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6.3: Homework

• Last updated
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• Page ID 70322

• Julie Harland
• MiraCosta College
• Submit homework separately from this workbook and staple all pages together. (One staple for the entire submission of all the unit homework)
• Start a new module on the front side of a new page and write the module number on the top center of the page.
• Answers without supporting work will receive no credit.
• Some solutions are given in the solutions manual.
• You may work with classmates but do your own work.

How would you explain to someone what the "absolute value of a number" means? Explain without using examples.

Simplify each of the following:

Write any and all numerals that have the given absolute value:

For each vector drawn, write the number that the vector represents:

Look again at the four vectors shown in exercise 4. If someone had drawn those vectors to find the difference of two numbers using the missing addend approach, write the four subtraction problems which generated those vectors.

Use vectors on the number line to add -8 + 5 + (-3) using vectors as actions. Mark and LABEL your number line with at least zero and a point on either side. Explain how to read the answer.

-8 + 5 + -3 = _____ since ______________________________________________

Use vectors on the number line to compute the 8 – 5 + -7 – -4 using vectors to DO and UNDO actions. Mark and LABEL your number line with at least zero and a point on either side. Explain how to read the answer.

______________________________________________8 – 5 + -7 – -4 = _____ since ______________________________________________.

Use the missing addend approach to perform the following subtractions. Label the vector on the number line.

Use red and green counters to add the following integers. For each problem, explain and show each of the steps involved.

For each problem, use positive and negative counters. First, state what the problem means, and then explain and show each step you need to take to find the answer. Each problem requires 3 or 4 steps.

For each of the following sets, determine if the set is closed under the operation given. Provide a counterexample if it is not closed.

Use your counters to do each of the following multiplication problems using the definition of multiplying two integers with positive and negative counters. Then, explain what the multiplication problem given means in terms of the counters, and explain and show each of the individual steps.

6.1 Exponential Functions

g ( x ) = 0.875 x g ( x ) = 0.875 x and j ( x ) = 1095.6 − 2 x j ( x ) = 1095.6 − 2 x represent exponential functions.

5.5556 5.5556

About 1.548 1.548 billion people; by the year 2031, India’s population will exceed China’s by about 0.001 billion, or 1 million people.

( 0 , 129 ) ( 0 , 129 ) and ( 2 , 236 ) ; N ( t ) = 129 ( 1 .3526 ) t ( 2 , 236 ) ; N ( t ) = 129 ( 1 .3526 ) t

f ( x ) = 2 ( 1.5 ) x f ( x ) = 2 ( 1.5 ) x

f ( x ) = 2 ( 2 ) x . f ( x ) = 2 ( 2 ) x . Answers may vary due to round-off error. The answer should be very close to 1.4142 ( 1.4142 ) x . 1.4142 ( 1.4142 ) x .

y ≈ 12 ⋅ 1.85 x y ≈ 12 ⋅ 1.85 x

about $3,644,675.88

e − 0.5 ≈ 0.60653 e − 0.5 ≈ 0.60653

$3,659,823.44

3.77E-26 (This is calculator notation for the number written as 3.77 × 10 − 26 3.77 × 10 − 26 in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)

6.2 Graphs of Exponential Functions

The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 0 , ∞ ) ; ( 0 , ∞ ) ; the horizontal asymptote is y = 0. y = 0.

The domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( 3 , ∞ ) ; ( 3 , ∞ ) ; the horizontal asymptote is y = 3. y = 3.

x ≈ − 1.608 x ≈ − 1.608

f ( x ) = − 1 3 e x − 2 ; f ( x ) = − 1 3 e x − 2 ; the domain is ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; the range is ( − ∞ , −2 ) ; ( − ∞ , −2 ) ; the horizontal asymptote is y = −2. y = −2.

6.3 Logarithmic Functions

• ⓐ log 10 ( 1 , 000 , 000 ) = 6 log 10 ( 1 , 000 , 000 ) = 6 is equivalent to 10 6 = 1 , 000 , 000 10 6 = 1 , 000 , 000
• ⓑ log 5 ( 25 ) = 2 log 5 ( 25 ) = 2 is equivalent to 5 2 = 25 5 2 = 25
• ⓐ 3 2 = 9 3 2 = 9 is equivalent to log 3 ( 9 ) = 2 log 3 ( 9 ) = 2
• ⓑ 5 3 = 125 5 3 = 125 is equivalent to log 5 ( 125 ) = 3 log 5 ( 125 ) = 3
• ⓒ 2 − 1 = 1 2 2 − 1 = 1 2 is equivalent to log 2 ( 1 2 ) = − 1 log 2 ( 1 2 ) = − 1

log 121 ( 11 ) = 1 2 log 121 ( 11 ) = 1 2 (recalling that 121 = ( 121 ) 1 2 = 11 121 = ( 121 ) 1 2 = 11 )

log 2 ( 1 32 ) = − 5 log 2 ( 1 32 ) = − 5

log ( 1 , 000 , 000 ) = 6 log ( 1 , 000 , 000 ) = 6

log ( 123 ) ≈ 2.0899 log ( 123 ) ≈ 2.0899

The difference in magnitudes was about 3.929. 3.929.

