0.6 Exponential and Logarithmic Functions
1. Review of Exponential Functions (Section 0.6)
Review: Rules of Exponents
i. a m/n n a m
ii. a "p 1p
iii. Ÿa p q a pq
a
iv. a p a q a pq
p
v. a q a p"q
a
Example Convert each exponential expression into fractional or root form:
Ÿ2 2 3/4
Ÿ3 4 "2/3
Ÿ1 3 "4
Ÿ1 3 "4 1
81
Ÿ2 2 3/4 4
23 4
Ÿ3 4 "2/3 8
3
1 42
3
1
16
Example Convert each expression into exponential form:
Ÿ1 Ÿ1 "2
3 x5
"2 " 2 x "5/2
3
3 x5
Ÿ2 4
Ÿ2 4
x7
2
x7
1 x 7/4
2
2
Example Express each value as b r form (try to make the value of b as small as possible):
Ÿ1 27
Ÿ2 3 1
Ÿ3 4 e 5
Ÿ4 64
81
32
Ÿ1 27 3 3 3 3/2
Ÿ2 3
1 32
3
1 2 "5/3
25
Ÿ3 4
e 5 e 5/4
Ÿ4 16 2 4 81
34
Ÿ3 27
3
2
3
Example Find the exact value of the expression:
Ÿ1 9 3/2
Ÿ1 9 3/2 Ÿ3 2 3/2
3 3 27
Ÿ2 64 "1/3
Ÿ2 64 "1/3 Ÿ2 6 "1/3
Ÿ3 2 "2 1
4
27
3
9 3
a. Definition: Let b 0 and b p 1. The function fŸx b x is called an exponential function, and b is
called the base and x is the exponent.
Note that gŸx x b is the generalized power function where b is the exponent and x is the base.
Example For each function, determine if it is an exponential function, a power or generalized power
function or neither.
1
4
a. fŸx x =
b. gŸx = x
c. hŸx = =
d. kŸx x x
fŸx x = is a generalized power function, gŸx = x is an exponential function.
hŸx = = is a constant and kŸx x x is neither a power function nor an exponential function.
b. A special exponential function: fŸx e x where e lim nv. 1 1n
c. The graph of an exponential function:
-3
-2
-1
15
15
10
10
5
5
0
1
x
2
3
-3
-2
y bx, b 1
x
-1
x
0
n
.
1
x
2
3
y bx, 0 b 1
1
are symmetric with respect to the y "axis.
b "x . So, the graphs of b x and 1
b
b
Note if b 1, then 0 1 1.
b
d. Domain and range of an exponential function: D b x "., . ; R b x 0, .
Observe that
e. Properties:
i. b 0 1
ii. For b 1, b x is an increasing function and lim xv " . b x 0
For 0 b 1, b x is a decreasing function and lim xv . b x 0
2. Review of Logarithmic Functions (Section 0.6)
a. Definition: Let b 0 and b p 1. The logarithm function with base b, fŸx log b x, is defined by
y log b x if and only if x b y .
When b e, log e x ln x. The relation of log b x, log a x and ln x :
log a x
log b x ,
log b x ln x
ln b
log a b
Observe that if x b r , then b r b y « y r.
Example Let fŸx log 2 x, gŸx log 1/3 x,
Ÿ1 fŸ4 , f
and hŸx log
1
16
Evaluate
Ÿ2 gŸ9 , g 1
Ÿ3 hŸ1 , hŸ5 27
Ÿ1 y fŸ4 log 2 4 log 2 2 2 « y 2
log 2 1 log 2 Ÿ2 "4 « y "4
yf 1
16
16
"2
« y "2
Ÿ2 y gŸ9 log 1/3 9 log 3 1
3
3
yg 1
«y3
log 1/3 1
27
3
2
5 x.
Ÿ3 y hŸ1 log
y hŸ5 log
51
55
log
log
5
5
5
0
«y0
5
2
«y2
b. Relation of b x and log b x
Recall: fŸx and gŸx are inverse functions they satisfy the following two conditions:
i. Ÿf ( g Ÿx fŸgŸx x;
ii. Ÿg ( f Ÿx gŸfŸx x.
