The document discusses exponential and natural logarithm functions. It provides examples of using exponential functions to calculate continuous compound interest on investments over time, and shows that continuous compounding yields slightly higher returns. It also discusses properties of the natural logarithm function and its relationship to the exponential function. An example calculates the area under a curve bounded by lines using the natural logarithm formula.

1. Section 6-4
e and Natural Logarithms

2. Begin by opening your books to page
389. Pair up with a partner to work on
the In-Class Activity. Make sure to
record your observations on your note
sheet.

3. e:

4. e:
Comes out to be approximately 2.7182818284590...

5. Example 1
If $10,000 is put into bonds that pay 7.35% interest
compounded continuously, find:
a. The annual yield

6. Example 1
If $10,000 is put into bonds that pay 7.35% interest
compounded continuously, find:
a. The annual yield
e −1
r

7. Example 1
If $10,000 is put into bonds that pay 7.35% interest
compounded continuously, find:
a. The annual yield
e −1= e
r .0735
−1

8. Example 1
If $10,000 is put into bonds that pay 7.35% interest
compounded continuously, find:
a. The annual yield
e −1= e
r .0735
− 1 ≈ .0762685367

9. Example 1
If $10,000 is put into bonds that pay 7.35% interest
compounded continuously, find:
a. The annual yield
e −1= e
r .0735
− 1 ≈ .0762685367
So the annual yield is about 7.63%

10. Example 1
If $10,000 is put into bonds that pay 7.35% interest
compounded continuously, find:
a. The annual yield
e −1= e
r .0735
− 1 ≈ .0762685367
So the annual yield is about 7.63%
b. The value of the investment after one year

11. Example 1
If $10,000 is put into bonds that pay 7.35% interest
compounded continuously, find:
a. The annual yield
e −1= e
r .0735
− 1 ≈ .0762685367
So the annual yield is about 7.63%
b. The value of the investment after one year
10,000 + 10,000(e .0735
− 1)

12. Example 1
If $10,000 is put into bonds that pay 7.35% interest
compounded continuously, find:
a. The annual yield
e −1= e
r .0735
− 1 ≈ .0762685367
So the annual yield is about 7.63%
b. The value of the investment after one year
10,000 + 10,000(e .0735
− 1) ≈ $10,762.69

13. Continuous Change Model

14. Continuous Change Model
When your principal P grows or decays continuously
and an annual rate r, the amount A(t) after t years is:

15. Continuous Change Model
When your principal P grows or decays continuously
and an annual rate r, the amount A(t) after t years is:
A(t) = Pe rt

16. Continuous Change Model
When your principal P grows or decays continuously
and an annual rate r, the amount A(t) after t years is:
A(t) = Pe rt
This means we have a model we can work with for
continuous compounding, just like the other types of
compounding we saw earlier in the year.

17. Example 2
Suppose $10,000 is put into a 5-year certificate of
deposit that pays 7.35% interest compounded
continuously.
a. What is the balance at the end of the period?

18. Example 2
Suppose $10,000 is put into a 5-year certificate of
deposit that pays 7.35% interest compounded
continuously.
a. What is the balance at the end of the period?
A(t) = Pe rt

19. Example 2
Suppose $10,000 is put into a 5-year certificate of
deposit that pays 7.35% interest compounded
continuously.
a. What is the balance at the end of the period?
A(t) = Pe = 10,000e
rt .0735(5)

20. Example 2
Suppose $10,000 is put into a 5-year certificate of
deposit that pays 7.35% interest compounded
continuously.
a. What is the balance at the end of the period?
A(t) = Pe = 10,000e
rt .0735(5)
≈ $14,441.20

21. Example 2
Suppose $10,000 is put into a 5-year certificate of
deposit that pays 7.35% interest compounded
continuously.
a. What is the balance at the end of the period?
A(t) = Pe = 10,000e
rt .0735(5)
≈ $14,441.20
b. How does this compare with the balance if the
interest were compounded annually?