It is not possible to take the logarithm of a negative number in the set of real numbers.

6.4 Graphs of Logarithmic Functions

( 2 , ∞ ) ( 2 , ∞ )

( 5 , ∞ ) ( 5 , ∞ )

The domain is ( 0 , ∞ ) , ( 0 , ∞ ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 0. x = 0.

The domain is ( − 4 , ∞ ) , ( − 4 , ∞ ) , the range ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the asymptote x = – 4. x = – 4.

The domain is ( 2 , ∞ ) , ( 2 , ∞ ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 2. x = 2.

The domain is ( − ∞ , 0 ) , ( − ∞ , 0 ) , the range is ( − ∞ , ∞ ) , ( − ∞ , ∞ ) , and the vertical asymptote is x = 0. x = 0.

x ≈ 3.049 x ≈ 3.049

x = 1 x = 1

f ( x ) = 2 ln ( x + 3 ) − 1 f ( x ) = 2 ln ( x + 3 ) − 1

6.5 Logarithmic Properties

log b 2 + log b 2 + log b 2 + log b k = 3 log b 2 + log b k log b 2 + log b 2 + log b 2 + log b k = 3 log b 2 + log b k

log 3 ( x + 3 ) − log 3 ( x − 1 ) − log 3 ( x − 2 ) log 3 ( x + 3 ) − log 3 ( x − 1 ) − log 3 ( x − 2 )

2 ln x 2 ln x

− 2 ln ( x ) − 2 ln ( x )

log 3 16 log 3 16

2 log x + 3 log y − 4 log z 2 log x + 3 log y − 4 log z

2 3 ln x 2 3 ln x

1 2 ln ( x − 1 ) + ln ( 2 x + 1 ) − ln ( x + 3 ) − ln ( x − 3 ) 1 2 ln ( x − 1 ) + ln ( 2 x + 1 ) − ln ( x + 3 ) − ln ( x − 3 )

log ( 3 ⋅ 5 4 ⋅ 6 ) ; log ( 3 ⋅ 5 4 ⋅ 6 ) ; can also be written log ( 5 8 ) log ( 5 8 ) by reducing the fraction to lowest terms.

log ( 5 ( x − 1 ) 3 x ( 7 x − 1 ) ) log ( 5 ( x − 1 ) 3 x ( 7 x − 1 ) )

log x 12 ( x + 5 ) 4 ( 2 x + 3 ) 4 ; log x 12 ( x + 5 ) 4 ( 2 x + 3 ) 4 ; this answer could also be written log ( x 3 ( x + 5 ) ( 2 x + 3 ) ) 4 . log ( x 3 ( x + 5 ) ( 2 x + 3 ) ) 4 .

The pH increases by about 0.301.

ln 8 ln 0.5 ln 8 ln 0.5

ln 100 ln 5 ≈ 4.6051 1.6094 = 2.861 ln 100 ln 5 ≈ 4.6051 1.6094 = 2.861

6.6 Exponential and Logarithmic Equations

x = − 2 x = − 2

x = − 1 x = − 1

x = 1 2 x = 1 2

The equation has no solution.

x = ln 3 ln ( 2 3 ) x = ln 3 ln ( 2 3 )

t = 2 ln ( 11 3 ) t = 2 ln ( 11 3 ) or ln ( 11 3 ) 2 ln ( 11 3 ) 2

t = ln ( 1 2 ) = − 1 2 ln ( 2 ) t = ln ( 1 2 ) = − 1 2 ln ( 2 )

x = ln 2 x = ln 2

x = e 4 x = e 4

x = e 5 − 1 x = e 5 − 1

x ≈ 9.97 x ≈ 9.97

x = 1 x = 1 or x = − 1 x = − 1

t = 703 , 800 , 000 × ln ( 0.8 ) ln ( 0.5 ) years  ≈ 226 , 572 , 993 years . t = 703 , 800 , 000 × ln ( 0.8 ) ln ( 0.5 ) years  ≈ 226 , 572 , 993 years .

