Applied Calculus Math 215 - University of Hawaii

24 CHAPTER 1. SOME BACKGROUND MATERIAL
For a given a (a>0anda=1)andb>0 we denote the unique solution
of the equation in (1.16) by loga(b). In other words:
Definition 1.15. If a and b are positive numbers, a = 1,thenloga(b) is the
unique number, such that
(1.17)
a log a (b) = b or exp a(log a(b)) = b.
Here are some sample logarithms for the base 2:
log2 4=2 log216 = 4 log2(1/8) = −3
√
log2 2=1/2
and for the base 10:
log 10 1=0 log 10 100 = 2 log 10(1/10) = −1.
Your calculator will give you good approximations for at least log10(x) for
any x>0.
Exercise 23. Find logarithms for the base 10:
(1) log10 5
(2) log10 100
(3) log10 π
(4) log10(1/4) (5) log10 25
(6) log10 1.
Mathematically speaking, we just defined a function. Let us express it
this way.
Definition 1.16. Let a be a positive number, a = 1. Mapping b to loga(b) defines a function, called the logarithm function with base a. It is defined
for all positive numbers, and its range is the set of real numbers.
Part of the graph of log2(x) is shown in Figure 1.10. In Figure 1.11 you
see the graph of a logarithm function with base a less than 1.
We also like to see for every real number y that
(1.18)
Setting b = a y in (1.17) we have that
log a(a y )=y or log a(exp a(y)) = y.
a log a (ay ) = a y .
The statement in (1.15) says that log a(a y )=y.
Taken together, (1.17) and (1.18) say that for every a>0, a = 1,we
have
This just means that
a loga (y) = y for all y>0and
loga(a x ) = x for all x ∈ (−∞, ∞).