Download Section 2.6 Logarithmic Functions

Section 2.6 Logarithmic Functions
One-to-One Functions
HORIZONTAL LINE TEST: A function is one-to-one if and only if no horizontal line intersects
its graph more than once.
DEFINITION: Let f be a one-to-one function with domain A and range B. Then its inverse
function f −1 has domain B and range A and is defined by
f −1 (y) = x
⇐⇒
f (x) = y
(∗)
for any y in B.
So, we can reformulate (∗) as
f −1 (f (x)) = x for every x in the domain of f
f (f −1 (x)) = x for every x in the domain of f −1
IMPORTANT: Do not confuse f −1 with
1
.
f
EXAMPLES:
1. Let f (x) = x3 , then f −1 (x) =
√
3
x, since
√
√
3
f −1 (f (x)) = x3 = x and f (f −1 (x)) = ( 3 x)3 = x
1
2. Let f (x) = 5x, then f −1 (x) = x, since
5
f
−1
1
(f (x)) = (5x) = x and f (f −1 (x)) = 5
5
1
1
x
5
=x
THEOREM: If f has an inverse function f −1 , then the graphs of y = f (x) and y = f −1 (x) are
reflections of one another about the line y = x; that is, each is the mirror image of the other
with respect to that line.
DEFINITION: Let a be a positive number with a 6= 1. The logarithmic function with base
a, denoted by loga , is defined by
loga x = y
ay = x
⇐⇒
So, loga x is the exponent to which the base a must be raised to give x.
4
4
4
y 2
y 2
y 2
0
-4
-2
0
0
2
4
-4
-2
0
0
2
x
4
0
2
x
4
6
8
10
x
-2
-2
-2
-4
-4
-4
BASIC PROPERTIES: f (x) = loga x is a continuous function with domain (0, ∞) and range
(−∞, ∞). Moreover,
loga (ax ) = x for every x ∈ R,
aloga x = x for every x > 0
REMARK: It immediately follows from property 1 that
loga a = 1,
loga 1 = 0
EXAMPLES:
1. log2 2 = 1, log2 4 = 2, log2 8 = 3, log2 16 = 4
2. log3 3 = 1, log3 9 = 2, log3 27 = 3, log3 81 = 4
1
1
1
−1
−2
3. log3
= log3 3 = −1, log3
= log3 3 = −2, log3
= log3 3−3 = −3
3
9
27
2
√
1
1
3
5 = log5 51/2 = , log7 7 = log7 71/3 =
2
3
1
1
1
= log11
= log11 11−1/5 = −
5. log11 √
5
1/5
11
5
11
4. log5
√
Common and Natural Logarithms
DEFINITION: The logarithm with base e is called the natural logarithm and has a special
notation:
loge x = ln x
BASIC PROPERTIES:
1. ln(ex ) = x for every x ∈ R.
2. eln x = x for every x > 0.
REMARK: It immediately follows from property 1 that
ln e = 1
DEFINITION: The logarithm with base 10 is called the common logarithm and has a special
notation:
log10 x = log x
BASIC PROPERTIES:
1. log(10x ) = x for every x ∈ R.
2. 10log x = x for every x > 0.
REMARK: It immediately follows from property 1 that
log 10 = 1
3
Laws of Logarithms
LAWS OF LOGARITHMS: If x and y are positive numbers, then
1. loga (xy) = loga x + loga y.
x
= loga x − loga y.
2. loga
y
3. loga (xr ) = r loga x where r is any real number.
EXAMPLES:
1. Use the laws of logarithms to evaluate log2 4.
Solution: We have
log2 4 = log2 22 = 2 log2 2 = 2 · 1 = 2
2. Use the laws of logarithms to evaluate log10 1000000.
Solution: We have
log10 1000000 = log10 106 = 6 log10 10 = 6 · 1 = 6
3. Use the laws of logarithms to evaluate log7
Solution: We have
log7
√
√
7 = log7 71/2 =
7.
1
1
1
log7 7 = · 1 =
2
2
2
1
4. Use the laws of logarithms to evaluate log5 √
.