6.7 Exponential and Logarithmic Models

f ( t ) = A 0 e − 0.0000000087 t f ( t ) = A 0 e − 0.0000000087 t

less than 230 years, 229.3157 to be exact

f ( t ) = A 0 e ln 2 3 t f ( t ) = A 0 e ln 2 3 t

6.026 hours

895 cases on day 15

Exponential. y = 2 e 0.5 x . y = 2 e 0.5 x .

y = 3 e ( ln 0.5 ) x y = 3 e ( ln 0.5 ) x

6.8 Fitting Exponential Models to Data

• ⓐ The exponential regression model that fits these data is y = 522.88585984 ( 1.19645256 ) x . y = 522.88585984 ( 1.19645256 ) x .
• ⓑ If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year.
• ⓐ The logarithmic regression model that fits these data is y = 141.91242949 + 10.45366573 ln ( x ) y = 141.91242949 + 10.45366573 ln ( x )
• ⓑ If sales continue at this rate, about 171,000 games will be sold in the year 2015.
• ⓐ The logistic regression model that fits these data is y = 25.65665979 1 + 6.113686306 e − 0.3852149008 x . y = 25.65665979 1 + 6.113686306 e − 0.3852149008 x .
• ⓑ If the population continues to grow at this rate, there will be about 25,634   25,634   seals in 2020.
• ⓒ To the nearest whole number, the carrying capacity is 25,657.

6.1 Section Exercises

Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.

When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal .

exponential; the population decreases by a proportional rate. .

not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function. .

The forest represented by the function B ( t ) = 82 ( 1.029 ) t . B ( t ) = 82 ( 1.029 ) t .

After t = 20 t = 20 years, forest A will have 43 43 more trees than forest B.

Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.

exponential growth; The growth factor, 1.06 , 1.06 , is greater than 1. 1.

exponential decay; The decay factor, 0.97 , 0.97 , is between 0 0 and 1. 1.

f ( x ) = 2000 ( 0.1 ) x f ( x ) = 2000 ( 0.1 ) x

f ( x ) = ( 1 6 ) − 3 5 ( 1 6 ) x 5 ≈ 2.93 ( 0.699 ) x f ( x ) = ( 1 6 ) − 3 5 ( 1 6 ) x 5 ≈ 2.93 ( 0.699 ) x

$ 10 , 250 $ 10 , 250

$ 13 , 268.58 $ 13 , 268.58

P = A ( t ) ⋅ ( 1 + r n ) − n t P = A ( t ) ⋅ ( 1 + r n ) − n t

$ 4,572.56 $ 4,572.56

continuous growth; the growth rate is greater than 0. 0.

continuous decay; the growth rate is less than 0. 0.

$ 669.42 $ 669.42

f ( − 1 ) = − 4 f ( − 1 ) = − 4

f ( − 1 ) ≈ − 0.2707 f ( − 1 ) ≈ − 0.2707

f ( 3 ) ≈ 483.8146 f ( 3 ) ≈ 483.8146

y = 3 ⋅ 5 x y = 3 ⋅ 5 x

y ≈ 18 ⋅ 1.025 x y ≈ 18 ⋅ 1.025 x

y ≈ 0.2 ⋅ 1.95 x y ≈ 0.2 ⋅ 1.95 x

APY = A ( t ) − a a = a ( 1 + r 365 ) 365 ( 1 ) − a a = a [ ( 1 + r 365 ) 365 − 1 ] a = ( 1 + r 365 ) 365 − 1 ; APY = A ( t ) − a a = a ( 1 + r 365 ) 365 ( 1 ) − a a = a [ ( 1 + r 365 ) 365 − 1 ] a = ( 1 + r 365 ) 365 − 1 ; I ( n ) = ( 1 + r n ) n − 1 I ( n ) = ( 1 + r n ) n − 1

Let f f be the exponential decay function f ( x ) = a ⋅ ( 1 b ) x f ( x ) = a ⋅ ( 1 b ) x such that b > 1. b > 1. Then for some number n > 0 , n > 0 , f ( x ) = a ⋅ ( 1 b ) x = a ( b − 1 ) x = a ( ( e n ) − 1 ) x = a ( e − n ) x = a ( e ) − n x . f ( x ) = a ⋅ ( 1 b ) x = a ( b − 1 ) x = a ( ( e n ) − 1 ) x = a ( e − n ) x = a ( e ) − n x .

47 , 622 47 , 622 fox

1.39 % ; 1.39 % ; $ 155 , 368.09 $ 155 , 368.09

$ 35 , 838.76 $ 35 , 838.76

$ 82 , 247.78 ; $ 82 , 247.78 ; $ 449.75 $ 449.75

6.2 Section Exercises

An asymptote is a line that the graph of a function approaches, as x x either increases or decreases without bound. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely large or extremely small.

g ( x ) = 4 ( 3 ) − x ; g ( x ) = 4 ( 3 ) − x ; y -intercept: ( 0 , 4 ) ; ( 0 , 4 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

g ( x ) = − 10 x + 7 ; g ( x ) = − 10 x + 7 ; y -intercept: ( 0 , 6 ) ; ( 0 , 6 ) ; Domain: all real numbers; Range: all real numbers less than 7. 7.

g ( x ) = 2 ( 1 4 ) x ; g ( x ) = 2 ( 1 4 ) x ; y -intercept: ( 0 , 2 ) ; ( 0 , 2 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

y -intercept: ( 0 , − 2 ) ( 0 , − 2 )

Horizontal asymptote: h ( x ) = 3 ; h ( x ) = 3 ; Domain: all real numbers; Range: all real numbers strictly greater than 3. 3.

