Download unit-3-exponential-logarithmic-functions_2017

Prof. M. Alonso
f (x )  log2 x
f (x )  e
x
 Define
exponential and logarithmic functions
 Graph both functions
 Solve exponential and logarithmic equations
 Apply the properties of logarithms
 Solve problems
 Definition
An exponential function is
f(x) = Abx
where b > 0 and b  1. A is a real number
The domain of this function are the real
numbers
Note the
variable is an
exponent
.
F(x) = 2x
H(x) = 102x+3
G(x) = (¼)x-1
Q(x) = (½)2x
P(x) = (3) –x
M(x) = (¾)4x
A
common base used in the sciences is the
base e also called natural base. e is a symbol
such as π which represents the number e =
2.718281828. . .
 Thus we have the exponential function
f (x )  e
x
find the value of e , use the ex button
(generally you’ll need to hit the 2nd function or
blue button first to get it depending on the
calculator). After hitting the ex, you then enter
the exponent 1 and push = or enter.
 If you have a scientific calculator that doesn’t
graph you may have to enter the 1 before
hitting the ex button . You should get
2.718281828
 To
Let us plot some points.
X
Y
-3 0.125
-2 0.25
-1 0.5
0
1
1
2
2
4
3
8
9
8
7
6
5
4
3
2
1
0
-1 0
Remember!
f(-3)= 2-3=.125
A negative exponent
means:
-4
-2
1 1
f  3  2  3 
2
8
3
2
4
 The
domain is the real numbers
 The range is (0, ∞ )
 Y - Intercept
f(0) = 20 = 1 , (0, 1)
 X - Intercepto 2x = 0
There are no x
intercepts because there is no x value that you can
put in the function to make it = 0.
 The graph is always increasing
9
8
7
6
5
4
3
2
1
-4
-2
0
-1 0
2
4
x
-3
-2
-1
0
1
2
3
1
2
 
y
8
4
2
1
.5
.25
.125
x
10
8
6
4
2
0
-4
-2
-2
0
2
4
1
f ( x)   
2
 Domain-
real numbers
 Range – (0, ∞)
 Y- intercept – (1, 0)
X
– intercept - There are no x intercepts
 The
graph is decreasing
x
The graphs can be increasing or decreasing.
By looking at the function we know if the
graph is increasing or decreasing.
 If b > 1 then the graph of f(x) = bx is
increasing..
 If 0 < b < 1, then the graph of f(x) = bx ss
decreasing.

 An
exponential equation is one in which a
variable occurs in the exponent:
3x = 9
52x+1 = 125x
6x = 7
42x-2 = 8
5 = 25x-2

Theorem: If bx = by then x = y.
This says that if both sides of the equation
have the same base, then the exponents are
equal
 Solution:
Use the theorem
 The left hand side is 3 to the something.
Re-write the right hand side, 9, as 3 to the
power 2
Thus, x = 2
3x = 9
3x = 32
Remember, if
the bases are
equal then the
exponents are
equal.
52x+ 1 = 125x
52x+1 = (53)x
Thus, 2x+1 = 3x
1 = 3x -2x
1 =x
Re-write 125 as
53.
42x-2 = 8
(22)2x-2 = 23
24x-4 = 23
Thus: 4x -4 = 3
4x = 7
x = 7/4
5 = 25
51/2 = (52)
x-2
51/2 = 5 2x-4
Thus:
½ = 2x - 4
½ + 4 = 2x
9/2 = 2x
9/4 = x
x-2
With the procedure used in the previous
cases we can not solve this equation since
there is no way to put the bases equal.
 Later we will use another technique to solve
this equation.


 Exponential
functions
have
many
applications, both in the Natural Sciences
and in Business Administration. We will see
some interesting applications.
 Compound
interest (or compounding interest)
is interest calculated on the initial principal
and also on the accumulated interest of
previous periods of a deposit.
 If
one invests a certain amount of money
P into a savings account that pays an
interest rate r and the interest is
calculated in certain periods, then after t
years the person gets the following
amount A of money:
A = P( 1 + r/n)nt
The letters in the formula A = P( 1 +r/n)nt
represent:
P
the principal, the money that you deposit
r
the annual rate of interest
n
the number of times that interest is
compounded per year ( if monthly, then n = 12; if
quarterly then n = 4.
t
number of years the amount is deposited
A
amount of money accumulated after t years,
A = P( 1 +r/n)nt
The formula is an exponential function
because the variable t is in the exponent.
 That is, the amount of money A that one will
have in the future depends on the time that
the principal is in the account.


If you invest $ 2000 in a 6.5% IRA compounded
monthly, after 5 years how much money does
the person have?
.065 

A = 2000  1+

12


(12)(5)
= 2000 (1 + .00541)60
= 2000 (1.0054166666666)60
= 2765.63
Use a calculator
Definition: The logarithmic function is defined
as:
f(x) = logbx
where b > 0, b  1.
interval (0,  ).
The domain is the
f(x) = log3 (x + 1)
g(x) = log x2
h(x) = log2( x -3 )
p(x) = log½ (3x3 + 2x - 4)
There are two bases that are used a lot when
working with logarithms: base 10 and base e.
When we refer to base 10 we only write log x,
that is, we do not indicate the base because it is
understood that the base is 10.
 When we refer to the base e, called natural
base, we write ln x; That is, ln x means logex.
 Both log x and ln x are keys that appear in any
scientific calculators.

You should check how your calculator
works.
 If we want to find log 100 we just have to
press the log key and then 100. The answer
is 2
 To find ln 3, press the ln button and
then 3 . The answer is ln 3 =
1.098612289.