3
5
Solution: We have
1
1
1
1
1
= log5 1/3 = log5 5−1/3 = − log5 5 = − · 1 = −
log5 √
3
5
3
3
3
5
or
1
1
1
1
1
1
log5 √
= log5 1/3 = log5 1 − log5 51/3 = 0 − log5 5 = − log5 5 = − · 1 = −
3
5
3
3
3
3
5
5. Use the laws of logarithms to evaluate log3 270 − log3 10.
Solution: We have
log3 270 − log3 10 = log3
270
10
= log3 27 = log3 33 = 3 log3 3 = 3 · 1 = 3
6. Use the laws of logarithms to evaluate log2 12 + log2 3 − log2 9.
Solution: We have
log2 12 + log2 3 − log2 9 = log2 (12 · 3) − log2 9 = log2
12 · 3
9
= log2 4 = log2 22
= 2 log2 2 = 2 · 1 = 2
4
EXAMPLES:
1. ln (x(x + 1)) = ln x + ln(x + 1)
x
2. ln
= ln x − ln(x + 2)
x+2
x(x + 1)
= ln x(x + 1) − ln(x + 2) = ln x + ln(x + 1) − ln(x + 2)
3. ln
x+2
x(x + 1)
= ln x(x + 1) − ln (x + 2)(x + 3)
4. ln
(x + 2)(x + 3)
= ln x + ln(x + 1) − ln(x + 2) + ln(x + 3)
= ln x + ln(x + 1) − ln(x + 2) − ln(x + 3)
√
x x+1
x(x + 1)1/2
5. ln √
= ln
3
(x + 2)1/3 (x + 3)5
x + 2(x + 3)5
= ln x(x + 1)1/2 − ln (x + 2)1/3 (x + 3)5
=
ln x + ln(x + 1)1/2 − ln(x + 2)1/3 + ln(x + 3)5
= ln x + ln(x + 1)1/2 − ln(x + 2)1/3 − ln(x + 3)5
= ln x +
1
1
ln(x + 1) − ln(x + 2) − 5 ln(x + 3)
2
3
Change of Base
IMPORTANT FORMULA: For any positive a and b (a, b 6= 1) we have
logb x =
loga x
loga b
In particular, if a = e or 10, then
logb x =
EXAMPLE: log4 8 = log22 23 =
log x
ln x
=
ln b
log b
3 log2 2
3·1
3
log2 23
=
=
=
2
log2 2
2 log2 2
2·1
2
5
Exponential Equations
An exponential equation is one in which the variable occurs in the exponent.
EXAMPLE: Solve the equation 2x = 7.
Solution 1: We have
2x = 7
log2 2x = log2 7
x log2 2 = log2 7
x = log2 7 ≈ 2.807
Solution 2: We have
2x = 7
ln 2x = ln 7
x ln 2 = ln 7
x=
ln 7
≈ 2.807
ln 2
EXAMPLE: Solve the equation 4x+1 = 3.
Solution 1: We have
4x+1 = 3
log4 4x+1 = log4 3
(x + 1) log4 4 = log4 3
x + 1 = log4 3
x = log4 3 − 1
Solution 2: We have
4x+1 = 3
ln 4x+1 = ln 3
(x + 1) ln 4 = ln 3
x+1=
x=
EXAMPLE: Solve the equation 3x−3 = 5.
6
ln 3
ln 4
ln 3
−1
ln 4
EXAMPLE: Solve the equation 3x−3 = 5.
Solution 1: We have
3x−3 = 5
x − 3 = log3 5
x = log3 5 + 3
Solution 2: We have
3x−3 = 5
ln 3x−3 = ln 5
(x − 3) ln 3 = ln 5
x−3=
x=
ln 5
ln 3
ln 5
+3
ln 3
EXAMPLE: Solve the equation 8e2x = 20.
Solution: We have
8e2x = 20
e2x =
20
5
=
8
2
2x = ln
x=
5
2
ln 25
1 5
= ln
2
2 2
EXAMPLE: Solve the equation e3−2x = 4.