As x → ∞ x → ∞ , f ( x ) → − ∞ f ( x ) → − ∞ ; As x → − ∞ x → − ∞ , f ( x ) → − 1 f ( x ) → − 1

As x → ∞ x → ∞ , f ( x ) → 2 f ( x ) → 2 ; As x → − ∞ x → − ∞ , f ( x ) → ∞ f ( x ) → ∞

f ( x ) = 4 x − 3 f ( x ) = 4 x − 3

f ( x ) = 4 x − 5 f ( x ) = 4 x − 5

f ( x ) = 4 − x f ( x ) = 4 − x

y = − 2 x + 3 y = − 2 x + 3

y = − 2 ( 3 ) x + 7 y = − 2 ( 3 ) x + 7

g ( 6 ) = 800 + 1 3 ≈ 800.3333 g ( 6 ) = 800 + 1 3 ≈ 800.3333

h ( − 7 ) = − 58 h ( − 7 ) = − 58

x ≈ − 2.953 x ≈ − 2.953

x ≈ − 0.222 x ≈ − 0.222

The graph of G ( x ) = ( 1 b ) x G ( x ) = ( 1 b ) x is the refelction about the y -axis of the graph of F ( x ) = b x ; F ( x ) = b x ; For any real number b > 0 b > 0 and function f ( x ) = b x , f ( x ) = b x , the graph of ( 1 b ) x ( 1 b ) x is the the reflection about the y -axis, F ( − x ) . F ( − x ) .

The graphs of g ( x ) g ( x ) and h ( x ) h ( x ) are the same and are a horizontal shift to the right of the graph of f ( x ) ; f ( x ) ; For any real number n , real number b > 0 , b > 0 , and function f ( x ) = b x , f ( x ) = b x , the graph of ( 1 b n ) b x ( 1 b n ) b x is the horizontal shift f ( x − n ) . f ( x − n ) .

6.3 Section Exercises

A logarithm is an exponent. Specifically, it is the exponent to which a base b b is raised to produce a given value. In the expressions given, the base b b has the same value. The exponent, y , y , in the expression b y b y can also be written as the logarithm, log b x , log b x , and the value of x x is the result of raising b b to the power of y . y .

Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation b y = x , b y = x , and then properties of exponents can be applied to solve for x . x .

The natural logarithm is a special case of the logarithm with base b b in that the natural log always has base e . e . Rather than notating the natural logarithm as log e ( x ) , log e ( x ) , the notation used is ln ( x ) . ln ( x ) .

a c = b a c = b

x y = 64 x y = 64

15 b = a 15 b = a

13 a = 142 13 a = 142

e n = w e n = w

log c ( k ) = d log c ( k ) = d

log 19 y = x log 19 y = x

log n ( 103 ) = 4 log n ( 103 ) = 4

log y ( 39 100 ) = x log y ( 39 100 ) = x

ln ( h ) = k ln ( h ) = k

x = 2 − 3 = 1 8 x = 2 − 3 = 1 8

x = 3 3 = 27 x = 3 3 = 27

x = 9 1 2 = 3 x = 9 1 2 = 3

x = 6 − 3 = 1 216 x = 6 − 3 = 1 216

x = e 2 x = e 2

14.125 14.125

2 . 7 0 8 2 . 7 0 8

0.151 0.151

No, the function has no defined value for x = 0. x = 0. To verify, suppose x = 0 x = 0 is in the domain of the function f ( x ) = log ( x ) . f ( x ) = log ( x ) . Then there is some number n n such that n = log ( 0 ) . n = log ( 0 ) . Rewriting as an exponential equation gives: 10 n = 0 , 10 n = 0 , which is impossible since no such real number n n exists. Therefore, x = 0 x = 0 is not the domain of the function f ( x ) = log ( x ) . f ( x ) = log ( x ) .

Yes. Suppose there exists a real number x x such that ln x = 2. ln x = 2. Rewriting as an exponential equation gives x = e 2 , x = e 2 , which is a real number. To verify, let x = e 2 . x = e 2 . Then, by definition, ln ( x ) = ln ( e 2 ) = 2. ln ( x ) = ln ( e 2 ) = 2.

No; ln ( 1 ) = 0 , ln ( 1 ) = 0 , so ln ( e 1.725 ) ln ( 1 ) ln ( e 1.725 ) ln ( 1 ) is undefined.

