Exponential functions
Recall that last class we defined
ln(x) =
Note: for t > 0,
ln(x) exists.
1
t
�
x
1
1
dt.
t
is continuous and continuous functions are integrable, so
We proved several properties about ln(x) last class.
Definition Let e be the number such that ln(e) = 1, i.e., e is the value
such that the area under 1t between 1 and e is exactly 1.
Slide on value of e being between 2 and 4.
Remark Note that
d
1
(ln(x)) = > 0 for x > 0,
dx
x
so ln(x) is strictly increasing. This implies that ln(x) is a one-to-one and
onto function, meaning it has an inverse function. Call this inverse function
exp(x).
def’n of inverse:
y = exp(x)
⇐⇒
ln(y) = x (�)
In other words, exp(ln(x)) = x and ln(exp(x)) = x.
Also, since domain of ln(x) is all positive real numbers and range is R, the
function exp(x) will have domain of R and range equal to all positive reals.
Slide showing graphs of ln(x) and exp(x).
Math 30G - Prof. Kindred - Lecture 14
Page 1
Recall that one of the important properties of ln x was that it takes products to sums. Naturally, we expect the inverse function to take sums to
products.
Claim: For all x, y ∈ R, we have
exp(x + y) = exp(x) exp(y).
Proof. Let z = exp(x) exp(y). From (�), we know that
z = exp(x + y) if and only if ln(z) = x + y,
so if we show that ln(z) = x + y, we’ll be done. We have
ln(z) = ln(exp(x) exp(y)) = ln(exp(x)) + ln(exp(y))
by property of ln
=x+y
since ln, exp are inverses.
�
It also can be shown that exp(−x) =
1
exp(x)
for all x ∈ R.
Remark We have that
exp(2) = exp(2 ln(e)) = exp(ln(e2)) = e2
����
1
and, more generally,
exp(q) = exp(q ln(e)) = exp(ln(eq )) = eq for any rational q.
What about irrational powers? We know
• exp(x) is a continuous function that agrees with ex for all rational
exponents x,
• and exp(x) is defined for all real numbers, rational or not!
Since the two functions agree on the rationals, we simply define ex for
irrational exponents to be the same value as exp(x).
Math 30G - Prof. Kindred - Lecture 14
Page 2
Definition For any x ∈ R, we define ex = exp(x).
Question
What is the derivative of ex?
d x
(e ) = ex.
dx
Proof. Let y = ex. Then
ln(y) = ln(ex) = ln(exp(x)) = x.
By differentiating both sides of the equation with respect to x, we have
1 dy
dy
·
= 1 =⇒
= y = ex .
y dx
dx
What about general exponential functions (like 3
Definition For b > 0 and a ∈ R, we define
π/2
or
�2
3
�√
�
2
)?
ba = ea ln b.
This definition explains why ex is called the exponential function — all
other exponential functions ax are derived from it. You can start to see
why Euler’s number e is considered one of the five fundamental constants!
Proposition. ln(xa) = a ln x for all a ∈ R and x > 0.
We previously proved this fact for rational values of a only.
Proof. Using the previous definition and the fact that ln and exp are inverses, we have
ln(xa) = ln(ea ln x) = ln(exp(a ln x)) = a ln x.
�
Math 30G - Prof. Kindred - Lecture 14
Page 3
Proposition. xa+b = xa xb for x > 0 and a, b ∈ R.
Proof. We have
xa+b = e(a+b) ln x = ea ln x+b ln x by definition and by distributivity
= exp(a ln x + b ln x) by definition
= exp(a ln x) exp(b ln x) by previous result
= ea ln xeb ln x
by definition
= xa xb .
Proposition. x−a =
�
1
for x > 0 and a ∈ R.
xa
Proof. We have
x−a = e−a ln x
by definition
= exp(−a ln x)
by definition
1
=
by previous result
exp(a ln x)
1
1
= a ln x = a
by definition.
e
x
�
Proposition.
d x
(b ) = bx ln b for b > 0 and x ∈ R.
dx
Proof. We have
d x
d
(b ) = (ex ln b)
by definition
dx
dx
d
= ex ln b
(x ln b)
by chain rule and fact that
dx
� �� �
d
x
dx (e )
= ex
ln b
= bx ln b
by definition.
�
Math 30G - Prof. Kindred - Lecture 14
Page 4
We now prove a general power rule that holds for all exponents a ∈ R.
Proposition.
d a
(x ) = axa−1 for a ∈ R.
dx
Proof. We have
d a
d
(x ) = (ea ln x)
by definition
dx
dx
d
= ea ln x
(a ln x) by chain rule,
dx
� �� �
d
x
(e
)
dx
= ex ,
d
dx (ln x)
=
1
x
a
x
= xa · ax−1 = axa−1 by previous propositions.
�
Proposition. (xa)b = xab for all a, b ∈ R.
Proof. We have
(xa)b = eb ln(x
a)
by definition
= eab ln x
= xab
by previous proposition
by definition.
�
We can define logarithms with different bases (other than e).
Definition For b > 0 and b �= 1, we define
logb x = y
⇐⇒
by = x.
Examples
• log2 64 = 6 since 26 = 64.
• log10 100 = 2 since 102 = 100.
Math 30G - Prof. Kindred - Lecture 14
Page 5
Nice fact
logb x =
ln x
.
ln b
That is, all logs are the same as the natural log up to a constant ( ln1b ).
We still do not have an exact value of e.
Find the Taylor series of ex at the point a = 0.
=⇒
e0
e = e + e (x − 0) + (x − 0)2 + · · ·
2!
2
3
x
x
=1+x+
+
+ ···
2!
3!
x
Thus,
e1 = 1 + 1 +
0
0
1
1
+ + · · · = 2.718281828459...
2! 3!
Also, e arises as the following limit:
�
e = lim 1 +
n→0
Math 30G - Prof. Kindred - Lecture 14
1
n
�n
Page 6
Value of e
Exponential functions
ln(4)
Math 30G, Calculus
Professor Kindred
October 5, 2012
ln(2)
area of square
ln(2) < w/side
length 1 = 1
ln(2) < 1 < ln(4)
=
ln(e)
Graph of exponential function exp(x)
The graph of exp(x) is the reflection of the graph of ln(x)
about the line y = x:
The graph of exp(x) is the reflection of the graph of ln x about the line y = x:
y
4
3
ln(x)
2
exp(x)
�3
1
�2
�1
1
2
3
4
x
�1
�2
exp(x)
�3
ln x
Example 4.5.1. Show that exp(0) = 1.
Solution. By definition exp(0) = y if and only if ln y = 0. Since we know ln 1 = 0,
of two
ln(4) > area
rectangles = 1 ·
1
1
+2· =1
2
4
=⇒ 2 < e < 4