Transcript Slide 1

Chapter 10
Exploring Exponential and
Logarithmic Functions
By Kathryn Valle
10-1 Real Exponents and
Exponential Functions
• An exponential function is any equation in the
form y = a·bx where a ≠ 0, b > 0, and b ≠ 1. b is
referred to as the base.
• Property of Equality for Exponential Functions: If
in the equation y = a·bx, b is a positive number
other than 1, then bx1 = bx2 if and only if x1 = x2.
10-1 Real Exponents and
Exponential Functions
• Product of Powers Property: To simplify two like
terms each with exponents and multiplied
together, add the exponents.
– Example: 34 · 35 = 39
5√2 · 5√7 = 5√2 + √7
• Power of a Power Property: To simplify a term
with an exponent and raised to another power,
multiply the exponents.
– Example: (43)2 = 46
(8√5)4 = 84·√5
10-1 Examples
• Solve:
128 = 24n – 1
27 = 24n – 1
7 = 4n – 1
8 = 4n
n=2
53n + 2 > 625
53n + 2 > 54
3n + 2 > 4
3n > 2
n > ²/³
10-1 Practice
1. Simplify each expression:
a. (23)6
b. 7√4 + 7√3
c. p5 + p3
d. (k√3)√3
2. Solve each equation or inequality.
a. 121 = 111 + n
b. 33k = 729
c. 343 = 74n – 1
d. 5n2 = 625
Answers: 1)a) 218 b) 7√4 + √3 c) p8 d) k3 2)a) n = 1 b) n = 2 c) n = 1 d) n = ±2
10-2 Logarithms and Logarithmic
Functions
• A logarithm is an equation in the from logbn = p
where b ≠ 1, b > 0, n > 0, and bp = n.
• Exponential Equation
Logarithmic Equation
n = bp
p = logbn
exponent or logarithm
base
number
– Example: x = 63 can be re-written as 3 = log6 x
³/2 = log2 x can be re-written as x = 23/2
10-2 Logarithms and Logarithmic
Functions
• A logarithmic function has the from y = logb,
where b > 0 and b ≠ 1.
• The exponential function y = bx and the
logarithmic function y = logb are inverses of
each other. This means that their composites
are the identity function, or they form an
equation with the form y = logb bx is equal to x.
– Example: log5 53 = 3
2log2 (x – 1) = x – 1
10-2 Logarithms and Logarithmic
Functions
• Property of Equality for Logarithmic Functions:
Given that b > 0 and b ≠ 1, then logb x1 = logb x2
if and only if x1 = x2.
– Example: log8 (k2 + 6) = log8 5k
k2 + 6 = 5k
k2 – 5k + 6 = 0
(k – 6)(k + 1) = 0
k = 6 or k = -1
10-2 Practice
1. Evaluate each expression.
a. log3 ½7
b. log7 49
c. log5 625
d. log4 64
2. Solve each equation.
a. log3 x = 2
b. log5 (t + 4) = log5 9t
d. log12 (2p2) – log12 (10p – 8)
e. log2 (log4 16) = x
c. logk 81 = 4
f. log9 (4r2) – log9(36)
Answers: 1)a) -3 b) 2 c) 4 d) 3 2)a) 9 b) ½ c) 3 d) 1, 5 e) 1 f) -3, 3
10-3 Properties of Logarithms
• Product Property of Logarithms:
logb mn = logb m + logb n as long as m, n, and
b are positive and b ≠ 1.
– Example: Given that log2 5 ≈ 2.322, find log2 80:
log2 80 = log2 (24 · 5)
= log2 24 + log2 5 ≈ 4 + 2.322 ≈ 6.322
• Quotient Property of Logarithms: As long as
m, n, and b are positive numbers and b ≠ 1,
then logb m/n = logb m – logb n
- Example: Given that log3 6 ≈ 1.6309, find log3 6/81:
log3 6/81 = log3 6/34 = log3 6 – log3 34
≈ 1.6309 – 4 ≈ -2.3691
10-3 Properties of Logarithms
• Power Property of Logarithms: For any real
number p and positive numbers m and b,
where b ≠ 1, logb mp = p·logb m
– Example: Solve ½ log4 16– 2·log4 8 = log4 x
½ log4 16– 2·log4 8 = log4 x
log4 161/2 – log4 82 = log4 x
log4 4 – log4 64 = log4 x
log4 4/64 = log4 x
x = 4/64
x = 1/16
10-3 Practice
1. Given log4 5 ≈ 1.161 and log4 3 ≈ 0.792, evaluate
the following:
a. log4 15
b. log4 192
c. log4 5/3
d. log4 144/25
2. Solve each equation.
2 log3 x = ¼ log2 256
3 log6 2 – ½ log6 25 = log6 x
½ log4 144 – log4 x = log4 4
1/3 log5 27 + 2 log5 x = 4 log5 3
Answers: 1)a) 1.953 b) 3.792 c) 0.369 d) 1.544 2)a) x = ± 2 b) x
= 8/5 c) x = 36 d) x = 3√3
a.
b.
c.
d.