6.4 Section Exercises

Since the functions are inverses, their graphs are mirror images about the line y = x . y = x . So for every point ( a , b ) ( a , b ) on the graph of a logarithmic function, there is a corresponding point ( b , a ) ( b , a ) on the graph of its inverse exponential function.

Shifting the function right or left and reflecting the function about the y-axis will affect its domain.

No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.

Domain: ( − ∞ , 1 2 ) ; ( − ∞ , 1 2 ) ; Range: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Domain: ( − 17 4 , ∞ ) ; ( − 17 4 , ∞ ) ; Range: ( − ∞ , ∞ ) ( − ∞ , ∞ )

Domain: ( 5 , ∞ ) ; ( 5 , ∞ ) ; Vertical asymptote: x = 5 x = 5

Domain: ( − 1 3 , ∞ ) ; ( − 1 3 , ∞ ) ; Vertical asymptote: x = − 1 3 x = − 1 3

Domain: ( − 3 , ∞ ) ; ( − 3 , ∞ ) ; Vertical asymptote: x = − 3 x = − 3

Domain: ( 3 7 , ∞ ) ( 3 7 , ∞ ) ; Vertical asymptote: x = 3 7 x = 3 7 ; End behavior: as x → ( 3 7 ) + , f ( x ) → − ∞ x → ( 3 7 ) + , f ( x ) → − ∞ and as x → ∞ , f ( x ) → ∞ x → ∞ , f ( x ) → ∞

Domain: ( − 3 , ∞ ) ( − 3 , ∞ ) ; Vertical asymptote: x = − 3 x = − 3 ; End behavior: as x → − 3 + x → − 3 + , f ( x ) → − ∞ f ( x ) → − ∞ and as x → ∞ x → ∞ , f ( x ) → ∞ f ( x ) → ∞

Domain: ( 1 , ∞ ) ; ( 1 , ∞ ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 1 ; x = 1 ; x -intercept: ( 5 4 , 0 ) ; ( 5 4 , 0 ) ; y -intercept: DNE

Domain: ( − ∞ , 0 ) ; ( − ∞ , 0 ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 0 ; x = 0 ; x -intercept: ( − e 2 , 0 ) ; ( − e 2 , 0 ) ; y -intercept: DNE

Domain: ( 0 , ∞ ) ; ( 0 , ∞ ) ; Range: ( − ∞ , ∞ ) ; ( − ∞ , ∞ ) ; Vertical asymptote: x = 0 ; x = 0 ; x -intercept: ( e 3 , 0 ) ; ( e 3 , 0 ) ; y -intercept: DNE

f ( x ) = log 2 ( − ( x − 1 ) ) f ( x ) = log 2 ( − ( x − 1 ) )

f ( x ) = 3 log 4 ( x + 2 ) f ( x ) = 3 log 4 ( x + 2 )

x = 2 x = 2

x ≈ 2 .303 x ≈ 2 .303

x ≈ − 0.472 x ≈ − 0.472

The graphs of f ( x ) = log 1 2 ( x ) f ( x ) = log 1 2 ( x ) and g ( x ) = − log 2 ( x ) g ( x ) = − log 2 ( x ) appear to be the same; Conjecture: for any positive base b ≠ 1 , b ≠ 1 , log b ( x ) = − log 1 b ( x ) . log b ( x ) = − log 1 b ( x ) .

Recall that the argument of a logarithmic function must be positive, so we determine where x + 2 x − 4 > 0 x + 2 x − 4 > 0 . From the graph of the function f ( x ) = x + 2 x − 4 , f ( x ) = x + 2 x − 4 , note that the graph lies above the x -axis on the interval ( − ∞ , − 2 ) ( − ∞ , − 2 ) and again to the right of the vertical asymptote, that is ( 4 , ∞ ) . ( 4 , ∞ ) . Therefore, the domain is ( − ∞ , − 2 ) ∪ ( 4 , ∞ ) . ( − ∞ , − 2 ) ∪ ( 4 , ∞ ) .

6.5 Section Exercises

Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, log b ( x 1 n ) = 1 n log b ( x ) . log b ( x 1 n ) = 1 n log b ( x ) .

log b ( 2 ) + log b ( 7 ) + log b ( x ) + log b ( y ) log b ( 2 ) + log b ( 7 ) + log b ( x ) + log b ( y )

log b ( 13 ) − log b ( 17 ) log b ( 13 ) − log b ( 17 )

− k ln ( 4 ) − k ln ( 4 )

ln ( 7 x y ) ln ( 7 x y )

log b ( 4 ) log b ( 4 )

log b ( 7 ) log b ( 7 )

15 log ( x ) + 13 log ( y ) − 19 log ( z ) 15 log ( x ) + 13 log ( y ) − 19 log ( z )

