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.. الطالبات
خديجه , مالك المسعد , نـدى الغـمـيـــز , حصه الرومي
الزهراني
..األستاذة
يــســــــرى
Exponential and logarithmic functions
Modeling with Exponential and Logarithmic Functions
Exponential Growth and Decay Models
The mathematical model for exponential growth or decay is given by
f (t) = A0ekt or A = A0e
kt.
•If k > 0, the function models the amount or size of a growing entity. A0
is the original amount or size of the growing entity at time t = 0. A is the amount at time t, and k is a constant representing the growth rate.
•If k < 0, the function models the amount or size of a decaying entity. A0 is the original amount or size of the decaying entity at time t = 0. A is
the amount at time t, and k is a constant representing the decay rate.
The graph below shows the growth of the Mexico City metropolitan area
from 1970 through 2000. In 1970, the population of Mexico City was 9.4
million. By 1990, it had grown to 20.2 million.
•Find the exponential growth function that models the data.
•By what year will the population reach 40 million?
20
15
10
5
25
30
1970 1980 1990 2000
Po
pula
tio
n (
mil
lio
ns)
Year
Example
a. We use the exponential growth model
A = A0ekt
in which t is the number of years since 1970. This means that 1970
corresponds to t = 0. At that time there were 9.4 million inhabitants, so we
substitute 9.4 for A0 in the growth model.
A = 9.4 ekt
We are given that there were 20.2 million inhabitants in 1990. Because
1990 is 20 years after 1970, when t = 20 the value of A is 20.2. Substituting
these numbers into the growth model will enable us to find k, the growth
rate. We know that k > 0 because the problem involves growth.
A = 9.4 ektUse the growth model with A0 = 9.4.
20.2 = 9.4 ek•20When t = 20, A = 20.2. Substitute these values.
Example cont
We substitute 0.038 for k in the growth model to obtain the exponential
growth function for Mexico City. It is A = 9.4 e0.038t where t is measured in
years since 1970.
Example cont
20.2/ 9.4 = ek•20Isolate the exponential factor by dividing both sides by 9.4.
ln(20.2/ 9.4) = lnek•20Take the natural logarithm on both sides.
0.038 = kDivide both sides by 20 and solve for k.
20.2/ 9.4 = 20kSimplify the right side by using ln ex = x.
b. To find the year in which the population will grow to 40 million, we
substitute 40 in for A in the model from part (a) and solve for t.
Because 38 is the number of years after 1970, the model indicates that the
population of Mexico City will reach 40 million by 2008 (1970 + 38).
A = 9.4 e0.038t This is the model from part (a).
40 = 9.4 e0.038t Substitute 40 for A.
Example cont
ln(40/9.4) = lne0.038t Take the natural logarithm on both sides.
ln(40/9.4)/0.038 =t Solve for t by dividing both sides by 0.038
ln(40/9.4) =0.038t Simplify the right side by using ln ex = x.
40/9.4 = e0.038t Divide both sides by 9.4.
•Use the fact that after 5715 years a given amount of carbon-14 will have
decayed to half the original amount to find the exponential decay model
for carbon-14.
•In 1947, earthenware jars containing what are known as the Dead Sea
Scrolls were found by an Arab Bedouin herdsman. Analysis indicated
that the scroll wrappings contained 76% of their original carbon-14.
Estimate the age of the Dead Sea Scrolls.
Solution
We begin with the exponential decay model A = A0ekt. We know that k < 0
because the problem involves the decay of carbon-14. After 5715 years
(t = 5715), the amount of carbon-14 present, A, is half of the original
amount A0. Thus we can substitute A0/2 for A in the exponential decay
model. This will enable us to find k, the decay rate.
Text Example
Substituting for k in the decay model, the model for carbon-14 is
A = A0e–0.000121t.
k = ln(1/2)/5715=-0.000121Solve for k.
1/2= ekt5715Divide both sides of the equation by A0.
Solution
A0/2= A0ek5715After 5715 years, A = A0/2
ln(1/2) = ln ek5715Take the natural logarithm on both sides.
ln(1/2) = 5715kln ex = x.