10-4 Common Logarithms
• Logarithms in base 10 are called common
logarithms. They are usually written without the
subscript 10.
– Example: log10 x = log x
• The decimal part of a log is the mantissa and
the integer part of the log is called the
characteristic.
– Example: log (3.4 x 103) = log 3.4 + log 103
= 0.5315 + 3
mantissa
characteristic
10-4 Common Logarithms
• In a log we are given a number and asked to find
the logarithm, for example log 4.3. When we are
given the logarithm and asked to find the log, we
are finding the antilogarithm.
– Example: log x = 2.2643
x = 10 2.2643
x = 183.78
– Example: log x = 0.7924
x = 10 0.7924
x = 6.2
10-4 Practice
1. If log 3600 = 3.5563, find each number.
a. mantissa of log 3600
b. characteristic of log 3600
c. antilog 3.5563
d. log 3.6
e. 10 3.5563
f. mantissa of log 0.036
2. Find the antilogarithm of each.
c. -1.793
d. 0.704 – 2
Answers: 1)a) 0.5563 b) 3 c) 3600 d) 0.5563 e) 3600 f)
0.5563 2)a) 314.775 b) 1.459 c) 0.016 d) 0.051
a. 2.498
b. 0.164
10-5 Natural Logarithms
• e is the base for the natural logarithms,
which are abbreviated ln. Natural logarithms
carry the same properties as logarithms.
• e is an irrational number with an
approximate value of 2.718. Also, ln e = 1.
10-5 Practice
1. Find each value rounded to four decimal places.
ln 6.94
ln 0.632
ln 34.025
ln 0.017
e.
f.
g.
h.
antiln -3.24
antiln 0.493
antiln -4.971
antiln 0.835
Answers: 1)a) 1.9373 b) -0.4589 c) 3.5271 d) -4.0745
e) 0.0392 f) 1.6372 g) 0.0126 h) 2.3048
a.
b.
c.
d.
10-6 Solving Exponential Equations
• Exponential equations are equations where the
variable appears as an exponent. These
equations are solved using the property of
equality for logarithmic functions.
– Example: 5x = 18
log 5x = log 18
x · log 5 = log 18
x = log 18
log 5
x = 1.796
10-6 Solving Exponential Equations
• When working in bases other than base 10, you
must use the Change of Base Formula which says
loga n = logb n
logb a
For this formula a, b, and n are positive numbers
where a ≠ 1 and b ≠ 1.
- Example: log7 196
log 196 change of base formula
log 7
a = 7, n = 196, b = 10
≈ 2.7124
10-6 Practice
1. Find the value of the logarithm to 3 decimal places.
a. log7 19
b. log12 34
c. log3 91
d. log5 48
2. Use logarithms to solve each equation. Round to
three decimal places.
a. 13k = 405
b. 6.8b-3 = 17.1
c. 5x-2 = 6x
d. 362p+1 = 14p-5
b) B = 4.481 c) x = -17.655 d) p = -3.705
Answers: 1)a) 1.513 b) 1.419 c) 4.106 d) 2.405 2)a) k = 2.341
10-7 Growth and Decay
• The general formula for growth and decay is
y = nekt, where y is the final amount, n is the
initial amount, k is a constant, and t is the
time.
• To solve problems using this formula, you will
apply the properties of logarithms.
10-7 Practice
1. Population Growth: The town of BloomingtonNormal, Illinois, grew from a population of 129,180
in 1990, to a population of 150,433 in 2000.
a. Use this information to write a growth equation for
Bloomington-Normal, where t is the number of years after
1990.
b. Use your equation to predict the population of
Bloomington-Normal in 2015.
c. Use your equation to find the amount year when the
population of Bloomington-Normal reaches 223,525.
10-7 Practice Solution
a. Use this information to write a growth equation for
Bloomington-Normal, where t is the number of
years after 1990.
y = nekt
150,433 = (129,180)·ek(10)
1.16452 = e10·k
ln 1.16452 = ln e10·k
0.152311 = 10·k
k = 0.015231
equation: y = 129,180·e0.015231·t
10-7 Practice Solution
b. Use your equation to predict the population of
Bloomington-Normal in 2015.
y = 129,180·e0.015231·t
y = 129,180·e(0.015231)(25)
y = 129,180·e0.380775
y = 189,044
10-7 Practice Solution
c. Use your equation to find the amount year
when the population of Bloomington-Normal
reaches 223,525.
y = 129,180·e0.015231·t
223,525 = 129,180·e0.015231·t
1.73034 = e0.015231·t
ln 1.73034 = ln e0.015231·t
0.548318 = 0.015231·t
t = 36 years
1990 + 36 = 2026