STUDY GUIDE FOR CHAPTER 5

```STUDY GUIDE FOR CHAPTER 5
In this chapter Student should be able to:
a) Solve exponential equations
b) Graph exponential function
c) Change a function from logarithmic to exponential form and from exponential to
logarithmic form
d) Apply lows of logarithms
e) Find the logarithms using calculator
f) Solve logarithmic equation
g) Solve problems involving compound interest
h) Solve problems involving exponential growth and decay.
5.1. INVERSE FUNCTION
Definition of one-to-one function:

If a  b are elements of a domain of f(x) and if f(a)  f(b) then f(x) is one-to-one
function.

A function in which elements from the domain always lead to different elements from
the range is called on-to-one function.
Examples: Is f(x) one-to-one function?


f(x) = 3x + 2
f ( x)  x 2  4
For every x there is only one y
For x = 2 and for x = - 2 there is only one y = 0
Horizontal Line test for one-to-one function:
Intersects with f(x) in no more then one
point if f(x) is one-to-one function’
Graph shows that parabola is not a one-to-one function.
Inverse Function:
If f(x) and g(x) are inverse function then:
f[g(x)] = x
and
g[f(x)] = x
If f(x) is given how to find g(x) = f -1(x):
Procedure for finding one-to-one function:
1.
2.
3.
4.
Check that f(x) is one-to-one function.
Solve for x [x = f -1(y)].
Exchange x and y [y = f -1(x) = g(x)].
Check that f[g(x)] = x and g[f(x)] = x
The domain of f(x) equals the range of f -1(x) and the range of f(x) equals the domain of f -1(x).
Practice: Find the inverse of:
1. y = f(x) = 4x – 5
y5
4
x5
y
 f 1 ( x )  g ( x )
4
x
a) Solve for x:
b) Exchange x and y:
c) Check that f[g(x)] = x
f [ g ( x)  4 
x5
5 x
4
and
g[ f ( x ) 
4x  5  5
x
4
y= x
y=(x + 5)/4
y=4x - 5
2. y = 2x + 1
3. y  6  x
4. y = - x2 + 2
5. y = 1/x
and
g[f(x)] = x
5.2. EXPONENTIAL FUNCTIONS
Properties of Exponents:



If x = y then ax = ay
If a &gt; 1 and x &lt; y then ax &lt; ay
If 0 &lt; a &lt; 1 and x &lt; y then ax &gt; ay
Graphing exponential function f(x) = ax





a) The points (0,1) and (1,a) are in the graph
b) If a &gt; 1 f(x) is increasing
c) If 0 &lt; a &lt; 1, f(x) is decreasing
d) x - axis is horizontal asymptote
e) Domain (No restriction), Range ( y &gt; 0)
Graph the following functions:

f(x) = 2x
5
y
y =2x
4
3
2
( 1 ,2 )
1
x
0
-5
-4
-3
-2
-1 -1 0
1
2
-2
-3
-4
-5









f(x) = 3 x
f(x) = - 3 x
f(x) = 3x+2
f(x) = 3 x + 2
f(x) = 3x+ 2 + 2
f(x) = 2 x + 1
f(x) = (1/2)x
f(x) = 2 –x
f(x) = - 2 x
3
4
5
Solve exponential equations using the property: if a x = a y then x = y:

(1/3) x = 81
Solution:
1
( ) x  36
3
1
1
( ) x  ( ) 6
3
3
x  6








(3-1) x = 34
(2/3)x = 9/4
23-x = 8
16 x + 2 = 8 1 - x
32t = 16 1 - t
(2/3)k - 1 = (81/16) k + 1
81 = x 4/3
5 2p + 1 = 25
Applications of Exponential Equations
Q 48 on page 368
Q 50 on page 368
Q 52 on page 368
Compound Interest
If P dollars are deposited in an account paying an annual rate of interest r, compounded m times
per year, then after t years the account will contain A dollars, where
A  P(1 
r mt
)
m
A – future value
P – present value
Example 1: How much should be deposited at 6 % interest compounded annually (m = 1) for
three years ( t = 3) in order to have a balance of \$20,000 in three years?
0.06 3
)
1
20,000  P(1.191016)
20,000  P(1 
P  16,792.39
Example 2: If only \$15,000 is deposited , what annual interest rate is required to increase to
\$20,000[10 %]
Example 3: Invest \$1 (P = 1) for one year ( t = 1) at the rate of 100 % (r = 1) and find A
compounded 1, 2, 5, 10, 25, 100, 500, 1000, 10000, 100000, 1000000..... times a year. Future
value should go toward fixed number e = 2.71828
Compounded annually:
1
P  1  (1  )1  2
1
Compounded twice a year:
1
P  1  (1  ) 2  2.25
2
Compounded ten times a year
P  1  (1 
1 10
)  2.5937
10
Compounded one hundred times a year
P  1  (1 
1 100
)  2.7048
100
Compounded one thousand times a year
P  1  (1 
1 1000
)
 2.7169
1000
Compounded one hundred thousand times a year
P  1  (1 
1
)100,000  2.71826
100,000
Compounded one million times a year
P  1  (1 
1
)1,000,000  2.71828
1,000,000
Compounded continuously
P  2.71828  e
Exponential Growth and Decay
y = yo e kt
y – future amount or number
yo – is amount or number at present time t = 0
k - constant
Work on Q. 56 on page 369
5.3. LOGARITMIC FUNCTION
How to find the inverse of a exponential function y = ax – Graph.
