Download 1206 Sec. 5.7 Notes

Mat h 12 06 Ca lc ul us – Sec . 5. 7: T he Lo gar it hm De fi ne d as a n Inte gral
I.
Revi e w
A. Relationship between Logarithms and Exponents
y = b ! x = logb y
x
B. Properties of Logarithms
For any numbers
b, m, n >0
1. log b ( mn ) = log b ( m ) + log b ( n )
! 1$
4. log b # & = ' log b ( m )
" m%
! m$
2. log b # & = log b ( m ) ' log b ( n )
" n%
5. log b ( b ) = 1
3. log b m r = r log b ( m )
6. log b (1) = 0
( )
C. Proof
Prove: If
x and y are positive numbers, then ln(xy)=ln(x)+ln(y)
PROOF: Let
f(x)=ln(ax), where a is a positive constant. Then
d
1 d
1
1
f ! ( x ) = ( ln ( ax )) =
" ( ax ) = " a = . Therefore, f(x) and lnx have the same
dx
ax dx
ax
x
derivative and so they must differ by a constant: ln(ax)=ln(x) +C. Putting x=1 in this
equation, we get lna=ln1+C=0+C=C. Thus ln(ax)= lnx+lna . If we replace the
constant a by any number y, we have ln(xy)=lnx+lny.
D. Common Abbreviations
1. log10(x)=log(x)
2. loge(x)=ln(x)  log base e is the Natural log function
II.
The Nat ural Lo gar ithm F uncti on
A. De f n of the Natural Logarithm Function:
B. Graph of
y= ln ( x ) , x > 0
x
1
ln ( x ) = ! dt ,
t
1
x>0
C. Interpretation of
1. If
ln(x)
x > 1 , then ln(x) is the area under the curve of y =
1
from t=1 to t=x.
t
1
2.
1
ln (1) = ! dt = 0
t
1
3. If
4.
0 < x < 1 , then ln(x) gives the negative of the area under the curve from t=x to
t=1.
ln(x) is not defined for x ! 0 .
5. The domain of
ln(x) is (0, ! ) and the range is (!"," ) .
6. Defn : The number
e is that number in the domain of the natural logarithm satisfying
ln(e)=1
7.
lim ( ln ( x )) ! " , lim+ ( ln ( x )) ! -"
x!"
x!0
8. The function is continuous, increasing, and one-to-one.
9. The graph is concave downward.
D. Example
By comparing areas, show that
1
3
< ln 2 <
2
4
E. The Derivative of
y=ln(x)
x
d "$ 1 %'
1
dt =
, x > 0 by the Fundamental Thm of Calculus
!
dx # 1 t &
x
d
1.
(ln ( x )) =
dx
d
(ln ( x )) = 1
dx
x
Therefore, for every positive value of x,
2. Applying the Chain Rule:
F. The Integral of
1. If
(
)
d
1
ln ( f ( x )) =
f ! ( x ) , f ( x ) >0
dx
f ( x)
1
! u du
u is a nonzero differentiable function,
1
! u du = !
du
= ln u + C .
u
2. Examples
e
a.
!
1
e
3
b.
!
1
3
c.
III.
dx
x
ln 2 (3x)
dx
x
x2
" 4 ! x 3 dx
The Nat ural E xp o ne nti al F uncti on
A. The Inverse of
ln(x) and the Function ex
x
1. The Function y= e
a. ln(x) is a 1-1 increasing function, therefore, it has an inverse,
b. De f n : For every real number x,
ex = ln-1(x).
ln-1(x).
c.
ln(x) has domain (0, ! ) and range (!"," ) , ! ex then will have domain
(!"," ) and range (0, ! ) .
ex
d. The graph of
is the reflection of
ln(x) across the line y=x.
e.
lim e x ! " and
f.
e= ln-11. e is not a rational number it is not even an algebraic number, it is a
x!"
lim e x = 0 .
x!#"
Transcendental number like π .
