e and Natural Logarithms

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