3 2 log ( x ) − 2 log ( y ) 3 2 log ( x ) − 2 log ( y )

8 3 log ( x ) + 14 3 log ( y ) 8 3 log ( x ) + 14 3 log ( y )

ln ( 2 x 7 ) ln ( 2 x 7 )

log ( x z 3 y ) log ( x z 3 y )

log 7 ( 15 ) = ln ( 15 ) ln ( 7 ) log 7 ( 15 ) = ln ( 15 ) ln ( 7 )

log 11 ( 5 ) = log 5 ( 5 ) log 5 ( 11 ) = 1 b log 11 ( 5 ) = log 5 ( 5 ) log 5 ( 11 ) = 1 b

log 11 ( 6 11 ) = log 5 ( 6 11 ) log 5 ( 11 ) = log 5 ( 6 ) − log 5 ( 11 ) log 5 ( 11 ) = a − b b = a b − 1 log 11 ( 6 11 ) = log 5 ( 6 11 ) log 5 ( 11 ) = log 5 ( 6 ) − log 5 ( 11 ) log 5 ( 11 ) = a − b b = a b − 1

2.81359 2.81359

0.93913 0.93913

− 2.23266 − 2.23266

x = 4 ; x = 4 ; By the quotient rule: log 6 ( x + 2 ) − log 6 ( x − 3 ) = log 6 ( x + 2 x − 3 ) = 1. log 6 ( x + 2 ) − log 6 ( x − 3 ) = log 6 ( x + 2 x − 3 ) = 1.

Rewriting as an exponential equation and solving for x : x :

6 1 = x + 2 x − 3 0 = x + 2 x − 3 − 6 0 = x + 2 x − 3 − 6 ( x − 3 ) ( x − 3 ) 0 = x + 2 − 6 x + 18 x − 3 0 = x − 4 x − 3 ​ x = 4 6 1 = x + 2 x − 3 0 = x + 2 x − 3 − 6 0 = x + 2 x − 3 − 6 ( x − 3 ) ( x − 3 ) 0 = x + 2 − 6 x + 18 x − 3 0 = x − 4 x − 3 ​ x = 4

Checking, we find that log 6 ( 4 + 2 ) − log 6 ( 4 − 3 ) = log 6 ( 6 ) − log 6 ( 1 ) log 6 ( 4 + 2 ) − log 6 ( 4 − 3 ) = log 6 ( 6 ) − log 6 ( 1 ) is defined, so x = 4. x = 4.

Let b b and n n be positive integers greater than 1. 1. Then, by the change-of-base formula, log b ( n ) = log n ( n ) log n ( b ) = 1 log n ( b ) . log b ( n ) = log n ( n ) log n ( b ) = 1 log n ( b ) .

6.6 Section Exercises

Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.

The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

x = − 1 3 x = − 1 3

n = − 1 n = − 1

b = 6 5 b = 6 5

x = 10 x = 10

No solution

p = log ( 17 8 ) − 7 p = log ( 17 8 ) − 7

k = − ln ( 38 ) 3 k = − ln ( 38 ) 3

x = ln ( 38 3 ) − 8 9 x = ln ( 38 3 ) − 8 9

x = ln 12 x = ln 12

x = ln ( 3 5 ) − 3 8 x = ln ( 3 5 ) − 3 8

no solution

x = ln ( 3 ) x = ln ( 3 )

10 − 2 = 1 100 10 − 2 = 1 100

n = 49 n = 49

k = 1 36 k = 1 36

x = 9 − e 8 x = 9 − e 8

n = 1 n = 1

x = ± 10 3 x = ± 10 3

x = 0 x = 0

x = 3 4 x = 3 4

x = 9 x = 9

x = e 2 3 ≈ 2.5 x = e 2 3 ≈ 2.5

x = − 5 x = − 5

x = e + 10 4 ≈ 3.2 x = e + 10 4 ≈ 3.2

x = 11 5 ≈ 2.2 x = 11 5 ≈ 2.2

x = 101 11 ≈ 9.2 x = 101 11 ≈ 9.2

about $ 27 , 710.24 $ 27 , 710.24

about 5 years

ln ( 17 ) 5 ≈ 0.567 ln ( 17 ) 5 ≈ 0.567

x = log ( 38 ) + 5 log ( 3 )    4 log ( 3 ) ≈ 2.078 x = log ( 38 ) + 5 log ( 3 )    4 log ( 3 ) ≈ 2.078

x ≈ 2.2401 x ≈ 2.2401

x ≈ − 44655 . 7143 x ≈ − 44655 . 7143

about 5.83 5.83

t = ln ( ( y A ) 1 k ) t = ln ( ( y A ) 1 k )

t = ln ( ( T − T s T 0 − T s ) − 1 k ) t = ln ( ( T − T s T 0 − T s ) − 1 k )

6.7 Section Exercises

Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.

