3.1 Exponential Functions
Exponential function
– any function whose equation contains a variable in the exponent.
[measures rapid increase or decrease (Example: epidemic growth)]
f(x) = bx
f – exponential function
-- domain is (-∞ , ∞)
b - constant base (b > 0, b  1)
f(x) = 2x
g(x) = 10x
Graphing Exponential Functions:
x
f (x) = 3x
g(x) = 3x+1
-2
3-2 = 1/9
3-2+1 = 3-1 = 1/3
-1
3-1 = 1/3
3-1+1 = 30 = 1
0
30 = 1
30+1 = 31 = 3
1
31 = 3
31+1 = 32 = 9
2
32
32+1
h(x) = 3x+1
g(x) = 3x+1
f (x) = 3x
x = any real number
Shift up c units
f(x) = bx + c
Shift down c units
f(x) = bx – c
Shift left c units
f(x) = bx+c
(-1, 1)
(0, 1)
=9
=
33
= 27
-5 -4 -3 -2 -1
1 2 3 4 5 6
Shift right c units
f(x) = bx-c
See P.193
All properties above are the same
Except #4 which is ‘decreasing’
3.2 Investment and Interest Examples
Compound Interest
Karen Estes just received an inheritance of $10000
And plans to place all money in a savings account
That pays 5% compounded quarterly to help her son
Go to college in 3 years. How much money will be
In the account in 3 years?
1. How much interest will she get?
A = amount in account after t years
2. How much money will she have
Use the formula:
P = principal or amount invested
at the end of 5 years
A = P(1 + r/n)nt
t = time in years
r = annual rate of interest
Use the formula:
n = number of times compounded
I = Prt
per year
• Annual
=> n = 1
I = (8000)(.06)(5)
• Semiannual
=> n = 2
= $2400 (interest earned in 5 years)
• Quarterly
=> n = 4
• Monthly
=> n = 12
In 5 years she will have a total of
• Daily
=> n = 365
A=P+I
= 8000 + 2400 = $10,400
As n increases (n  ∞), A does NOT increase indefinitely.
It gets closer and closer to a fixed number “e”, and gives us
the continuous compound interest formula: A = Pert
Simple Interest
Juanita deposited $8000
in a bank for 5 years at a
simple interest rate of 6%
Compound Interest related to Simple Interest
If a principal of P dollars is borrowed for a period of
t years at a perannum interest rate r, the interest I charged is:
I = Prt
Annually -> Once per year
Quarterly -> Four times per year
Daily -> 365 times per year
Semiannually -> Twice per year
Monthly -> 12 times per year
Example: A credit union pays interest of 8% per annum compounded quarterly on a
Certain savings plan. If $1000 is deposited in such a plan and the interest is
Left to accumulate, how much is in the account after 1 year?
I = (1000)(.08)(1/4) = $20 => New principal now is: $1020
I = (1020)(.08)(1/4) = $20.40 => New principal now is: $1040.40
I = (1040.40)(.08)(1/4) = $20.81 => New principal now is: $1061.21
I = (1061.21)(.08)(1/4) = $21.22 => New principal now is: $1082.43
Compound Interest Formula: The amount, A, after t years due to a principal P
Invested at an annual interest rate r compounded n times per year is:
A = P  (1 + r/n)nt
The Natural Base e
An irrational number, symbolized by
the letter e, appears as the base in many
applied exponential functions. This
irrational number is approximately
equal to 2.72. More accurately,
The number e is called the natural base.
The function f (x) = ex is called the
natural exponential function.
f (x) = 3x f (x) = ex
Analogous to the continuous
compound interest formula, A = Pert
The model for continuous growth and decay
is: A(t) = A0ekt
A(t) = the amount at time t
A0 = A(0), the initial amount
k
= rate of growth (k>0) or decay (k < 0)
In year 2000, population of the world was
about 6 billion; annual rate of growth 2.1%.
Estimate the population in 2030 and 1990.