Text Example cont
Exponential Functions
Exponential FunctionThe w/ base a for a > 0
xaxf
* For a 1,
- Domain: (–, )
- Range: (0, )
xaxf
for a > 1 xaxf
for 0 < a < 1 xaxf
Natural Exponential Function
with base e
xexf
Compound Interest
-A(t) = amount after t years-P = Principal-r = interest rate-n = # of times interest is compounded per year-t = number of years
nt
nrPtA )1(
Continually Compounded Interest
-A(t) = amount after t years
-P = Principal
-r = interest rate
-t = number of years
rtPetA
Ex 1:
If $350,000 is invested at a rate of 5½% per year, find the amount of the investment at the end of 10 years for
the following compounding methods:
Quarterly -
- Monthly
- Continuously
Exponential Growth
n(t) = population at time t
no = initial size of population
r = rate of growth
t = time
rtoentn
Logarithmic Functions
The Log Function is the inverse of the Exponential Function, so…
f x ax
f1x loga x
If
Then
And… for
Domain: (0, )
Range: (–, )
f x loga x
xayxy
a log
a is positive with a 1
exponential form
logarithmic form
base exponent base exponent
Common Logarithm(base 10)
xx 10loglog
Natural Logarithm(base e)
xx elogln
xeyx y ln
Properties of Logs
1)
2)
3)
4)
01log a
1log aa
xaxa log
xaxa
log
Properties of Natural Logs
1)
2)
3)
4)
01ln 1ln e
xex ln
xe x ln
Exponential and Logarithmic Functions
Exponential Functions
These functions model rapid growth or decay:
- # of users on the Internet
16 million (1995) 957 million (late 2005)
- Compound interest
- Population growth or decline
Exponential Functions
Comparison
- Linear Functions
Rate of change is constant
- Exponential Functions
Change at a constant percent rate.
The Exponential Function
y = abx
b is the base:
- It must be greater than 0
- It cannot equal 1.
x can be any real number
Identify Exponential Functions
* Which of the following are exponential functions?
y = 3x y = x3
y = 2(7)x y = 2(-7)x
yes
yes
no
no
Identify the Base
* Identify the base in each of the following.
y = 3x y = 2(7)x
y = 3axy = 4x - 3
Evaluate Exponential Functions
y = 3x for x = 4
y = 2(7)x for x = 3
y = -2(4x) for x = 3/2
Graph Exponential Functions (b > 1)
* Graph y = 2x for x = -3 to 3
x y
-3
-2
-1
0
1
2
3
1/8
1/4
1/2
1
2
4
8
Graph of y = 2^x
0
1
2
3
4
5
6
7
8
9
-4 -3 -2 -1 0 1 2 3 4
x
y
Graph Exponential Functions (0< b < 1)
* Graph y = (1/2)x for x = -3 to 3
x y
-3
-2
-1
0
1
2
3
8
4
2
1
1/2
1/4
1/8
Graph of y = (1/2)^x
0
1
2
3
4
5
6
7
8
9
-4 -3 -2 -1 0 1 2 3 4
x
y
Summary
y = abx
- x can be any value
- The resulting y value will always be positive.
- The y-intercept is always (0,1)
- When b > 1, as x increases, y increases.
- When 0 < b < 1, as x increases, y decreases.
Practice
* Using Microsoft Excel:
- Graph the function y = 3x for x = -3 to 3
(in 0.5 increments)
- Graph the function y = (1/3)x for x = -3 to 3
(in 0.5 increments)
Practice: Graph using Excel
- Result:Graphing Exponential Functions
0
5
10
15
20
25
30
-3 -2 -1 0 1 2 3
Exponential Functions
Suppose you are a salaried employee, that is, you are paid afixed sum each pay period no matter how many hours youwork. Moreover, suppose your union contract guaranteesyou a 5% cost-of-living raise each year. Then your annual
salary is an increasing function of the number of years youhave been employed, because your annual salary will
increase by some amount each year. However, theamount of the increase is different from year to year,
because as your salary increases, the amount of your 5%raise increases too. This phenomenon is known as
compounding.
Example
Assume your starting salary is $28,000 per year. Let S(t) be your annual salaryafter full years of employment. Therefore, S(0) is interpreted to mean your
initial salary of $28,000. How can we evaluate S(1), your salary after 1 year ofemployment? Since your salary is increasing by 5% each year, this means
S(1) is 5% more than S(0). In other words, S(1) is 105% of S(0). Thus, we canevaluate S(1) as shown here, by changing the percentage 105% to a decimal
number:S(1) = 105% of S(0) = 1.05 × S(0) = 1.05 × 28000S(2) = 105% of S(1) = 1.05 × S(1) = 1.052 × 28000S(3) = 105% of S(2) = 1.05 × S(2) = 1.053 × 28000S(4) = 105% of S(3) = 1.05 × S(3) = 1.054 × 28000S(5) = 105% of S(4) = 1.05 × S(4) = 1.055 × 28000
Example
Graph of Exponential Functions
Graph of Exponential Functions
Exponential functions have symbol rules of the formf (x) = c ⋅bx
b: base or growth factor -- must be positive real number but cannot be 1, i.e. b > 0 and b ≠ 1
c: coefficient greater than 0the domain of f is (−∞, ∞)the range of f is (0, ∞)
Exponential Functions
Natural Exponential Function
f (x) = ex f (x) = e− x
Example
Example
Example
Find the exponential function whosegraph as shown below.
Find the exponential function whosegraph as shown below.