Solution: We have
e3−2x = 4
3 − 2x = ln 4
−2x = ln 4 − 3
x=
3 − ln 4
ln 4 − 3
=
≈ 0.807
−2
2
7
Logarithmic Equations
A logarithmic equation is one in which a logarithm of the variable occurs.
EXAMPLE: Solve the equation ln x = 8.
Solution: We have
ln x = 8
eln x = e8
x = e8
EXAMPLE: Solve the equation log2 (x + 2) = 5.
Solution: We have
log2 (x + 2) = 5
2log2 (x+2) = 25
x + 2 = 25
x = 25 − 2 = 30
EXAMPLE: Solve the equation log7 (25 − x) = 3.
Solution: We have
log7 (25 − x) = 3
7log7 (25−x) = 73
25 − x = 73
x = 25 − 73 = −318
EXAMPLE: Solve the equation 4 + 3 log(2x) = 16.
8
EXAMPLE: Solve the equation 4 + 3 log(2x) = 16.
Solution: We have
4 + 3 log(2x) = 16
3 log(2x) = 12
log(2x) = 4
2x = 104
x=
104
= 5000
2
EXAMPLE: Solve the equation log(x + 2) + log(x − 1) = 1.
Solution: We have
log(x + 2) + log(x − 1) = 1
log[(x + 2)(x − 1)] = 1
(x + 2)(x − 1) = 10
x2 + x − 2 = 10
x2 + x − 12 = 0
(x + 4)(x − 3) = 0
x+4=0
or
x = −4
x−3=0
x=3
We check these potential solutions in the original equation and find that x = −4 is not a
solution (because logarithms of negative numbers are undefined), but x = 3 is a solution.
EXAMPLE: Solve the following equations
(a) log(x + 8) + log(x − 1) = 1
(b) log(x2 − 1) − log(x + 1) = 3
9
EXAMPLE: Solve the following equations
(a) log(x + 8) + log(x − 1) = 1
(b) log(x2 − 1) − log(x + 1) = 3
Solution:
(a) We have
log(x + 8) + log(x − 1) = 1
log[(x + 8)(x − 1)] = 1
(x + 8)(x − 1) = 10
x2 + 7x − 8 = 10
x2 + 7x − 18 = 0
(x + 9)(x − 2) = 0
x+9=0
or
x−2=0
x = −9
x=2
We check these potential solutions in the original equation and find that x = −9 is not a
solution (because logarithms of negative numbers are undefined), but x = 2 is a solution.
(b) We have
log(x2 − 1) − log(x + 1) = 3
log
x2 − 1
=3
x+1
x2 − 1
= 103
x+1
(x − 1)(x + 1)
= 1000
x+1
x − 1 = 1000
x = 1001
EXAMPLE: Find the solution of the equation, correct to two decimal places.
(a) 10x+3 = 62x
(b) 5 ln(3 − x) = 4
(c) log2 (x + 2) + log2 (x − 1) = 2
10
EXAMPLE: Find the solution of the equation, correct to two decimal places.
(a) 10x+3 = 62x
(b) 5 ln(3 − x) = 4
(c) log2 (x + 2) + log2 (x − 1) = 2
Solution:
(a) We have
10x+3 = 62x
ln 10x+3 = ln 62x
(x + 3) ln 10 = 2x ln 6
x ln 10 + 3 ln 10 = 2x ln 6
x ln 10 − 2x ln 6 = −3 ln 10
x(ln 10 − 2 ln 6) = −3 ln 10
x=
−3 ln 10
≈ 5.39
ln 10 − 2 ln 6
(b) We have
5 ln(3 − x) = 4
ln(3 − x) =
4
5
3 − x = e4/5
x = 3 − e4/5 ≈ 0.77
(c) We have
log2 (x + 2) + log2 (x − 1) = 2
log2 (x + 2)(x − 1) = 2
(x + 2)(x − 1) = 4
x2 + x − 2 = 4
x2 + x − 6 = 0
(x − 2)(x + 3) = 0
x−2=0
or
x=2
x+3=0
x = −3
Since x = −3 is not from the domain of log2 (x + 2) + log2 (x − 1), the only answer is x = 2.
11