## Lecture Notes: Friday, October 23, 1998

Sixth Homework
Assignment - for discussion at recitations Tuesday October 20, and
Thursday October 22. The sixth homework also includes your first
**MAPLE ASSIGNMENT**.

Seventh Homework
Assignment - for discussion at recitations Tuesday October 27, and
Thursday October 29.

### Chapter 5

**Section 1 (Exponential Functions)**
**Section 2 (Logarithmic Functions)**

**Logarithmic Functions**

For c > 0, a = log_{b}(c) if and only if b^a = c (here,
as above, b > 0, b not equal to 1)

In the text, log with no other indication of base means log to the
base 10 and "ln" means log to the base e. However, in MAPLE both ln
and log mean base e. To get base 10 in MAPLE you use log[10],
so log[10](1000) = 3.

Properties:

log_{b}(x*y) = log_{b}(x) +
log_{b}(y)
log_{b}(x/t) = log_{b}(x) -
log_{b}(y)

log_{b}(x^y) = y*log_{b}(x)

log_{b}(1) = 0

log_{b}(b) = 1

**Inverse functions.** g and f are inverse functions if g(f(x))
= x and f(g(x)) = x. In this case, (a,b) is on the graph of f if and
only if (b,a) is on the graph of g, so the graphs are reflections of
each other about the line y = x.

log_{b}(x) and b^x are inverse functions.

e^(ln(x)) = x for x > 0. ln(e^x) = x for all x. So exp(x) and
ln(x) are inverse functions. We looked at the graphs in MAPLE.

Problems:

5.1 # 24
5.2 # 14, 24, 26, 40, 44

Finally, the first MAPLE homework problem suggests the use of
induction. It is not required that you give a proof by induction.
However, I would like you to know about induction. If you can do the
proof by induction, that would be great.

**MATHEMATICAL INDUCTION** says this: Suppose you have a
collection of statements P(n), one for each positive integer n.
Suppose

- P(1) is true and
- P(k+1) is true whenever P(k) is true.

Then P(n) is true for all n.

*Typical Example:* I claim 1+ ... + n = n(n+1)/2. This is our
statement P(n).

Proof by induction:

- P(1) just says 1 = 1*2/2 which is clearly true.
- Assume P(k) is true. Then
1+...+k + (k+1) = k(k+1)/2 + (k+1) (by the induction
assumption)
=(k+1)(k+2)/2 (by algebra)

But this is the statement P(k+1). Thus P(k) true implies
P(k+1) true.

Then induction says that P(n) is true for all n.

Lecture Notes Directory

Course Guide