Download pptx - Chapter 2 Functions and Graphs Section 6 Logarithmic Functions (Part I)

Chapter 2

Functions and Graphs

Section 6

Logarithmic Functions

(Part I)

2Barnett/Ziegler/Byleen Business Calculus 12e

Learning Objectives for Section 2.6 Logarithmic Functions

The student will be able to:• Identify the graphs of one-to-one functions.• Use and apply inverse functions.• Evaluate logarithms.• Rewrite log as exponential functions and vice versa.


One to One Functions

Definition: A function f is said to be one-to-one if no x or y values are represented more than once.• One-to-one:

• Not one-to-one:

The graph of a one-to-one function passes both the vertical and horizontal line tests.


Which Functions Are One to One?

-30

-20

-10

0

10

20

30

40

-4 -2 0 2 40

2

4

6

8

10

12

-4 -2 0 2 4

One-to-one

NOT One-to-one


Definition of Inverse Function

If f is a one-to-one function, then the inverse of f is the function formed by interchanging the x and y coordinates for f. Thus, if (a, b) is a point on the graph of f, then (b, a) is a point on the graph of the inverse of f.• Let • Then

The domain of f becomes the range of . The range of f becomes the domain of . Note: If a function is not one-to-one then f does not have

an inverse.

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Finding the Inverse Function

Given the equation of a one-to-one function f, you can find algebraically by exchanging x for y and solving for y.

Example: Find

Barnett/Ziegler/Byleen Business Calculus 12e

𝑦=− 2 x −3 𝑥=−2 y −3𝑥+3=−2 y𝑥+3− 2

= y

𝑓 −1 (𝑥 )=𝑥+3− 2

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Graphs of f and f-1

The graphs of and are reflections of each other over the line

If you know how to graph then simply take a few key points and switch their x and y coordinates to help you graph .

Or find the equation of algebraically first, then graph it.


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Graphs of f and f-1

Graph and (from the previous example) on the same coordinate plane.


𝑓 (𝑥 )=−2 x− 3

𝑓 −1 (𝑥 )=−12𝑥−

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𝑓

𝑓 −1

𝑓 −1 (𝑥 )=𝑥+3− 2

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Graphs of f and f-1

The graph of is shown. Graph .


𝑦=𝑥

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Exponential functions are one-to-one because they pass the vertical and horizontal line tests.



𝑦=2𝑥


Inverse of an Exponential Function

Start with the exponential function: Now, interchange x and y:

Solving for y:

The inverse of an exponential function is a log function.

𝑓 (𝑥 )=2𝑥 𝑓 −1 (𝑥 )=𝑙𝑜𝑔2𝑥

𝑦=𝑙𝑜𝑔2𝑥

𝑥=2𝑦𝑦=2𝑥

This is called a logarithmic function .


Logarithmic Function

The inverse of an exponential function is called a logarithmic function. For b > 0 and b 1,

𝑓 (𝑥)=𝑏𝑥 𝑓 −1 (𝑥 )=𝑙𝑜𝑔𝑏 𝑥𝐷𝑜𝑚𝑎𝑖𝑛 : (− ∞ , ∞ )𝑅𝑎𝑛𝑔𝑒 : (0 , ∞ ) 𝑅𝑎𝑛𝑔𝑒 : (− ∞ ,∞ )

𝐷𝑜𝑚𝑎𝑖𝑛 : ( 0 , ∞ )

𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 h𝐿𝑜𝑔𝑎𝑟𝑖𝑡 𝑚𝑖𝑐


Graphs

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Transformations

Parent function: Children:

• Shifted up 2

• Shifted right 5

• Shifted down 3 and left 7


𝑦=𝑙𝑜𝑔𝑏𝑥+2

𝑦=𝑙𝑜𝑔𝑏(𝑥−5)

𝑦=𝑙𝑜𝑔𝑏 (𝑥+7 ) −3

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Log Notation

Common Log• log base 10• When no base is specified, it’s base 10•

Natural Log• log base e


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Simple Logs

Evaluate each log expression without a calculator:


10−321−27100

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Log Exponential

Think of the word “log” as meaning “exponent on base b” To convert a log equation to an exponential equation:

• What’s the base?• What’s the exponent?• Write the equation


𝑦=𝑙𝑜𝑔327

𝟑𝒚𝟑𝒚=𝟐𝟕

𝑙𝑜𝑔3 27=𝑦


Converting a log into an exponential expression: 1.

2.

Log Exponential

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Exponential Log

To convert an exponential equation to a log equation:

• What’s the base?• What’s the exponent?• Write the equation • Check:


16=2𝑦

𝟐𝒚𝒚=𝒍𝒐𝒈𝟐𝟏𝟔

𝒚=𝒍𝒐𝒈𝟐𝟏𝟔


Exponential Log

Converting an exponential into a log expression:1.

2.


Solving Simple Equations

Convert each log to an exponential equation and solve for x:

1.

2.

𝑥3=1000𝑥=3√1000𝑥=10

𝑥=777665=𝑥

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Using Your Calculator

Use your calculator to evaluate and round to 2 decimal places:


𝑙𝑛15≈ 2.71 𝑙𝑜𝑔15≈ 1.18

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Chapter 2


Section 6


(Part II)



The student will be able to:• Use log properties.• Solve log equations.• Solve exponential equations.