B. Equations Involving
ln(x) and ex
1. Inverse Equations for ex and
ln(x)
; x>0
a.
eln(x) =x
b.
ln(ex)=x
;
!x
2. Useful Operating Rules
a. To remove logarithms from an equation, exponentiate both sides.
b. To remove exponentials, take the logarithm of both sides.
C. Laws of Exponents
For all numbers
x, x1, and x2.
1. e x1 ! e x2 = e x1 + x2
2. e" x =
1
ex
3.
e x1
= e x1 " x2
e x2
( )
4. e x1
x2
( )
= e x1 x2 = e x2
x1
D. The Derivative of
ex
1. Given the function y=ex we will use logarithmic differentiation to find
y = ex
dy
.
dx
ln y = ln (e x )
ln y = x
1 dy
= 1
y dx
dy
= y
dx
dy
= ex
dx
2. Rules
d x
e ) = ex
(
dx
a.
b. If f (x) is any differentiable fn of x, then
3. Examples
Differentiate the following.
a.
y = e4x+1
E. The Integral of
1. Since
(
)
d f (x )
( )
e
= e f x ! f "( x )
dx
b.
y = ln e x + eln ( 6 x )
3
ex
ex is its own derivative, it should be its own antiderivative: ! e dx = e + C
x
2. Extending the rule using u-substitution:
3. Examples
a.
! 15e
b.
! x cos(x
5x+3
dx
2
2
)esin x dx
! e du
u
= eu + C
x
IV.
The G e ne ral E xp o ne nti al F uncti o n
A. Definition: For any numbers
B. Derivatives Involving
ax
a>0 and x, the exponential function with base a is a x = e x ln a .
ax
a>0 and u is a differentiable function of x, then a u is a differentiable function of x
! du $
d f (x )
d u
( )
and
a ) = a u ( ln a) # & . i.e.,
a
= a f x ( ln a) ! f ' ( x )
(
" dx %
dx
dx
1. If
(
2. Derivation
a x = e x ln a , a > 0
d x
d x ln a
a ) =
(
(e )
dx
dx
d x
d
a ) = e x ln a ! ( x ln a)
(
dx
dx
d x
(a ) = e x ln a (ln a)
dx
d x
(a ) = a x (ln a )
dx
3. Examples: Differentiate the following.
1.
y = 72 x
2.
f (t ) = 5t
3
C. Integrals Involving
1.
2. Examples
!5
b.
4y
! y3
x
ax
au
+C
ln a
u
! a du =
a.
!7t +8
dx
2
+1
dy
)
IV.
Logar ithm s with Ba se
a
A. Definition: For any positive number
a x ! 1 , the logarithm of x with base a, denoted by
log a x is the inverse function a x .
a x and log a x
B. Inverse Equations for
1.
a loga x = x ; x > 0
2.
log a (a x ) = x ; !x
C. Converting Bases of Logarithms - The Evaluation of
log a x =
log a x :
ln x
ln a
Proof:
For x > 0 and a ! 1,
a loga x = x
, Inverse Equation
ln (a loga x ) = ln( x )
, Take natural log of both sides
(log a x )( ln a)
log a x =
= lnx , log property
lnx
ln a
D. Derivatives Involving
, divide by ln a
log a x
1. Proof
Let y =
log a x . Then a y = x
dy
a y ( ln a)
= 1
differentiating implicitly wrt x
dx
dy
1
= y
dx
a (ln a )
dy
1
y
=
substituting the original eq: a = x
dx
x (ln a )
2. If
u is a positive differentiable function of x, then
" %
( )
d
d
(log a f (x )) = 1 ! 1 f "( x ) = f " x
(log a u) = 1 ! 1 $# du '& OR
dx
ln a f ( x)
f ( x ) ln a
dx
ln a u dx
3. Example
a. y = log 3 (7x + 1)
b.
y = log 2 (4x 3 )
E. Integrals Involving
log a x
1. When integrating functions involving
integrate
2. Examples
a.
b.
!
!
log 3 x
dx
x
( log 2 (3x))4 dx
x
log a x , convert log a x to base e and then