4
f (x) = 2x
(1, 3)
2000=year 0, 2030=year 30, year 1990=year -10
3
(1, e)
2
(1, 2)
A(30) = 6e(.021)(30) = 11.265663 (billion people)
A(-10) = 6e(.021)(-10) = 4.8635055 (billion people)
(0, 1)
** Note these are just estimates. The actual
population in 1990 was 5.28 billion
-1
1
Exponential Growth & Decay
(Law of Uninhibited Growth/Decay)
A0 is the original amount of ‘substance’ (drug, cells, bacteria, people, etc…)
Solving Exponential Equations
Solve: 2
3 x -1
25 = 32, so
23x-1 = 25
3x – 1 = 5
3x = 6
X = 2
 32
Solve: e
2 x 1
e2x-1 = e-4x
e3x
e2x-1 = e-7x
2x – 1 = -7x
-1 = -9x
X = 1/9
 
1
 3 x  e x
e
4
3.3 Logarithmic Functions
A logarithm is an exponent such that for b > 0, b  1 and x > 0
y = logb x
if and only if
by = x
Logarithmic equations
1) 2 = log5 x
2) 3 = logb 64
3) log3 7 = y
4) y = loge 9
Corresponding exponential forms
1) 52 = x
2) b3 = 64
3) 3y = 7
4) ey = 9
Evaluate the Question Needed for
Logarithmic Evaluation
Expression
Logarithmic Expression
Evaluated
log2 16
2 to what power is 16?
log2 16 = 4 because 24 = 16.
log3 9
3 to what power is 9?
log3 9 = 2 because 32 = 9.
log25 5
25 to what power is 5?
log25 5 = 1/2 because 251/2 = 5.
Logarithmic Properties
Logb b = 1
Logb 1 = 0
1 is the exponent to which b must be raised to obtain b. (b1 = b).
0 is the exponent to which b must be raised to obtain 1. (b0 = 1).
logb bx = x
b logb x = x
The logarithm with base b of b raised to a power equals that power.
b raised to the logarithm with base b of a number equals that number.
Graphs of f (x) = 2x and g(x) = log2 x [Logarithm is the inverse of the exponential function]
x
-2
-1
0
1
2
3
x
1/4
1/2
1
2
4
8
f (x) = 2x
1/4
1/2
1
2
4
8
g(x) = log2 x
-2
-1
0
1
2
3
Reverse coordinates.
y=x
f (x) =
2x
6
5
4
3
f (x) = log2 x
2
-2
-1
-1
-2
2
3 4
5
6
Properties of f(x) = logb x
•Domain = (0, ∞)
•Range = (-∞, +∞)
•X intercept = 1 ; No y-intercept
•Vertical asymptote on y-axis
•Decreasing on 0<b<1; increasing if b>1
•Contains points: (1, 0), (b, 1), (1/b, -1)
•Graph is smooth and continuous
Common Logs and Natural Logs
A logarithm with a base of 10 is a ‘common log’
3 because 103 = 1000
log10 1000 = ______
If a log is written with no base it is assumed to be 10.
log 1000 = log10 1000 = 3
A logarithm with a base of e is a ‘natural log’
0
loge 1 = ______
because e0 = 1
If a log is written as ‘ln’ instead of ‘log’ it is a natural log
ln 1 = loge 1 = 0
Doubling Your Investments
How long will it take to double your
Money if it earns 6.5% compounded
continuously?
At what rate of return
compounded continuously
would your money double in
5 years?
Recall the continuous compound
interest formula: A = Pert
A = Pert
If P is the principal and we want P to
double, the amount A will be 2P.
2P = Per(5)
2 = er(5)
ln 2 = ln e5r
2P = Pe(.065)t
2 = e(.065)t
Ln 2 = ln e(.065)t
Ln 2 = .065t
.6931 = 5r
T = ln 2 = .6931 ≈ 10.66 ≈ 11 years
.065 .065
So, annual interest rate
needed is: 13.86%
r = .6931/5
R = .1386
Newton’s Law of Cooling
T = Ts + (T0 – Ts)e-kt
T = Temperature of object a time t
Ts = surrounding temperature
T0 = T at t = 0
A McDonald’s franchise discovered that when coffee is poured from a CoffeeMaker
whose contents are 180° F into a noninsulated pot, after 1 minute, the Coffee cools to
165° F if the room temperature is 72° F. How long should employees wait before
pouring coffee from this noninsulated pot into cups to deliver it to customers at 125°F?