The exponential function fŸx b x and the logarithm function gŸx log b x are inverse functions:
Let log b x y. Then we have b y x
Ÿf ( g Ÿx fŸgŸx fŸlog b x b log b x b y x
Ÿg ( f Ÿx gŸfŸx log b Ÿb x x
c. The graph of a logarithm function: the graph of y log b x is symmetric to the graph of y b x with
respect to the line y x :
3
2
1
-3
-2
-1
1
x
2
3
-1
-2
-3
– y b x , -.-. y log b x,
d. Domain and range of a logarithm function:
Since x b y 0, D log b x 0, . . R log b x ... y x
"., .
e. Properties:
i. log b b 1, log b 1 0
ii. lim xv0 log b x "., lim xv. x .
iii. log b b fŸx fŸx iv. b log b fŸx fŸx v. log b Ÿxy log b x log b y
vi. log b xy log b x " log b y
vii. log b Ÿx y y log b x
Observe that the property iv. says that any number x can be written as an exponential form with base b.
Example Rewrite the exponential 2 x , 5 x ,
x
2 x Ÿe ln 2 e x ln 2 ,
3
x
2
5
5 x Ÿe ln 5 e x ln 5 ,
x
with base e.
2
5
x
e xŸln 2"ln 5 Example Rewrite the expression as one logarithm:
5
2
ln x 3
2
ln y " 4 ln z
5 ln x 3 ln y " 4 ln z ln x 5/2 ln y 3/2 " ln z 4
2
2
x5y3
z4
ln 5 ln x 3 ln 4 ln x 5 y 3 " ln z 4 ln
2
2
3
Example Express ln
x2 z2
y5
3
ln
as a sum/difference and scalar multiplication.
x2 z2
y5
ln x 2/3 ln z 2 " ln y 5/2 2 ln x 2 ln z " 5 ln y
2
3
Example Find a function of the form fŸx ae bx if we know fŸ0 2 and fŸ3 1.
fŸ0 ae 0 a 2
fŸ3 2e 3b 1, e 3b 1 , 3b ln 1
2
2
" 13 ln 2 x
fŸx 2e
ln 1 " ln 2 " ln 2, b " 1 ln 2
3
Example Solve the equation for x.
1
Ÿ1 e " 2 x 3
1
Ÿ1 e " 2 x 3, "
Ÿ2 1 1
3
1
2
Ÿ2 1 1 2 4x1 5
3
Ÿ3 4 lnŸ2x 1 1
Ÿ4 lnŸx " lnŸx 1 2
x ln 3, x "2 ln 3
2 4x1 5,
1
3
2 4x1 4, 2 4x1 12, Ÿ4x 1 ln 2 ln 12, 4x 1 ln 12 , 4x ln 12 " 1
ln 2
ln 2
x 1 ln 12 " 1
4 ln 2
Ÿ3 4 lnŸ2x 1 1, lnŸ2x 1 1 , 2x 1 e 1/4 , 2x e 1/4 " 1, x 1 Ÿe 1/4 " 1 2
4
2
2
2
2
,
x
x
e
x
"
e
0
lnŸx "
lnŸx
1 2,
lnŸxŸx
1 2,
x
Ÿ4 x
"1 o 1 " 4Ÿ1 Ÿ"e 2 "1 o 1 4e 2
2
2
Example The population of a city can be described as PŸt P 0 e k t where t 0 represents 1990. If we
know the population of the city was 10,000 in 1990 and was doubled in 1999, find the
population of the city in 2003.
PŸ0 P 0 e 0 P 0 10, 000, PŸt 10000e k t
PŸ9 10000e 9k 20000, e 9k 2, 9k ln 2, k 1 ln 2, PŸt 10, 000e
9
1
PŸ13 10000e 9 ln 2 Ÿ13 27215. 8
The population of the city is about 27215 in 2003.
4
1
9
ln 2 t
Example Given a half-life of 3 hours, by what percentage will the amount of morphine in the bloodstream
have decreased in 1 day?
A half-life of 3 hours means that the amount of morphine in the bloodstream is 1/2 of the original
amount in 3 hours. Let MŸt M 0 e kt . Then
1 M 0 M 0 e 3k . e 3k 1 , 3k ln 1 ln 1 " ln 2 " ln 2, k " 1 ln 2
2
3
2
2
1
MŸt M 0 e " 3 ln 2 t , MŸ24 M 0 e "8 ln 2 pM 0 , p e "8 ln 2 0. 003 906 25 0. 39%
There will be 0.39% of morphine in the bloodstream after 1 day.
5