Doubling time is a measure of growth and is thus associated with exponential growth models. The doubling time of a substance or quantity is the amount of time it takes for the initial amount of that substance or quantity to double in size.

An order of magnitude is the nearest power of ten by which a quantity exponentially grows. It is also an approximate position on a logarithmic scale; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ by a great amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as saying that the mass of Saturn is about 10 2 10 2 times, or 2 orders of magnitude greater, than the mass of Earth.

f ( 0 ) ≈ 16.7 ; f ( 0 ) ≈ 16.7 ; The amount initially present is about 16.7 units.

exponential; f ( x ) = 1.2 x f ( x ) = 1.2 x

logarithmic

about 1.4 1.4 years

about 7.3 7.3 years

4 4 half-lives; 8.18 8.18 minutes

M = 2 3 log ( S S 0 ) log ( S S 0 ) = 3 2 M S S 0 = 10 3 M 2 S = S 0 10 3 M 2 M = 2 3 log ( S S 0 ) log ( S S 0 ) = 3 2 M S S 0 = 10 3 M 2 S = S 0 10 3 M 2

Let y = b x y = b x for some non-negative real number b b such that b ≠ 1. b ≠ 1. Then,

ln ( y ) = ln ( b x ) ln ( y ) = x ln ( b ) e ln ( y ) = e x ln ( b )             y = e x ln ( b ) ln ( y ) = ln ( b x ) ln ( y ) = x ln ( b ) e ln ( y ) = e x ln ( b )             y = e x ln ( b )

A = 125 e ( − 0.3567 t ) ; A ≈ 43 A = 125 e ( − 0.3567 t ) ; A ≈ 43 mg

about 60 60 days

A ( t ) = 250 e ( − 0.00822 t ) ; A ( t ) = 250 e ( − 0.00822 t ) ; half-life: about 84 84 minutes

r ≈ − 0.0667 , r ≈ − 0.0667 , So the hourly decay rate is about 6.67 % 6.67 %

f ( t ) = 1350 e ( 0.03466 t ) ; f ( t ) = 1350 e ( 0.03466 t ) ; after 3 hours: P ( 180 ) ≈ 691 , 200 P ( 180 ) ≈ 691 , 200

f ( t ) = 256 e ( 0.068110 t ) ; f ( t ) = 256 e ( 0.068110 t ) ; doubling time: about 10 10 minutes

about 88 88 minutes

T ( t ) = 90 e ( − 0.008377 t ) + 75 , T ( t ) = 90 e ( − 0.008377 t ) + 75 , where t t is in minutes.

about 113 113 minutes

log ( x ) = 1.5 ; x ≈ 31.623 log ( x ) = 1.5 ; x ≈ 31.623

MMS magnitude: 5.82 5.82

N ( 3 ) ≈ 71 N ( 3 ) ≈ 71

6.8 Section Exercises

Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic model best describes populations.

Regression analysis is the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined, select the appropriate regression analysis command from the STAT then CALC menu.

The y -intercept on the graph of a logistic equation corresponds to the initial population for the population model.

P ( 0 ) = 22 P ( 0 ) = 22 ; 175

p ≈ 2.67 p ≈ 2.67

y -intercept: ( 0 , 15 ) ( 0 , 15 )

about 6.8 6.8 months.

About 38 wolves

About 8.7 years

f ( x ) = 776.682 ( 1.426 ) x f ( x ) = 776.682 ( 1.426 ) x

f ( x ) = 731.92 e -0.3038 x f ( x ) = 731.92 e -0.3038 x

When f ( x ) = 250 , x ≈ 3.6 f ( x ) = 250 , x ≈ 3.6

y = 5.063 + 1.934 log ( x ) y = 5.063 + 1.934 log ( x )

When f ( 10 ) ≈ 2.3 f ( 10 ) ≈ 2.3

When f ( x ) = 8 , x ≈ 0.82 f ( x ) = 8 , x ≈ 0.82

f ( x ) = 25.081 1 + 3.182 e − 0.545 x f ( x ) = 25.081 1 + 3.182 e − 0.545 x

When f ( x ) = 68 , x ≈ 4.9 f ( x ) = 68 , x ≈ 4.9

f ( x ) = 1.034341 ( 1.281204 ) x f ( x ) = 1.034341 ( 1.281204 ) x ; g ( x ) = 4.035510 g ( x ) = 4.035510 ; the regression curves are symmetrical about y = x y = x , so it appears that they are inverse functions.

f − 1 ( x ) = ln ( a ) - ln ( c x - 1 ) b f − 1 ( x ) = ln ( a ) - ln ( c x - 1 ) b

Review Exercises

exponential decay; The growth factor, 0.825 , 0.825 , is between 0 0 and 1. 1.

y = 0.25 ( 3 ) x y = 0.25 ( 3 ) x

$ 42 , 888.18 $ 42 , 888.18

continuous decay; the growth rate is negative.

domain: all real numbers; range: all real numbers strictly greater than zero; y -intercept: (0, 3.5);

g ( x ) = 7 ( 6.5 ) − x ; g ( x ) = 7 ( 6.5 ) − x ; y -intercept: ( 0 , 7 ) ; ( 0 , 7 ) ; Domain: all real numbers; Range: all real numbers greater than 0. 0.