T0 = 180 and Ts = 72
T = 72 + (180 – 72) e-kt
T = 72 + 108e-kt
Since T = 165 when t = 1
165 = 72 + 108e-k(1)
93 = 108e-k
93 = 108e-k
Ln .8611 = ln e-k
-.1495 = -k
K = .1495
T = 72 + 108e-.1495t
125 = 72 + 108e-.1495t
53 = 108e-.1495t
.4907 = e-.1495t
ln .4907 = ln e-.1495t
-.7119 = -.1495t
t = 4.76
Employees should wait abou
to deliver the coffee at the de
temperature.
Now,
3.4 Properties & Rules of Logarithms
Basic Properties
Logb b = 1
1 is the exponent to which b must be raised to obtain b. (b1 = b).
Logb 1 = 0
0 is the exponent to which b must be raised to obtain 1. (b0 = 1).
Inverse Properties
logb bx = x
The logarithm with base b of b raised to a power equals that power.
b logb x = x
b raised to the logarithm with base b of a number equals that number.
For M>0 and N > 0
Product Rule
logb(MN) = logb M + logb N
Quotient Rule
logb M = logb M - logb N
N
Power Rule
p
logb M = p logb M
Logarithmic Property Practice
Quotient Rule
Product Rule
logb(MN) = logb M + logb N
1) log3 (27 • 81) =
logb M = logb M - logb N
N
1) log8 23 =
x
2) log (100x) =
2) Ln
3) Ln (7x) =
Power Rule
p
logb M = p logb M
1) log5 74
2) Log
=
(4x)5 =
3) Ln x2 =
4) Ln x =
e5
11
=
Expanding Logarithms
logb(MN) = logb M + logb N
logb M = logb M - logb N
N
p
logb M = p logb M
1) Logb (x2 y )
2)
x
36y4
log6 3
3) log5
x
25y3
4) log2 5x2
3
Condensing Logarithms
logb(MN) = logb M + logb N
logb M = logb M - logb N
N
p
logb M = p logb M
Note: Logarithm coefficients
Must be 1 to condense.
(Use power rule 1st)
1) log4 2 + log4 32
4) 2 ln x + ln (x + 1)
2) Log 25 + log 4
5) 2 log (x – 3) – log x
3) Log (7x + 6) – log x
6) ¼ logb x – 2 logb 5 – 10 logb y
The Change-of-Base Property
log a M
log b M =
log a b
Example:
Evaluate log3 7
Most calculators only use:
• Common Log [LOG] (base 10)
• Natural Log [LN] (base e)
It is necessary to use the change
Of base property to convert to
A base the calculator can use.
log10 7
log 3 7 =
log10 3
ln 7
log 3 7 =
ln 3
log3 7 = 1.77
Matching Data to an Exponential Curve
Find the exponential function of the form f(x) = aebx that passes through the
points (0,2) and (3,8)
F(0) = 2
2 = aeb(0)
2 = ae0
2 = a(1)
a=2
f(3) = 8
8 = aeb(3)
8 = 2e3b
4 = e3b
ln 4 = ln e3b
ln 4 = 3b
b = (ln4)/3
f(x) = 2e((ln4)/3)x
3.5 Exponential & Logarithmic Equations
Step 1: Isolate the exponential expression.
Step 2: Take the natural logarithm on both sides of the equation.
Step 3: Simplify using one of the following properties:
ln bx = x ln b
or
ln ex = x.
Step 4: Solve for the variable using proper algebraic rules.
Examples (Solve for x):
log4(x + 3) = 2.
log 2 (3x-1) = 18
54x – 7 – 3 = 10
3x+2-7 = 27
More Equations to Try
Solve: log3 4  2log3 x
Solve: log2  x  2  log2 1  x   1
Solve: 3x  7
Solve: 5  2 x  3
Solve: 2
x 1
2 x 3
5
Solve: 9  3  6  0
x
x