Transcript Exponential Logarithmic Functions and

Exponential
and
Logarithmic
Functions
By: Hendrik Pical to Revition
Exponential and Logarithmic Functions
By Hendry P 2011
Last Updated: January 30, 2011
With your Graphing Calculator
graph each of the following
y = 2x
y = 3x
y = 5x
y = 1x
Determine what is happening when the
base is changing in each of these graphs.
By Hendry P 2011
y = 3x
y = 2x
x
y = 2x y = 3x
-2
1/4
1/9
-1
½
1/3
0
1
1
1
2
3
2
4
9
3
8
27
By Hendry P 2011
y=
5x
y = 3x
y = 2x
x
y = 5x y = 1x
-2
1/25
1
-1
1/5
1
0
1
1
1
5
1
2
25
1
3
125
1
By Hendry P 2011
y = 1x
5x
y=
y = 10x
Determine
where each of
the following
would lie?
y=10x
y=4x
y = (3/2)x
By Hendry P 2011
y = 3x
y = 4x
y = 2x
y = (3/2)x
y = 1x
By Hendry P 2011
f(x) =
By Hendry P 2011
x
2
f(x) =
By Hendry P 2011
x-3
2
f(x) = 2x+2 3
By Hendry P 2011
f(x) =
By Hendry P 2011
x-4
-(2)
–2
Compound Interest
You deposit $5000 into an account that pays 4.5 %
interest. What is the balance of the account after
10 years if the interest is compounded quarterly?
A  P1 

r nt
n
A = Final amount = unknown
P = Principal = $5000
r = rate of interest = .045
n = number of times compounded per year = 4
t = number of years compounded = 10
By Hendry P 2011
Compound Interest
You deposit $5000 into an account that pays 4.5 %
interest. What is the balance of the account after
10 years if the interest is compounded quarterly?
A  P1 
A = unknown
P = $5000
r = .045
n=4
t = 10
By Hendry P 2011

r nt
n
A  50001 
A  $ 7821.88

0.045 410
4
Compound Interest
You deposit $5000 into an account that pays 4.5 %
interest. What is the balance of the account after
10 years if the interest is compounded quarterly?
weekly?
r nt
n
5210
0
.
045
A = unknown
52
A  P1 

A  50001 
P = $5000
r = .045
n = 452
t = 10
By Hendry P 2011
A  $ 7840.04

By Hendry P 2011
With your Graphing Calculator
graph each of the following
y = (1/2)x
y = (1/3)x
y = 1x
Determine what is happening when the
base is changing in each of these graphs.
By Hendry P 2011
y = (1/3)x
y = (½)x
x
y = (½)x y = (1/3)x
-2
4
9
-1
2
3
0
1
1
1
½
1/3
2
¼
1/9
3
1/8
1/27
By Hendry P 2011
Jeff Bivin -- LZHS
y = 5x
y = 3x
y = 2x
y = 1x
f(x) =
By Hendry P 2011
Jeff Bivin -- LZHS
-x
2
=
x
(1/2)
f(x) =
By Hendry P 2011
x-3
(½) -
2=
-x+3
(2)
-2
By Hendry P 2011
A new Number
e


1
n!
0
1 1 1 1 1
1
e        
0! 1! 2! 3! 4! 5!
1 1 1 1
1
1
e    

 
1 1 2 6 24 120
We could use a spreadsheet to determine an approximation.
By Hendry P 2011
A new Number
e


