In the previous sections we have considered independently the
functions
that solved two differential equations. The "natural logarithm"
function,
ln
,
was the solution to the differential equation L'(t) = 1/t with L(1) = 0
, while the "natural exponential" function, exp or e, was the
solution
to the differential equation P'(t) = P(t) with P(0) = 1. You probably
have
seen these two functions before in your pre-calculus course work where
they were closely related as the next theorem shows.
Theorem VI.C.1: (a) For all t > 0, exp(ln(t)) = t.
(b)
For all x, ln(exp(x)) = x.
Proof: (a) Let F(t) = exp(ln(t))/t for t > 0.
Then using the quotient and the chain rules we see that
F'(t) = [t^{ .}exp(ln(t))^{ .}1/t - exp(ln(t))^{
.}1]/t^{ 2} = 0 for all t > 0.
Thus F(t) is a constant and since F(1)= exp(ln(1))/1 =
exp(0)/1
= 1 we have
exp(ln(t))/t = 1 or exp(ln(t)) = t for all t > 0.
(b) Let F(x) = ln(exp(x)) - x for all x
.
Then using the chain rule we see that F'(x)= 1/exp(x)
^{.}
exp(x)-1= 0 for all x .
Thus F(x) is a constant. Since F(0) = ln(exp(0)) - 0=
ln(1)= 0 we have ln(exp(x)) - x = 0 for all x or
ln(exp(x))
= x for all x .EOP
We can visualize the relationship described in the theorem in several
ways.
Using mapping diagrams as in Figure VI.C.i and as in the GeoGebra figure, we see that the
arrows
used to visualize the ln function as a mapping reverses
the
arrows used to visualize the exp function and vice versa.
In functional terms, the composition of ln with exp is
the
identity
function,
as
is also the composition of exp with ln,
i.e.,
ln _{°}
exp = id = exp _{°}
ln,
when the domain is suitably chosen. For both these reasons we say
that
the natural logarithm and natural exponential function are inversely
related
as functions.
The cartesian
graphical way to view this inverse relation as in the GeoGebra figure is to note that if b =
exp(a)
then ln(b) = ln(exp(a)) = a . Thus if (a,b) is a point on the graph of
the natural exponential function, then (b,a) is a point on the graph of
the natural logarithm function. Similarly if a = ln(b) , b > 0 then
exp(a)
= exp(ln(b)) = b which shows that if (b,a) is a point on the graph of
the
natural logarithm function then (a,b) is a point on the graph of the
natural
exponential function. See Figure VI.C.ii. We summarize this by saying
exp(a)
= b if and only if b = ln(a) , or using the notation e^{ a} =
exp(a),
e^{ a} = b if and only if b = ln(a).
This last statement justifies calling the natural logarithm "the
logarithm
with the base e" and writing in more conventional terms ln(a) = log_{
e}(a). For example, e^{ 0} = 1 so, ln(1) = 0 and e^{1}=e
so
ln(e)
=
1.
In general ln(e^{ a}) = a , which merely restates
part (b) of the theorem.
Applications: The inverse relation of these two functions is
most convenient for finding ways to express the solutions to
differential
equations that are in exponential form as the following examples
illustrate.
Example VI.C.1: Suppose f(x) = A e^{ kx}
. Find A and k when f(0) = 20 and f(1) = 40. Find t so
that
f(t)
= 100.
Solution: First we merely substitute the given values into
the
formula for f: 20 = f(0) = Ae^{ k(0)} = A so
A=20,
while 40=f(1)=A e^{ k(1)} =20 e^{ k}. So 2 = e^{
k}.
Using the inverse relation between ln and exp in this equation means
precisely that k = ln(2). Thus f(x) = 20 e^{ (ln(2) x)}.
To find when f(t) = 100 we need to solve 100 = 20 e^{
(ln(2)
t)}, or 5 = e^{ (ln(2) t)}. Using the inverse properties
once
again, we have ln(5) = t ln(2) so
t = ln(5) / ln(2) » 2.32.
Example VI.C.2: Suppose f(x) = Ae^{ kx}
and f(3) = 2f(0). If f(1) = 40, find f(5).
Solution: f(0) = A so we have been given 2 A = A e^{
k3} in the statement f(3) = 2f(0). Thus 2 = e^{
3k
},
so by the inverse relation of ln and exp we have that 3k = ln(2) or k =
(1/3) ln(2).
The second equation f(1) = 40 becomes 40 = Ae^{ k}.