Properties of Logarithms

If b, M, and N are positive real numbers, b 1, and p and x are real numbers, then

5. logb

MN logb

M logb

N

6. logb

M

Nlog

bM log

bN

7. logb

M p p logb

M

8. logb

M logb

N iff M N

1. logb(1) 0

2. logb(b) 1

3. logbbx x

4. blogb x x

9. h𝐶 𝑎𝑛𝑔𝑒𝑜𝑓 𝑏𝑎𝑠𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 : log𝑏 𝑥=log 𝑥log𝑏

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Using Properties Rewrite each expression by using the appropriate log

property:

•


¿ 𝑙𝑜𝑔2205

¿ 𝑙𝑜𝑔2 4¿2¿ 𝑥𝑙𝑜𝑔5 25

¿ log10+log 𝑥¿1+ log𝑥2 𝑥+1=5 𝑥=2

¿ 𝑥+1

¿2 𝑥

¿log 19log 3

≈ 2.68


Solving Log Equations

Solve for x: log

4x 6 log

4x 6 3

log 4 (𝑥+6 ) (𝑥−6 )=3

log 4 (𝑥2− 36 )¿3

43=𝑥2−3664=𝑥2− 36

100=𝑥2

𝑥=± 10𝑥=10

x can’t be -10 because you can’t take the log of a negative number.


Solving Log Equations

Solve for x. Obtain the exact solution of this equation in terms of e.

ln (x + 1) – ln x = 1

ex = x + 1

ex - x = 1

x(e - 1) = 11

1x

e

𝑙𝑛(𝑥+1𝑥 )=1

𝑒1=(𝑥+1𝑥 )

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Solving Exponential Equations

Method 1:• Convert the exponential equation to a log equation.• Then evaluate.


9𝑥=2𝑥=𝑙𝑜𝑔92

𝑥=log 2log 9

𝐱≈𝟎 .𝟑𝟏𝟓𝟓

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Method 2:• Isolate the exponential part on one side, then take the

log or ln of both sides of the equation.• Then evaluate.


log 9𝑥= log 2

x ∙ log 9=log 2𝑥=

log 2log 9

𝐱≈𝟎 .𝟑𝟏𝟓𝟓

9𝑥=2

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Solve and round answer to 4 decimal places:


5𝑒𝑥=2

𝑒𝑥=25

𝑙𝑛𝑒𝑥=𝑙𝑛25

𝑥 ∙ 𝑙𝑛𝑒=𝑙𝑛25

𝑥=𝑙𝑛25

𝒙≈ −𝟎 .𝟗𝟏𝟔𝟑

1

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Chapter 2


Section 6


(Part III)



The student will be able to:• Solve applications involving logarithms.


Application: Finance

How long will it take money to double if compounded monthly at 4% interest?

𝐴=𝑃 (1+ 𝑟𝑛 )

𝑛𝑡

2𝑃=𝑃 (1+ 0.0412 )

(12 ∙𝑡 )

2=(1+ 0.0412 )

(12∙ 𝑡 )

ln 2=ln (1+ 0.0412 )

(12 ∙𝑡 )

ln 2=12 t ∙ ln(1+ 0.0412 )

❑

ln 2

12∙ ln (1+0.0412 )

❑=t

𝑡≈ 17.4You can take the log or

the ln of both sides.It will take about 17.4 yrs for the

money to double.


Application: Finance

Suppose you invest $1500 into an account that is compounded continuously. At the end of 10 years, you want to have a balance of $6500. What must the annual percentage rate be?

𝐴=𝑃 𝑒𝑟𝑡

6 500=1500𝑒(𝑟 ∙10)

𝑟 ≈ . 147

65001500

=𝑒(𝑟 ∙10)

ln133

=ln𝑒 (𝑟 ∙ 10)❑

ln133

=10𝑟 ∙ ln𝑒

ln133

10=𝑟

The annual percentage rate must be 14.7%


Application: Archeology

Recall from Lesson 2-5 that Carbon-14 decays according to the model:

Estimate that age of a fossil if 15% of the original amount of C-14 is still present.

0.15=1 ∙𝑒(−0.000124 ∙𝑡 )

𝑡≈ 15,299ln 0.15=ln𝑒 (−0.000124 ∙𝑡 )❑

ln 0.15=− 0.000124 𝑡 ∙ ln𝑒

ln 0.15−0.000124

=𝑡

The fossil would be 15,299 years old.

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Application: Sound Intensity

Sound intensity is measured using the formula:

I = sound intensity in watts per

= intensity of sound just below the threshold of hearing =

N = number of decibels


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Solve for N:


𝐼=𝐼 0 ∙10𝑁 /10

𝐼𝐼0

=10𝑁 /10

)

𝑙𝑜𝑔𝐼𝐼0

= (𝑁 /10 ) log 10

𝑙𝑜𝑔𝐼𝐼0

= (𝑁 /10 )

𝑁=10 ∙ 𝑙𝑜𝑔𝐼𝐼 0

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Use the formula from the previous example to find the number of decibels for the sound of heavy traffic which has a sound intensity of


𝑁=10 ∙ 𝑙𝑜𝑔𝐼𝐼 0

𝑁=10 ∙ 𝑙𝑜𝑔 10−8

10− 16

)

𝑁=10 ∙ 𝑙𝑜𝑔108

𝑁=10 ∙ 8 ∙ log 10𝑁=80 The sound of heavy traffic is

about 80 decibels.


Logarithmic Regression

When the scatter plot of a data set indicates a slowly increasing or decreasing function, a logarithmic function often provides a good model.

We use logarithmic regression on a graphing calculator to find the function of the form y = a + b*ln(x) that best fits the data.


Example of Logarithmic Regression

A cordless screwdriver is sold through a national chain of discount stores. A marketing company established the following price-demand table, where x is the number of screwdrivers in demand each month at a price of p dollars per screwdriver.

x p = D(x)

1,000 912,000 733,000 644,000 565,000 53

Find a log regression

equation to predict the price per

screwdriver if the demand reaches

6,000.



x p = D(x)

1,000 912,000 733,000 644,000 565,000 53

𝑦=256.47 − 24.04¿



Xmax=6500TraceUp arrowEnter 6000

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