22. Example 2
Suppose $10,000 is put into a 5-year certificate of
deposit that pays 7.35% interest compounded
continuously.
a. What is the balance at the end of the period?
A(t) = Pe = 10,000e
rt .0735(5)
≈ $14,441.20
b. How does this compare with the balance if the
interest were compounded annually?
A = P(1+ r ) t

23. Example 2
Suppose $10,000 is put into a 5-year certificate of
deposit that pays 7.35% interest compounded
continuously.
a. What is the balance at the end of the period?
A(t) = Pe = 10,000e
rt .0735(5)
≈ $14,441.20
b. How does this compare with the balance if the
interest were compounded annually?
A = P(1+ r ) = 10,000(1.0735)
t 5

24. Example 2
Suppose $10,000 is put into a 5-year certificate of
deposit that pays 7.35% interest compounded
continuously.
a. What is the balance at the end of the period?
A(t) = Pe = 10,000e
rt .0735(5)
≈ $14,441.20
b. How does this compare with the balance if the
interest were compounded annually?
A = P(1+ r ) = 10,000(1.0735) ≈ $14,256.41
t 5

25. Example 2
Suppose $10,000 is put into a 5-year certificate of
deposit that pays 7.35% interest compounded
continuously.
a. What is the balance at the end of the period?
A(t) = Pe = 10,000e
rt .0735(5)
≈ $14,441.20
b. How does this compare with the balance if the
interest were compounded annually?
A = P(1+ r ) = 10,000(1.0735) ≈ $14,256.41
t 5
About $185 more

26. The Exponential Function with Base e:

27. The Exponential Function with Base e:
A function of the form f(x) = ex

28. The Exponential Function with Base e:
A function of the form f(x) = ex
Natural Logarithm:

29. The Exponential Function with Base e:
A function of the form f(x) = ex
Natural Logarithm:
The logarithm to the base e; written as ln x

30. The Exponential Function with Base e:
A function of the form f(x) = ex
Natural Logarithm:
The logarithm to the base e; written as ln x
In other words, ln x = loge x

31. The Exponential Function with Base e:
A function of the form f(x) = ex
Natural Logarithm:
The logarithm to the base e; written as ln x
In other words, ln x = loge x
This means that the Exponential Function with Base e
and the Natural Logarithm are inverses of each other.

32. Example 3
Consider the region bounded by the following graphs:
y = , the x-axis, the line x = a, the line x = b, x > 0.
1
x
Using calculus, it can be proven that the area of that
region is ln b - ln a. What then is the area bounded by
the graphs:
y= 1
x
, the x-axis, the line x = 1, the line x = 7, x > 0?

33. Example 3
Consider the region bounded by the following graphs:
y = , the x-axis, the line x = a, the line x = b, x > 0.
1
x
Using calculus, it can be proven that the area of that
region is ln b - ln a. What then is the area bounded by
the graphs:
y= 1
x
, the x-axis, the line x = 1, the line x = 7, x > 0?
ln7 − ln1

34. Example 3
Consider the region bounded by the following graphs:
y = , the x-axis, the line x = a, the line x = b, x > 0.
1
x
Using calculus, it can be proven that the area of that
region is ln b - ln a. What then is the area bounded by
the graphs:
y= 1
x
, the x-axis, the line x = 1, the line x = 7, x > 0?
ln7 − ln1 ≈ 1.945910149

35. Example 3
Consider the region bounded by the following graphs:
y = , the x-axis, the line x = a, the line x = b, x > 0.
1
x
Using calculus, it can be proven that the area of that
region is ln b - ln a. What then is the area bounded by
the graphs:
y= 1
x
, the x-axis, the line x = 1, the line x = 7, x > 0?
ln7 − ln1 ≈ 1.945910149 units 2

36. Homework

37. Homework
p. 394 #1-20