At
this point we can solve for A theoretically by dividing both sides of
the
last equation by e^{ k}, giving A = 40 / e^{ k}. We
continue
with this form for A to find f(5) = A e^{ k5} = (40 / e^{
k}) e^{ 5k} = 40 e^{ 4k} = 40 e^{ 4/3 ln(2)}.
Recognizing this result is not the easiest number to compute, let's
notice that
e ^{4/3 ln(2) }= (e^{ln(2)})^{4/3 }=
2^{ 4/3}
because of the properties of exponents and the inverse relation of ln
and
exp.
So f(5) = 40 (2)^{ 4/3}.
The connection of the natural logarithm to other
core
exponential and logarithmic functions.
The natural logarithm function can be defined as a definite integral:
`ln(x)
= int _1^x
1/t dt` . Using
this
definition, the natural logarithm can provide the mathematical
foundation
for all other logarithmic and exponential functions. The
following
chart demonstrates the major connections of these functions in one "Big
Picture."
Summary.....The Big Picture : Log's and
Exponential
Functions
Def'n.:
`ln(x) = int_1^x
1/t dt`
[ x>0]
Def'n.: exp (x) = y`-=` ln(y)
= x exp(1) `-=` e
[so
ln(e) = 1]. exp(x) `-=` e^{x}
Def'n.: For b >0, b^{x}`-=`e^{x}^{ln(b)}.
Note: ln(b^{x}) = x ln(b)
Def'n.: For b >0, log _{b} (x) = y `-=`
x= b^{y}
ln(1) = 0
ln(x) > 0 for x >1
ln(x) < 0 for 0< x <1
exp(0) = e^{0}=1
e^{x}> 1 for x > 0
0< e^{x} < 1 for x < 0
b^{0}= 1
For b > 1: b^{x }> 1 for x
>
0
0< b^{x }< 1 for x < 0
For 0< b < 1: b^{x }> 1 for x
<
0
0< b^{x }< 1 for x > 0
log _{b} (1) = 0
For b > 1:
log _{b} (x)>0 for x >1
log _{b} (x)<0 for 0< x
<1
For 0< b < 1:
log _{b} (x)<0 for x >1
log _{b} (x)>0 for 0< x <1
ln(A*B) = ln(A) + ln(B)
e^{A}e^{B }=e^{A+B}
b^{A}b^{B }= b^{A+B}
log _{b} (A*B) = log _{b} (A)
+ log _{b}
(B)
ln(A/B) = ln(A) - ln(B)
e^{A }/ e^{B }= e^{A-B}
b^{A }/b^{B }= b^{A-B}
log _{b} (A/B) = log _{b} (A
) - log _{b}
(B)
ln(A^{p/q}) =p/q ln(A)
(e^{x})^{ p/q }= e^{(p/q)*x}
(b^{x})^{ p/q }= b^{(p/q)*x}
log _{b} (A^{p/q} )= p/q log _{b}
(A)
ln'(x) = D ln(x) = 1/x [So ln is continuous and increasing for x
>0]
exp'(x) = D(e^{x}) = e^{x}
D(b^{x}) = ln(b) b^{x}
log _{b}'(x) = D
log _{b}
(x) = 1/( x ln(b))
`int 1/u du = ln|u| + C`
`int
e^u
du
=
e^u + C`
` int b^u
du = b^u / ln(b)
+ C`
Not relevant!
As `x -> oo , ln
(x) -> oo`.
As `x -> oo, e^x -> oo`.
b >0: As `x -> oo, b^x -> oo`
b <0:As`x
-> oo, b^x -> 0`
For b > 1:As `x -> oo, log_b
(x) -> oo`.
For 0< b < 1:As `x ->
oo, log_b (x) ->
- oo`.
For b > 1:As `x -> 0^+, log_b
(x) -> - oo`.
For 0< b < 1: As `x ->
0^+, log_b (x) -> oo`
Exercises VI.C
Find the first derivative for the functions in problems 1-4 using
logarithmic differentiation:
f(x) = 7^{ x}
f(x) = 3 ^{sin(x)}
y = sin(x) 5^{ x}
`g(t) = t^pi`
The numbers `e^pi` and `pi^e` have been of interest to
mathematicians for many years.
Consider
the function f(x) = e^{ x} - x^{
e}.
Use calculus to determine the local extreme values of f
and when f is increasing or decreasing. With this information
determine
whether `f(pi)` is positive or negative.
Which is larger, `e^pi` or `pi^e`?
Find the first derivative for the functions in problems 6-9.