LECTURE MATH OCT.27,2011
Today:
Properties of the Natural Logarithm
Chapter 4 Review
Exponential Growth and Decay
Ln(x)
Previously we introduced the natural log, ln(x) as the inverse of the exponential function
and established some of its properties
By definition, ln(x)=y exactly when e^y=x
The doman of ln(x) is (o, infinity)
D[ln(x)]/dx=1/x
By the Chain Rule, d[ln(x)]/dx=g(x)/g(x)
We also drew the graph y=ln(x) by reflecting the graph y=e^x through the line y=x. Now
we sketch the curve y=ln(x) directly
Ln(x) has domain (0, infinity) and the following limits
Lim x0=-infinity
Lim xinfinity ln(x)=infinity
Now, solve for ln(x)=0
Ln(x)=0
X=e^0
X=1
Further,
F(x) is <0 if x<1
>0 if 1<x
Ln(x) has domain (0, infininty), a vertical asymptote at x=0 and infinite limit as x
inifinity
Ln(x) has one x-intercept at x=1
Ln(x) is negative in (0,1) and positive in (1, infinity)
Ln(x), (ln(x))=1/x
Since (ln(x))=1/x and ln(x) has domain (0, infinity) and (ln(x)) is positive on the domain
of ln(x)
ln(x) is increasing on its domain (0, infinity)
ln(x), (ln(x)) = 1/x, (ln(x)) = 1/x2
Since (ln(x)) = 1/x2 and ln(x) has domain (0,), (ln(x)) is negative on the domain of
ln(x).
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ln(x) is concave down on its domain (0,).
Summary of f(x)=ln(x)
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.
?
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ln(x ) has domain (0, ), a vertical asymptote at x = 0 and infinite limit as x
ln(x ) has one x -intercept at x = 1.
ln(x ) is negative in (0, 1) and positive in (1, ).
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f(x)=1/x
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ln(x) is increasing on its domain (0,).
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f(x)=1/x2
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ln(x) is concave down on its domain (0,)
Example:
Differentiate f(x)=ln(x)
Note, that x is positive for all lxl cannot equal 0, so the domain of ln(lxl) is the set of x
such that x cannot equal 0. Since ln(lxl) is not defined for x=0, ln(lxl) is not differentiable
at x=0. Otherwise,
Lxl=
-x, if x<0
X, if 0<x
And we can apply the chain rule for each case
1. When x<0, d[ln(lxl)]/dx=d[ln(-x)]/dx=1/-x(-x)=-1/-x=1/x
2. When x>0, d[ln(lxl)]/dx=d[ln(x)]=1/x
Laws of Logarithms:
The Laws of Exponentiation that we have already isolated can be translated into the
language of logarithms
Ln(ab)=ln(a)+ln(b)
Ln(1/a)=-ln(a)
Ln(a/b)=ln(a)-ln(b)
Ln(a^b)=blna
Reason:
1. ln(ab)=ln(a)+ln(b)
e^ln(ab)
=ab
=e^ln(a)e^ln(b)
=e^ln(a)+ln(b)
Since e^x is increasing, ln(ab)=ln(a)+ln(b)
2. ln(1/a)=-ln(a)
e^ln(1/a)
=1/a
=1/e^ln(a)
=e^-ln(a)
So, ln(1/a)=-ln(a)
3. ln(a/b)=ln(a)-ln(b) Follows from the previous two
4. ln(a^b)=blna
ln(a^b)
=ln(e^ln(a))^b)
=ln(e^bln(a))
=bln(a)
Example:
1. Simplify the expression eln(x^2)+3ln(y)
eln(x^2)+3ln(y)
=(eln(x^2))(e3ln(y))
=x2(eln(y))3
=x2y3
2. Differentiate f(x)=ln(x(x+2)(x2+1))
One could differentiate immediately with the chain rule and iterated applications of the
product rule
OR
With foresight, we can use the properties of the log and differentiate
Ln(x(x+2)(x2+1))=ln(x)+ln(x+2)+ln(x2+1)
(ln(x(x+2)(x2+1))
=(ln(x)=ln(x+2)+ln(x2+1))
=1/x+(1/x+2)(x+2)+(1/(x2+1))(x2+1)
=1/x+1/(x+2)+(2x/(x2+1))
The Power General Rule
When we first introduced the power rule, we checked it for exponents 2, 3, and
indicated how to verify it for roots and powers. We have been using it for all powers ever
since. Now, we can use the logarithm to verify that (xr)=rxr-1
(Xr)
=(eln(x^r))
=(eln(x^r))(ln(xr))
=xr(rln(x))
=xr(r(1/x))
=rxr-1
Chapter 4 Recap:
Topics of Chapter 4:
Exponentiation
o Laws of exponents
o Exponential Functions
The number e and the function ex
Differentiation of ex
o The differential equation y=ky
The natural logarithm function ln(x)
Differentiation of ln(x)
Laws of logarithms
CHAPTER 5 EXPONENTIAL GROWTH
Many applications involve solving the differential equation
Y=ky
For k a constant
We have shown that the solutions to this equation are the functions
Y=Cekx
For constant C
Notation.
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When k > 0, y = Cekx describes exponential growth and k is called the growth constant.
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When k < 0, y = Cekx describes exponential decay and k is called the decay constant.
Example
The number of bacteria cells, N(t), increases at a rate proportional to N(t). Suppose that
N(0) = N0 and solve for N(t).
Solution.
N(t) satisfies the equation N(t) = kN(t). So, N(t) = Cekt
N (0) = Cek0=CN (0) = N0 given initial condition
C = N0 N(t) = N0ekt
Example: The amount of a radioactive isotope, A(t), decreases by radioactive decay at a
rate proportional to A(t). That is
A(t) =
A(t)where is a positive constant. Solve for t such that A(t) =
(1/2)A(0),
the half-life of the isotope.
Solution. Again, A(t) = Cet and C = A(0), the amount of material at t = 0. Now, we
solve for the half-life
A(0)e
t
= 1/2A(0)
t
e
=1/2
t = ln(1/2)=-ln(2)
ln(1) = ln(2) 2
t=ln(2)/
C
C
Example: Carbon Dating. 12 is the common stable isotope of carbon and 14 is a
14C
radioactive isotope. The decay constant for
is .00012 years1 .
Thus, the half-life of
14C
is
ln(2)/.00012
5775 years.
14C
12C
Example: The ratio of the natural occurrence of the isotopes
and
in the
12
atmosphere or in a living plant or animal is 14C/12C 10 . When an organism stops
respirating, it no longer stays in equilibrium with the atmosphere.
Let A(t) be the amount of
equilibrium.
14C
and B(t) be the amount of
12C
in a specimen taken out of
A(t)=A(0)e-0.00012t
B(t) = B(0)
The ratio of the two, R(t) satisfies
R(t)=A(t )/B(t)
=A(0)e
.00012t
/B(0)
00012t
=R(0)e
=10-12 e-0.00012t
One can determine the time t at which a organism died by measuring the ratio of
12C
in its remains, to the extent that is possible.
14C
to
Example: Determine the percentage (or the original amount) of 14C that remains in the
wood of a 4500 year antique chest
Solution: A(t)=A(0)e-0.00012t
A(4,500)
=A(0)e-.00012-4,500
This is about A(0)(0.58)
So, approximately 58% of the original 14C remains