0
By Hendry P 2011
1
n!
 2.718
y = 3x
Graph y =
x
y = 2x y = 3x
-2
¼
1/9
-1
½
1/3
0
1
1
1
2
3
2
4
9
3
8
27
By Hendry P 2011
x
e
y = ex
y = 2x
Graph:
y = ex+2
x+2=0
x = -2
By Hendry P 2011
y = ex+2
y = ex
Compound Interest-continuously
You deposit $5000 into an account that pays 4.5 %
interest. What is the balance of the account after 10
years if the interest is compounded continuously?
A  Pe
rt
A = Final amount = unknown
P = Principal = $5000
r = rate of interest = .045
t = number of years compounded = 10
By Hendry P 2011
Compound Interest-continuously
You deposit $5000 into an account that pays 4.5 %
interest. What is the balance of the account after 10
years if the interest is compounded continuously?
A  Pe
0.04510
A  5000e
A  $ 7841.56
rt
A = unknown
P = $5000
r = .045
t = 10
By Hendry P 2011
Bacteria Growth
You have 150 bacteria in a dish. It the constant of
growth is 1.567 when t is measured in hours. How
many bacteria will you have in 7 hours?
y  ne
kt
y = Final amount = unknown
n = initial amount = 150
k = constant of growth = 1.567
t = time = 7
By Hendry P 2011
Bacteria Growth
You have 150 bacteria in a dish. It the constant of
growth is 1.567 when t is measured in hours. How
many bacteria will you have in 7 hours?
y  ne
1.5677
y  150e
y  8,706,977.678
kt
y = unknown
n = 150
k = 1.567
t=7
By Hendry P 2011
8,706,977.678 bacteria
By Hendry P 2011
y = 2x
x = 2y
x
y
x
y
-2
1/4
1/4
-2
-1
½
½
-1
0
1
1
0
1
2
2
1
2
4
4
2
3
8
8
3
By Hendry P 2011
y = 2x
x = 2y
How do we
solve this
exponential
equation
for the variable y?
By Hendry P 2011
LOGARITHMS
exponential
b A
m
logarithmic
log b ( A)  m
b>0
A>0
By Hendry P 2011
exponential
logarithmic
b A
log b ( A)  m
3  9
log 3 (9)  2
m
2
53  125
3
2 

1 5
2
log 2 18    3
1
8
 32
x

2
By Hendry P 2011
log 5 (125)  3
log 1 (32)   5
2
y
log 2 ( x)  y
Evaluate
log 5 (25)  u
5  25
u
5 5
u
2
u2
log 5 (25)  2
By Hendry P 2011
Evaluate
log 3 (81)  u
3  81
u
3 3
u
4
u4
log 3 (81)  4
By Hendry P 2011
Evaluate
log 2 321   u
2 
1
32
2 
1
25
u
u
2 2
u
5
u  5
log 2  321    5
By Hendry P 2011
Evaluate
log 7 (7)  u
7 7
u
7 7
u
1
u 1
log 7 (7)  1
By Hendry P 2011
Evaluate
log 8 (1)  u
8 1
u
8 8
u
0
u0
log 8 (1)  0
By Hendry P 2011
Evaluate
log n (n )  u
5
n n
u
5
u5
log n (n 5 )  5
By Hendry P 2011
y = 2x
x = 2y
y=log2x
x
y
x
y
-2
1/4
1/4
-2
-1
½
½
-1
0
1
1
0
1
2
2
1
2
4
4
2
3
8
8
3
By Hendry P 2011
x = 2y
x
y = log2x
1/4
-2
y = log2x
½
-1
y = log3x
y = log5x
1
0
2
1
4
2
8
3
By Hendry P 2011
x = (½)y
x
y = log½x
1/4
2
½
1
1
0
2
-1
4
-2
8
-3
By Hendry P 2011
y = log½x
Solve for x
log2(x+5) = 4
24 = x + 5
16 = x + 5
11 = x
By Hendry P 2011
Solve for x
logx(32) = 5
x5 = 32
x5 = 25
x = 2
By Hendry P 2011
Evaluate
log3(25) = u
3u = 25
3u = 52
??????
By Hendry P 2011
By Hendry P 2011
Change of Base Formula
log b x
log a x 
log b a
log 10 x
log a x 
log 10 a
By Hendry P 2011
Evaluate
log3(25)
log 10 25

log 10 3
= 2.930
By Hendry P 2011
Evaluate
log 10 568
log5(568)  log 5
10
= 3.941
By Hendry P 2011
Properties of Logarithms
•
•
•
•
Product Property
Quotient Property
Power Property
Property of Equality
By Hendry P 2011
Product Property
a a  a
m
n
multiplication
m n
addition
log b (m  n)  log b (m)  log b (n)
multiplication
By Hendry P 2011
addition
Product Property
log 2 (16  4)  log 2 (16)  log 2 (4)
log 2 (2  2 )  log 2 (2 )  log 2 (2 )
4
2
4
log 2 (2 ) 
6
6
By Hendry P 2011