17 x = 4913 17 x = 4913

log a b = − 2 5 log a b = − 2 5

x = 64 1 3 = 4 x = 64 1 3 = 4

log ( 0 .000001 ) = − 6 log ( 0 .000001 ) = − 6

ln ( e − 0.8648 ) = − 0.8648 ln ( e − 0.8648 ) = − 0.8648

Domain: x > − 5 ; x > − 5 ; Vertical asymptote: x = − 5 ; x = − 5 ; End behavior: as x → − 5 + , f ( x ) → − ∞ x → − 5 + , f ( x ) → − ∞ and as x → ∞ , f ( x ) → ∞ . x → ∞ , f ( x ) → ∞ .

log 8 ( 65 x y ) log 8 ( 65 x y )

ln ( z x y ) ln ( z x y )

log y ( 12 ) log y ( 12 )

ln ( 2 ) + ln ( b ) + ln ( b + 1 ) − ln ( b − 1 ) 2 ln ( 2 ) + ln ( b ) + ln ( b + 1 ) − ln ( b − 1 ) 2

log 7 ( v 3 w 6 u 3 ) log 7 ( v 3 w 6 u 3 )

x = log ( 125 ) log ( 5 ) + 17 12 = 5 3 x = log ( 125 ) log ( 5 ) + 17 12 = 5 3

x = − 3 x = − 3

x = ln ( 11 ) x = ln ( 11 )

a = e 4 − 3 a = e 4 − 3

x = ± 9 5 x = ± 9 5

about 5.45 5.45 years

f − 1 ( x ) = 2 4 x − 1 3 f − 1 ( x ) = 2 4 x − 1 3

f ( t ) = 300 ( 0.83 ) t ; f ( t ) = 300 ( 0.83 ) t ; f ( 24 ) ≈ 3.43     g f ( 24 ) ≈ 3.43     g

about 45 45 minutes

about 8.5 8.5 days

exponential

y = 4 ( 0.2 ) x ; y = 4 ( 0.2 ) x ; y = 4 e -1.609438 x y = 4 e -1.609438 x

about 7.2 7.2 days

logarithmic; y = 16.68718 − 9.71860 ln ( x ) y = 16.68718 − 9.71860 ln ( x )

Practice Test

About 13 dolphins.

$ 1,947 $ 1,947

y -intercept: ( 0 , 5 ) ( 0 , 5 )

8.5 a = 614.125 8.5 a = 614.125

x = ( 1 7 ) 2 = 1 49 x = ( 1 7 ) 2 = 1 49

ln ( 0.716 ) ≈ − 0.334 ln ( 0.716 ) ≈ − 0.334

Domain: x < 3 ; x < 3 ; Vertical asymptote: x = 3 ; x = 3 ; End behavior: x → 3 − , f ( x ) → − ∞ x → 3 − , f ( x ) → − ∞ and x → − ∞ , f ( x ) → ∞ x → − ∞ , f ( x ) → ∞

log t ( 12 ) log t ( 12 )

3 ln ( y ) + 2 ln ( z ) + ln ( x − 4 ) 3 3 ln ( y ) + 2 ln ( z ) + ln ( x − 4 ) 3

x = ln ( 1000 ) ln ( 16 ) + 5 3 ≈ 2.497 x = ln ( 1000 ) ln ( 16 ) + 5 3 ≈ 2.497

a = ln ( 4 ) + 8 10 a = ln ( 4 ) + 8 10

x = ln ( 9 ) x = ln ( 9 )

x = ± 3 3 2 x = ± 3 3 2

f ( t ) = 112 e − .019792 t ; f ( t ) = 112 e − .019792 t ; half-life: about 35 35 days

T ( t ) = 36 e − 0.025131 t + 35 ; T ( 60 ) ≈ 43 o F T ( t ) = 36 e − 0.025131 t + 35 ; T ( 60 ) ≈ 43 o F

exponential; y = 15.10062 ( 1.24621 ) x y = 15.10062 ( 1.24621 ) x

logistic; y = 18.41659 1 + 7.54644 e − 0.68375 x y = 18.41659 1 + 7.54644 e − 0.68375 x

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Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Algebra and Trigonometry
• Publication date: Feb 13, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/algebra-and-trigonometry/pages/chapter-6

© Dec 8, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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