer, PE Chabot College Mathematics
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
1log log 7log
3 b b bx y z
71
log3 b
xy
z
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 44
Bruce Mayer, PE Chabot College Mathematics
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?
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 45
Bruce Mayer, PE Chabot College Mathematics
Example Cesium-137 ½-Life
SOLUTION: Start with the math model
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: Divide both sides by A0:
A t A0ekt
2.20095.0 keAA
ke 2.295.0
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 46
Bruce Mayer, PE Chabot College Mathematics
Example Cesium-137 ½-Life
Now take the ln of both Sides
Using the Base-to-Power Rule
Find by Algebra
Now set the amount, A, to ½ of A0
kk ee 2.22.2 ln95.0ln95.0ln
kee ke
k 2.295.0lnlogln95.0ln 2.22.2
HLtHL eA
AtA 0233.0
00
2
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 47
Bruce Mayer, PE Chabot College Mathematics
Example Cesium-137 ½-Life
After dividing both sides by A0
Taking the ln of Both Sides
Solving for the HalfLife
State: The HalfLife of Cesion-137 is approximately 29.7 years
HLte 0233.05.0
HLt te HL 0233.06931.0ln5.0ln 0233.0
73.290233.0
6931.0
HLt
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 48
Bruce Mayer, PE Chabot College Mathematics
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:
1ln
At
r P
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 49
Bruce Mayer, PE Chabot College Mathematics
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?
FamiliarizeIn the Compounding Eqn replace P with 25,000, r with 0.08, A with $100,000, and then simplify.
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 50
Bruce Mayer, PE Chabot College Mathematics
Example Compound Interest
Solution
17.33t
Substitute.
Divide.
Approximate using a calculator.
1 100,000ln
0.08 25,000t
1ln 4
0.08t
State AnswerThe account balance will reach $100,000 in about 17.33 years.
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 51
Bruce Mayer, PE Chabot College Mathematics
Example Compound Interest
Check: 1
17.33 ln0.08 25,000
A
1.3864 ln25,000
A
1.3864 ln ln 25,000A 1.3864 ln 25,000 ln A
11.513 ln A11.513e A
100,007.5 A
Because 17.33 was not the exact time, $100,007.45 is reasonable for the Chk
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 52
Bruce Mayer, PE Chabot College Mathematics
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
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 53
Bruce Mayer, PE Chabot College Mathematics
All Done for Today
Napier’sMasterWorkYear 1619
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 54
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot Mathematics
Appendix
–
srsrsr 22
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 55
Bruce Mayer, PE Chabot College Mathematics
ConCavity Sign Chart
a b c
−−−−−−++++++ −−−−−− ++++++
x
ConCavityForm
d2f/dx2 Sign
Critical (Break)Points Inflection NO
InflectionInflection
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 56
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 57
Bruce Mayer, PE Chabot College Mathematics
BMayer@ChabotCollege.edu • MTH15_Lec-19_sec_4-2_Logarithmic_Fcns.pptx 58
Bruce Mayer, PE Chabot College Mathematics