4
2

6
2
Quotient Property
m
division
a
mn

a
n
a
subtraction
log ( )  log b (m)  log b (n)
m
b n
division
By Hendry P 2011
subtraction
Quotient Property
32

log 2 4   log 2 (32)
 log 2 (4)
log 2 8  log 2 (2 )  log 2 (2 )
5
log 2 (2 ) 
3
3
By Hendry P 2011

5
2

3
2
Power Property
a 
m n
a
mn
p logb(m p )
logb(mp ) = p•logb(m)
By Hendry P 2011
Power Property
   7 log
log 2 2
By Hendry P 2011
7
2
7 
7 1
7
7
(2)
Property of Equality
if
log b ( A)  log b (C )
then
By Hendry P 2011
AC
Expand
3
2
log 5 ( x y )
product property
log 5 ( x )  log 5 ( y )
power property
3 log 5 ( x)  2 log 5 ( y)
By Hendry P 2011
3
2
Expand
log 5
quotient property
product property
power property
By Hendry P 2011
 
x5 y 3
z4
log 5 ( x y )  log 5 ( z )
5
3
4
log 5 ( x )  log 5 ( y )  log 5 ( z )
5
3
4
5 log 5 ( x)  3 log 5 ( y)  4 log 5 ( z )
Expand
log
quotient property
 
x7
5 y2 z5
log 5 ( x )  log 5 ( y z )
7

2 5
product property
log 5 ( x )  log 5 ( y )  log 5 ( z )
distributive property
log 5 ( x )  log 5 ( y )  log 5 ( z )
power property
By Hendry P 2011
7
7
2
2
5

5
7 log 5 ( x)  2 log 5 ( y)  5 log 5 ( z )
Condense
5 log 3 x  6 log 3 y  2 log 3 z
power property
product property
quotient property
By Hendry P 2011
log 3 x  log 3 y  log 3 z
5
6

5
log 3 x y
6
log 3
  log
 
x5 y 6
z2
3
z
2
2
Condense
1
2
log 10 x  2 log 10 y  4 log 10 z
Power property
group / factor
product property
quotient property
By Hendry P 2011
1
2
log 10 x  log 10 y  log 10 z
1
2
2

log 10 x  log 10 y  log 10 z
1
2
2

log 10 x  log 10 y z
log 10  y 2 z 4 


1
x2
2 4
 log
4
4

 
x
10 y 2 z 4

Condense
3 log e x  2 log e y  5 log e z  4 log e w
Power property
log e x  log e y  log e z  log e w
re-organize
log e x  log e z  log e y  log e w
group
3
3
log
product property
quotient property
By Hendry P 2011
2
5
log e
2
  log y  log
x z   log y w 
x  log e z
3
e
5
log
5
2
e
3 5
 
x3 z 5
e y 2 w4
2
e
4
e
w
4
4
4

By Hendry P 2011
Solve for x
log 3 3x  9  log 3 x  3
Property of
Equality
3x  9  x  3
2 x  12
x  6
By Hendry P 2011
Solve for x
log 3 3x  9  log 3 x  3
check
x  6
log 3 3(6)  9  log 3 6  3
log 3 18  9  log 3 6  3
log 3 9  log 3 9
checks!
By Hendry P 2011
Solve for x
log 3 3x  9  log 3 x  3
3x  9  x  3
2 x  12
6 
By Hendry P 2011
x  6
Solve for n
log 4 7  log 4 n  2  log 4 6n
Condense
left side
log 4 7(n  2)  log 4 6n
Property of
Equality
7 n  14  6n
n  14
By Hendry P 2011
Solve for n
log 4 7  log 4 n  2  log 4 6n
check
n  14
log 4 7  log 4 14  2  log 4 6(14)
log 4 7  log 4 12  log 4 84
log 4 7(12)  log 4 84
log 4 84  log 4 84
checks!
By Hendry P 2011
Solve for n
log 4 7  log 4 n  2  log 4 6n
log 4 7(n  2)  log 4 6n
7 n  14  6n
14 
By Hendry P 2011
n  14
Solve for x
log 2 x  1  log 2 x 1  3
Condense
left side
Convert to
exponential
form
log 2 ( x  1)( x 1)  3
2  ( x  1)( x  1)
3
8  x 1
2
9  x
3  x
By Hendry P 2011
2
Solve for x
log 2 x  1  log 2 x 1  3
check x  3
log 2 3  1  log 2 3 1  3
log 2 4  log 2 2  3
2 1  3
3  3
checks!
By Hendry P 2011
check x   3
log 2  3  1  log 2  3 1  3
log 2  2  log 2  4  3
fails
The argument
must be positive
Solve for x
log 2 x  1  log 2 x 1  3
log 2 ( x  1)( x 1)  3
2  ( x  1)( x  1)
3
8  x 1
2
9  x
3  x
By Hendry P 2011
2
3 
Solve for x


