Download Lecture13_Exponentials and Logs

Lecture 13:
Exponentials and Logs
Laws of Exponents
a a a
Notation:
2
a a a a
3
a a ... a a
n times
a
2
a
3
=
aa aaa
=
Fundamental Rule
( m n )
m n
a a  a
a
5
n
Corollary rule
n
(m n)
m
m m
m
( a )  a a ... a = a
m a’s n times = mn a’s
5
3
3*5
( 7 )  7
Conventions consistent with law of
exponents
1
a  a
1 n
n+1
a a  a
0
a  1
0 n
n+0
a a  a
So
( -1 ) 1
a

a
0
a  1
( -1 ) 1
0
a
a  a
So
( -1 ) 1
a

a
Fractional Exponents
Definition:
a
1
 
n
is a number x so that
n
x  a
n
Formula works
Other Notation
  1 
n
   
 
  n  
n
a
  a
a
= a
is a positive number such that
a
1
 
2
 a
2
x  a
Further convention: rational
exponents
x
a
 
b
  1 
   
  b  
  x

a
a
1
 
b
 ( x )
Can cause problems:
- if x is negative and b is even
( -1 )
- if fraction not in lowest terms
Will not cause problems if x > 0
3
 
2
( -1 )
6
 
4
x
What about
e.g.
7
( 2)
( 2)
a
Cannot be written as
with a, b whole numbers
b
However
2  1.414213562
14 141 1414 14142
1, ,
,
,
, etc
10 100 1000 10000
are approximating
2
ai
2  lim
bi
i  
( 2)
7
 lim 7
i  
14 141 1414 14142
1, ,
,
,
, etc
10 100 1000 10000
7, 7
 14 


 10 
,7
 141 


 100 
,7
 1414 


 1000 
,7
 14142 


 10000 
, etc
7., 15.2453, 15.5449, 15.6663, 15.6724
From computer
15.67289086





ai 


bi 
This means that
If a is any positive number then there is a function
1 a
f(x) a
a 1
x
Makes sense for any a>0
f( x y ) f( x ) f( y )
( x y )
x y
a
 a a
f( x y ) f( x )
f( 0 ) 1
y
(x y)
x
a
 ( a )
0
a  1
y
f( x ) a
x
has an inverse
y a
x
Each Horizontal above the x-axis line meets the graph exactly once
x loga( y )
Any x determines a unique y > 0 and any y > 0 determines a unique x.
Fundamental Identities
loga( x )
x
 x
x
loga( a ) x
From these we have the basic calculations
log(1) = 0
loga( x ) loga( y ) loga( x y )
y
loga( x ) y loga( x )
Change of Base
If we know how to calculate logs in one base we can calculate them
in any base.
loga( x )
a
 x
loga( b )
a
 b
b
 loga( x ) 




 log ( b ) 
a


 x
so
So
x
 loga( b ) loga( x ) 






log
(
b
)
a


logb( x )
 x
loga( x )
loga( b )
Laws of Exponents for Logs
By
( x y )
x y
a
 a a
( loga( x ) loga( y ) )
loga( x ) loga( y )
a
a
 a
( loga( x ) loga( y ) )
a
 x y
( loga( x ) loga( y ) )
loga( x y )
a
 a
so
loga( x ) loga( y ) loga( x y )
Simple Interest
$1000 is borrowed for 5 years at 7% per year.
(a)What amount is due after 5 years and, (b) at what rate
is the amount changing at that time?
A( t ) P( 1 r t )
A( t ) 1000 (1+.07*t)
A( 5 ) 1000 (1+.07*5)
(b)
A '( t ) Pr
A '( 5 ) 1000*.07
Compound Interest
$1000 is borrowed for 10 years at 9% per year compounded quarterly.
(a) To what has the principal accumulated after 10 years and
(b) at what rate is the amount due changing after 10 years
r 

A( t ) P  1 
m

(m t )
1



A( t ) 1000  1 .09 
4


A( 10 ) 2435.188965
(4 t )
mt 

 

r

ln( A( t ) ) ln P  1  
m 
 
r 


 m ln 1 
A(t)
m

A ' (t)
A '( 10 ) 4 A( 10 ) ln( 1000*(1+.09/4) )
A '( 10 ) 216.7377493
Continuous Interest
$1000 is borrowed at 9% compounded continuously for 10 years.
a. To what amount has the $1000 accumulated after 10 years
b. At what rate is the total accumulation changing after 10 years
A( t ) P e
(r t )
A( 10 ) 1000 e
.09*10
A(10) 2459.603111
A ' ( t ) r P e
(r t )
A ' ( 10 ) .09 1000 e
.09*10
A ' (10) 221.3642800
Connection With Calculus
If a is a positive number then there is a number A
(which depends only on a) such that
d
dx
a
x
=
Aa
x
There is one and only one number such that the
corresponding “A” is 1. That is the number we call “e”. The
“e” is in honor of Euler.
d
dx
e
x
=
e
x
ln(x) and its derivative
The inverse of f(x) =
e
x
is denoted ln(x) and is called the “natural” logarithm function
We have
ln( x )
e
 x
Take the derivative of both sides
e
ln( x )
ln( x ) ' 1
1
ln( x ) '
e
ln( x )
ln( x ) '
1
x