 Find
this logarithms of:
 Log1000
What is a logarithm?
 Log 100
 Log 10
 Log 1
a quantity
 Log .01
representing the
 Log .001
power to which a
fixed number (the
base) must be raised
to produce a given
number.
Logarithm base 10
value
Note
Log 1000
3
103=1000
Log 100
2
102=100
Log 10
1
101=10
Log 1
0
100=1
Log .01
-2
10-2=.01
Log .001
-3
10-3=.001
It
follows that a logarithm is an
exponent. That is, a logarithm is
the exponent to which the base
must be raised to obtain the
number. Therefore, the
equivalent form is
logb N  e
Logarithm form
b N
e
Exponential
form

y = log2 x is equivalent to 2y = x .
EXPONENT
y = log2 x
2y = x
BASE
NUMBER
x
y
¼
Log2(¼)=-2
½
Log2 (½)=-1
1
Log21 = 0
2
Log22= 1
4
Log24= 2
8
Log28 = 3
Powers of 2
Remember what a
logarithm is!!
Domain is
the interval
( 0,  ).
4
3
2
1
0
-1
-2
0
2
4
6
8
10
F(x) = 2x
g(x) = log2x
 Compare
these two graphs and their value tables,
what do you observe?
F(x) = 2x
g(x) = log2x
 Find
log2 (1/64)
 Solution:
 Write
log2 (1/64) = N
 Change to an equivalent form: 2N = 1/64
 Solve this exponential equation
 N = -6
Thus, log
2
( 1/64) = -6
 2N
= 1/64
 We have an exponential equation
2
N
2
N
 Thus,
1
=
64
-6
=2
N = -6. The answer is:
log2
1
= -6
64
 Find
log381
log3 81= N
3 = 81
N
3 =3
N
4
Thus, N = 4. The answer is log381 = 4
Let A, C and N be real numbers
 1.
logb 1 = 0
 2. logbb = 1
 3. logb (AC) = logb A + logb C
 4. logb (A/C) = logb A - logb C
 5. log b AN = N logb A
1.
2.
3.
Simplify logarithms
Solve exponential equations
Solve logarithmic equations
We use property
#3.
 Simplify
 log3
log3x + log3x2
x + log3 x2 = log3 (x)(x2 ) = log3 x3
If logb2 = .23 and logb5 = .42 Find
1.
2.
3.
4.
Logb10
Logb8
Logb ½
Logb25b
Use logarithm
properties!!!!
Logb10 = Logb(2)(5)
= Logb2 +Logb5
= .23 + .42
= .65
Add
Write 10 as a product
Usa property #3
Substitute
Logb8 = Logb23
= 3Logb2
= 3(.23)
= .69
Write 8 as a power of 2
Use property #5
Substitute
Multiply
Logb(1/2) = Logb1 - Logb2
Use property #4
= 0 - .23 Substitute
= -.23
Add
Logb25b = Logb25 + logbb
= Logb52 + 1
= 2Logb5 + 1
= 2(.42) + 1
= 1.84
Use property #3
Use #5 and #2
Substitute
Add
Solve log
3
(x - 2) + log
3
(x - 4) = 2
log
3
(x - 2) + log
log
log
3
3
(x - 4) = 2
(x - 2)(x - 4) = 2
(x2 –6x + 8) = 2
Use property
# 3
Multiply
(x2 –6x + 8) = 32
Change to
exponential
form
x2
Quadratic
equation
3
–6x + 8 – 9 = 0
x2 –6x –1 = 0
Use quadratic formula to solve
x=
=
=
=
-b ±
b2 - 4ac
2a
-(-6) ±
6±
6±
(-6) 2 - 4(1)(-1)
2(1)
36 + 4
2
40
2
6 ± 2 10
=
2
2(3 ± 10 )
=
2
= 3 ± 10
The only solution is
3
10
 Solve
 Since
2x = 5
5 cannot be written as a power of 2,
we need another procedure to solve the
equation.
 To
solve an exponential equation, take the
log of both sides ( base 10 or base e) and
solve for the variable.

2x = 5
log 2x = log 5
Solve
2x = 5
log 2x = log 5
x log 2 = log 5
xlog2 log5
=
log2
log2
Use property # 5
Solve for x
log5
x=
= 2.3219
log2
Use
the
calculator
 Solve
3x-1 = 7 2x
log3x-1 = log7 2x
(x -1)log3 = 2xlog7 Use property # 5
xlog3-log3 = 2xlog7
xlog3-2xlog7 = log3
x(log3-2log7) = log3
Common Factor x
x(log3-2log7)
log3
=
log3-2log7
log3-2log7
x =-.3933
 Exponential
functions are used to model
carbon date artifacts, population growth,
exponential decay, and compound interest.
 Let’s
look at a general form for population
models. Most of the time, we start with an
equation that looks like
P  P0 e
 P0
is the initial population
 k is the growth rate
 t time
kt
 Suppose
that the population of a certain
country grows at an annual rate of 3%. If the
current population is 5 million, what will the
population be in 10 years?
Solution:
P(t) = 5e0.03(10) = 6.7492… million
 In
the same country as
in the previous
example, how long will
it take the population
to reach 10 million?
 Solution:
 10
= 5e0.03t
isolate the
 10/5 = e0.03t exponential
 2 = e0.03t
 ln 2 = ln e0.03t
apply the natural log
function
ln
2 = ln e0.03t
Ln 2 =.03t lne
Property #5
and #2
Ln2 = .03t
Ln2/.03 = t
23.10 = t , thus 23 years