log 3 x  x  3
Convert to
exponential
form
2
 2
3  x x3
2
9  x x3
2
2
0  x x6
0  ( x  3)( x  2)
x   3 or x  2
2
By Hendry P 2011
Solve for x


log 3 x  x  3
2
check x   3


log 3 (3) 2  (3)  3
check x  2
 2
log 3 9  3  3  2
log 3 9  2
2  2
checks!
By Hendry P 2011
 2


log 3 (2) 2  2  3
 2
log 3 4  2  3  2
log 3 9  2
2  2
checks!
Solve for x


log 3 x  x  3
2
 2
3  x x3
2
9  x x3
2
2
0  x x6
0  ( x  3)( x  2)
x   3 or x  2
2
  3, 2 
By Hendry P 2011
Solve for x
3 x 7
log  5   log  19 
(3x  7)  log( 5)  log( 19)
3x log( 5)  7 log( 5)  log( 19)
3x log( 5)  log( 19)  7 log( 5)
3  log( 5)
3  log( 5)
log(19)  7 log(5) 
x
3log(5) 
By Hendry P 2011
x  2.943
Solve for x
4x
2
log  5   log  7

(4 x)  log( 5)  (2)  log( 7)
4  log( 5)
4  log( 5)
x
 2 log(7 ) 
 4 log(5 ) 
x  0.605
By Hendry P 2011
Solve for x
2 x 1
5x
log 11   log  9

(2 x  1)  log( 11)  (5 x)  log( 9)
2 x log( 11)  1 log( 11)  5 x log( 9)
2 x log( 11)  5 x log( 9)   log( 11)
x2 log( 11)  5 log( 9)   log( 11)
x
By Hendry P 2011
 log(11)
2 log(11)  5 log(9) 
x  0.387
Solve for x
x2
x 1
log  3   log  5 
( x  2)  log( 3)  ( x  1)  log( 5)
x log( 3)  2 log( 3)  x log( 5)  1log( 5)
x log( 3)  x log( 5)  log( 5)  2 log( 3)
xlog( 3)  log( 5)  log( 5)  2 log( 3)
x
By Hendry P 2011
log(5 )  2 log(3) 
log(3)  log(5 ) 
x  1.151
Solve for x
3 x2
9 x 8
  log  3
log  7

(3x  2)  log( 7)  (9 x  8)  log( 3)
3x log( 7)  2 log( 7)  9 x log( 3)  8 log( 3)
3x log( 7)  9 x log( 3)  2 log( 7)  8 log( 3)
x3 log( 7)  9 log( 3)  2 log( 7)  8 log( 3)
 2 log(7 )  8 log(3) 
3 log(7 )  9 log(3) 
By Hendry P 2011
x
x  1.209
Solve for x
3 x 2
ln  15   ln  e

ln( 15)  (3x  2)  ln(
1 e)
ln( 15)  3x  2
ln( 15)  2  3x
ln(15)  2
3
 x
1.569  x
By Hendry P 2011
Solve for x
5 x 6
7  3e
ln 
7
3
  ln  e
5 x6

1 e)
ln( 73 )  (5 x  6)  ln(
ln( 73 )  5 x  6
ln( 73 )  6  5 x
ln( 73 )  6
5
By Hendry P 2011
 x
 1.031  x
Solve for x
3 x 1
5  7 2
log 
5
7
  log  2
3 x 1

log( 75 )  (3x  1)  log( 2)
log( 75 )  3x log( 2)  1 log( 2)
log( 75 )  log( 2)  3x log( 2)
log(75 )  log(2 ) 
3 log(2 ) 
By Hendry P 2011
 x
0.172  x