Logarithmic Functions
Logarithmic Functions
Consider the exponential function f shown here withbase b = 2 and initial value c = 1
Suppose we want to find the input number for thatmatches the output values 8 and 15, in other
words, we want to solve the equation
Logarithmic Functions
Let's introduce a new function designed to help usexpress solutions to equations like the two shownhere, which are solved by finding particular inputnumbers for the exponential function f. We give
this new function a special label:
Logarithmic Functions
helps us express inputs for the function f. Thus,for example, we evaluate , because f(3)=
. Likewise, we evaluate
In general,
That is exponential function and logarithmic function are inverseof each other.
Common and Natural Logarithms
• A common logarithm is a logarithm with base 10,log10.
• A natural logarithm is a logarithm with base e, ln.
Properties of Logarithms
Graphs of Logarithmic Functions
Graphs of Logarithmic Functions
Graphs of Logarithmic Functions
Laws of Logarithms
Change of Base
Compound Interest
If P is a principal of an investment with an interest rfor a period of t years, then the amount A of theinvestment is
Modeling with Exponential andLogarithmic Functions
Exponential Growth Model
A population that experiences exponential growthincreases according to the model
where
population at time tinitial size of populationrelative rate of growthtime
Radioactive Decay Model
If m0 is the initial mass of a radioactive substancewith half-life h, then the remaining mass ofradioactive at time t is modeled by
where
Total differential of Q using logs
68
dLL
dKK
dAA
QdQ
dLL
dKK
dAAQ
dQ
LdKdAdQd
LKAQ
LAKQ
1
1
lnlnlnln
lnlnlnln
Exponent example
69
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1
1
K and Lfor Solve
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3)
5 )2
1)
38)-337 pp. 5, (Example
firm a of decisionsInput 11.6(c)
P-rKLαπ
P-wKLαπ
-wL-rKKLPπ
.αKLQ
PQ-wL-rKR-Cπ
αα
K
αα
L
αα
αα
Maximization conditions
71
years 251004
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%10)(let
4
1
2
1
21
2
1
2
2
).(t
r
rt
rt
rt
rt
bottle/38.148$
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bottle/18.12$
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251
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V
e.$V
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eA(t)
eA(t)
k
keA(t)
))((.
rt
.
))(((.
rtt½
72
10.6(c) Timber cutting problemplot of .5(t)-.5ln(2)=r, r=.05, t=48
10.7(b) Rate of growth of a combination of
functions; Example 3 consumption & pop.
vvuuvuvu
vuu
z
z
z
z
rsrsrvu
vr
vu
ur
vrtgurtftf
tfrif
tgtfvu
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d
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d
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dz
dt
dr
vuz
tgvtfuvuz
)('and,)('then,)(
)(')(
))(')('(1
))()((1
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tf
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dt
tfd
ln
Trigonometric, Logarithmic,
and Exponential Functions
In this tutorial, we review trigonometric, logarithmic, and exponential functions with a focus on those
properties which will be useful in future math and science applications.
Trigonometric Functions
Geometrically, there are two ways to describe trigonometric functions:
Trigonometric FunctionsGeometrically, there are two ways to describe trigonometric
functions: Polar Anglex=cos y=sin Measure in radians: =radiusarc length For
example,180=rr= radians Radians=180degrees
Graph Exponential Functions(0< b < 1)
:
Graph y = (1/2)x for x = -3 to 3
x y
-3
-2
-1
0
1
2
3
8
4
2
1
1/2
1/4
1/8
Graph of y = (1/2)^x
0
1
2
3
4
5
6
7
8
9
-4 -3 -2 -1 0 1 2 3 4
x
y
Summaryy = abx
•x can be any value
•The resulting y value will always be
positive.
•The y-intercept is always (0,1)
•When b > 1, as x increases, y
increases.
•When 0 < b < 1, as x increases, y
decreases.
Substituting for k in the decay model, the model for carbon-14 is
A = A0e–0.000121t.
k = ln(1/2)/5715=-0.000121 Solve for k.
1/2= ekt5715 Divide both sides of the equation by A0.
Solution
A0/2= A0ek5715 After 5715 years, A = A0/2
ln(1/2) = ln ek5715 Take the natural logarithm on both sides.
ln(1/2) = 5715k ln ex = x.
Text Example cont
Solution
The Dead Sea Scrolls are approximately 2268 years old plus the number of
years between 1947 and the current year.
A = A0e-0.000121t This is the decay model for carbon-14.
0.76A0 = A0e-0.000121t A = .76A0 since 76% of the initial amount remains.
0.76 = e-0.000121t Divide both sides of the equation by A0.
ln 0.76 = ln e-0.000121t Take the natural logarithm on both sides.
ln 0.76 = -0.000121t ln ex = x.
Text Example cont.
t=ln(0.76)/(-0.000121) Solver for t.
Finally we hope you can
understand the logarithmic
function , and we hope you like
our project =)
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