Download Exponentials and logarithms to the base e

1.2. Exponentials and Logarithms.
Exponential and logarithmic functions arise in a many applications. Exponential functions are ones where
the independent variable is in the exponent. For example,
y = 6  2x
y = 102x + 4
In this section we look at exponential and logarithmic functions, particularly their derivatives and integrals.
1.2.1 Exponentials and logarithms to the base e.
When we take derivatives and do integrals of exponentials and logarithms it is convenient if they are
expressed in base e. This is similar to when we take derivatives and do integrals of trigonometric functions
it is convenient if angles are expressed in radians.
e is the number
e = 2.718281828459…
follow any particular pattern. It is defined by the following
formula
2
3
100
1000
10000
1 n
e = lim 1 + 
n
n
As we take larger and larger values of n the quantity 1 +

1 n
has to be quite large before 1 +  is close to e.
 n
y = ex = f(x)
-1
Here is a table of values and graph of this function.
-2
-3
20
0.5
1.5
15
- 0.5
10
5
3
2
1
1
2
3
1.2.1 - 1
(1.01)100  2.70481
(1.001)1000  2.716921
(1.0001)10000  2.71815
1n
gets closer and closer to e. As you can see n
n
x
0
1
2
3
We are interested in the function
(2)
n
1
Like  it is an unending decimal whose digits don't seem to
(1)
1 + 1
 n
1
1 + 1 = 2
1


2
2
1 + 1 = 3 = 2.25
 2 2
3
3
1 + 1 = 4  2.37037
 3 3
n
y = ex
e0 = 1
e1 = e  2.7
e2  (2.718)2  7.4
e3  (2.718)3  20.1
1
e-1 =  0.37
e
1
-2
e = 2  0.14
e
1
e-3 = 3  0.05
e
e1/2 = e  0.16
e3/2 = e e  4.5
1
1
e-1/2 =

 0.6
e 0.16
Here are some properties of the function y = ex.
Proposition 1. For all x and y one has
e0 = 1
ex > 0
ex is increasing with x
(3)
ex+y = exey
e-x =
1
ex
(4)
ex-y =
ex
ey
(5)
exy = (ex)y
Proof. These formulas are special case of basic properties of exponents. If a is any positive real number
1
ax
then one has a0 = 1, ax+y = axay, a-x = x , ax-y = y and axy = (ax)y. //
a
a
The natural logarithm function y = ln x is the inverse function to the function x = ey. Thus
(6)
ln x = that number y such that ey = x
Note that x must be positive for ln x to be defined. Here are a few special values.
ln 1 = 0
since e0 = 1
ln e = 1
since e1 = e
ln e2 = 2
since e2 = e2
1
ln  = -1
e 
since e-1 =
1
2
since e1/2 =
ln
e =
1
e
e
To find y = ln x for most numbers x one usually has to settle for an approximation obtained by solving
numerically the equation x = ey for y. For example, to find y = ln 2, one has to solve the equation ey = 2, at
least in principle. It turns out there are other methods. When one does this one gets ln 2  0.693.
1.1 - 2
At the right is a table of values and graph of this function. The following
x
1
2
3
0.5
1.5
0.1
proposition has some properties of the function y = ln x.
Proposition 2. For all x > 0, y > 0 and real numbers z one has
(7)
eln x = x
(8)
ln (ez) = z
(9)
ln(xy) = ln x + ln y
2
(10)
1
ln  = - ln x
x 
1
(11)
x
ln  = ln x - ln y
y
2
4
y = ln x
0
0.693
1.1
- 0.693
0.4
- 2.3
6
8
10
1
(12)
ln(xz) = z (ln x)
Proof. (7) follows from the fact that y = ln x is that number such that ey = x. (8) follows from the fact that z
is that number such that ez = ez. To prove (9) – (12), let u = ln x and v = ln y. Then eu = x and ev = y. By
(3) one has eu+v = euev = xy. Now ln(xy) is that number w such that ew = xy. However, we just saw eu+v = xy.
So ln(xy) = u + v = ln x + ln y which proves (9). By (4) one has e-u = 1/eu = 1/x. Now ln(1/x) is that
number w such that ew = 1/x. However, we just saw e-u = 1/x. So ln(1/x) = - u = - ln x which proves (10).
By (4) one has eu-v = eu/ev = x/y. Now ln(x/y) is that number w such that ew = x/y. However, we just saw eu-v
= x/y. So ln(x/y) = u - v = ln x - ln y which proves (11). By (4) one has euz = (eu)z = xz. Now ln(xz) is that
number w such that ew = x. However, we just saw euz = xz. So ln(xz) = zu = z(ln x) which proves (12). //
With the ln function one can solve a number of equations where the unknown is in the exponent. In general,
if the unknown is in the exponent, try taking ln of both sides of the equation.
Example 1. Find x so that 32x+1 = 4.
Solution. Take ln of both sides of the equation to get ln(32x+1) = ln 4. Use (12) to get (2x+1)(ln 3) = ln 4.
1 ln 4  1  1.38  1
1
So 2x + 1 = (ln 4)/(ln 3) which implies x = 
-1 
-1  (1.26 – 1)  (0.26)  0.13.
2  ln 3  2  1.10  2
2
Similarly, to solve certain equations involving ln you exponentiate both sides of the equation.
Example 2. Find x so that ln(2x + 1) = 4.
Solution. Raise e to both sides of the equation to get eln(2x + 1) = e4. Use the fact that eln u = u. This gives
e4 - 1 54.6 - 1 53.6
2x + 1 = e4 or 2x = e4 - 1 or x =


 26.8.
2
2
2
People often express exponential functions with other bases in terms of base e. This makes it easier to take
derivatives and do integrals.
1.1 - 3
Example 3. Suppose y = 1.3x. Express y as y = ekx for some constant k.
Solution. Since eln u = u it follows that 1.3 = eln(1.3). So y = 1.3x = (eln(1.3))x = e[ln(1.3)]x  e0.262x. See graph at
right.
In general,
(13)
ax = e(ln a)x
Problem 1. Simplify e(ln 2
+ ln x)
.
2
Problem 2. Solve for y: e(x )e2x+1 = ey.
1.1 - 4