1. It’s one-to-one function
2. How to solve for x ?
Introduction of Logarithmic functions
y = ax Solving for x:
f(x) = a x
x = log a y
f –1(x) = log a x
Exchange x and y:
y = log a x
Meaning of a logarithm:
Exponential form
25 = 5x
Logarithmic form:
x = log 5 25
“Five raised to what exponent is 25”
Log function is an Inverse of exponential function
Write eq. in logarithmic or exponential form:
a)
b)
c)
d)
e)
f)
x=ay
(1/2)-4 = 16
51=5
(3/4)0 = 1
9 2.7 = z
0.315=2 –5






y = loga x
-4 = log &frac12; 16
1 = log 5 5
0 = log &frac34; 1
2.7 = log 9 z
-5 = log 2 0.3125
Properties of Logarithms





log a xy = log a x + log a y
log a x/y = log a x - log a y
log a x r = r log a x
log a a = 1
log a 1 = 0
Application of properties

log a (m n q/p2) = ?
=log a ( m n q ) – log a p 2 =
= log a m + log a n + log a q – 2 log a p


log6 3 m2  ?
x y
log a
?
z3
x3 y5
?
zm

logb

log a x + log a y – log a m = ?

2 log m a – 3 log m (b2 ) = ?

(log b k – log b m ) – log b a =
n
 log b
k
k
 log b a  log b
m
ma
Use graphing calculator to graph logarithmic functions:

f(x) = log 10 ( x – 1)

f(x) = log 10 x - 1
5.4 EVALUATING LOGARITHMS
Common logarithm:
log x = log 10 x
Natural number e = 2.73…
Natural logarithm:
log e x = ln x
Use calculator to evaluate ( to the nearest ten-thousands )










log 1247 = ?
log0.0069 = ?
ln0.7 = ?
ln580 = ?
ln650 = ?
log312.5 = ?
log 10 = ?
ln 1 = ?
ln0.049 = ?
ln4.44 = ?
Change of base theorem:
log b x log x ln x


log b a log a ln a

log a x 

log 9 12 = log 12/log 9 = ?

log2.9 7.5 = log 7.5/log 2.9 = ?

log 5 17 =

log 2 0.1 =
5.5 EXPONENTIAL AND LOGARITHMIC EQUATIONS
Property: When x &gt; 0, y &gt; 0, a &gt; 0 ( a  1 ) and if log a x = log a y then x = y
Solve exponential equations:

5 x = 14
ln( 5 x )  ln 14
x ln( 5)  ln( 14)
x
ln 14
 1.64
ln 5

3 4x – 1 = 4 x – 2

3 2x – 5 = 13

e 2 – y = 12

e 2x e 5x = e 14
Solve logarithmic equations:

ln (5 + 4y ) – ln (3 + y) = ln 3
5  4y
 ln 3
3 y
5  4y
3
3 y
5  4 y  3(3  y )
ln
4 y  3y  9  5
y4

ln x – ln (x + 1) = ln 5

ln ( k - 5) + ln ( k+ 2) = ln (14 k)

ln e x – ln e 3 = ln e 5
5.6 EXPONENTIAL GROWTH AND DECAY
In many applications:
 Biology
 Economics
 Social studies
 Physics
a quantity changes at a rate proportional to the amount present
At any time (instant) t, the amount present is:
y  y o e kt
y is future amount
yo is present amount (at t = 0)
k – constant (k &gt; 0 growth, k &lt; 0 decay)
Example: A sample of 500 g of plutonium 341, decays according to the function:
A(t )  Ao e 0.053t
where t is time in years.
A(t) is future amount
Ao id present amount
Find the amount of the sample remaining after:
a) 4 years
b) 8 years
c) 20 years
d) Find the half life
Solution:
a)
b)
c)
t4
A(4)  500  e 0.0534  500  e 0.212  404 g
t 8
A(8)  500  e 0.0538  327 g
t  20
A(20)  500  e 0.05320  173g
The HALF-LIFE of quantity that decays exponentially is the time it takes for half of a given
amount to decay.
500
 250
2
250  500  e 0.053t
A(t ) 
d)
1
 e 0.053t
2
1
ln( )  0.053  t
2
1
ln( )
2  13 years
t
 0.053
Example: By Newton’s law of cooling, the temperature of a body at time t after being introduced
into environment having constant temperature To is:
T (t )  To  C  e kt
where C and k are constants. If C = 100 and k = 0.1, and t is time measured in minutes, how long
will it take a hot cup of coffee to cool to a temperature of 25 oC in a room at 20 oC.
A(t )  25o C
To  20o C
25  20  100  e 0.1t
5  100  e 0.1t
1
 e 0.1t
20
1
ln( )  0.1t
20
t  30 min
Example: How long it take for \$5000 to grow to \$8400 at an interest rate of 6 % if interest is
compounded a) semiannually and b) continuously.
a)
r
A  P (1  ) mt
m
0.06 2t
8400  5000(1 
)
2
84
 1.025  e 2 t
50
84
ln( )  2  t  ln( 1.025)
50
t  8.8 years
b)
A  P  e rt
8400  5000  e 0.06t
84
 e 0.06t
50
84
ln( )  0.06  t
50
t  8.64 years
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