64 introduction to logarithm

The Logarithmic Functions

There are three numbers in an exponential notation.The Logarithmic Functions

4 3 = 64


the base4 3 = 64


the exponent

the base4 3 = 64


the exponent

the base

the output

4 3 = 64

There are three numbers in an exponential notation.

Given the above expression, we say that “(base) 4 raised to the exponent (power) 3 gives 64”.


the exponent

the base

the output

4 3 = 64


the exponent

the base

the output

4 3 = 64

Given the above expression, we say that “(base) 4 raised to the exponent (power) 3 gives 64”. The focus of the above statement is that when 43 is executed, the output is 64.


the exponent

the base

the output

4 3 = 64

However if we are given the output is 64 from raising 4 to a power,



the exponent

the base

the output

4 3 = 64

However if we are given the output is 64 from raising 4 to a power,

the power

the base the output

4 = 643



the exponent

the base

the output

4 3 = 64

However if we are given the output is 64 from raising 4 to a power, then the needed power is called log4(64)

the power = log4(64)

the base the output

4 = 643



the exponent

the base

the output

4 3 = 64

However if we are given the output is 64 from raising 4 to a power, then the needed power is called log4(64) which is 3.


the base the output

4 = 643



the exponent

the base

the output

4 3 = 64

However if we are given the output is 64 from raising 4 to a power, then the needed power is called log4(64) which is 3.


the base the output

4 = 643

or that log4(64) = 3 and we say that “log–base–4 of 64 is 3”.


The Logarithmic FunctionsJust as the sentence “Bart's dad is Homer.” contains the same information as “Homer's son is Bart.”,

The Logarithmic FunctionsJust as the sentence “Bart's dad is Homer.” contains the same information as “Homer's son is Bart.” The expression“64 = 43” contains the same information as“log4(64) = 3”.


The expression “64 = 43” is called the exponential form and “log4(64) = 3” is called the logarithmic form of the expressed relation.


The expression “64 = 43” is called the exponential form and “log4(64) = 3” is called the logarithmic form of the expressed relation. In general, we say that “log–base–b of y is x” or logb(y) = x


The expression “64 = 43” is called the exponential form and “log4(64) = 3” is called the logarithmic form of the expressed relation. In general, we say that “log–base–b of y is x” or logb(y) = x if y = bx (b > 0).


The expression “64 = 43” is called the exponential form and “log4(64) = 3” is called the logarithmic form of the expressed relation. In general, we say that “log–base–b of y is x” or logb(y) = x if y = bx (b > 0).

the power = logb(y)

the base the output

b = yx


The expression “64 = 43” is called the exponential form and “log4(64) = 3” is called the logarithmic form of the expressed relation. In general, we say that “log–base–b of y is x” or logb(y) = x if y = bx (b > 0),i.e. logb(y) is the exponent x.

the power = logb(y)

the base the output

b = yx

The Logarithmic FunctionsWhen working with the exponential form or the logarithmic expressions, always identify the base number b first.

The Logarithmic FunctionsWhen working with the exponential form or the logarithmic expressions, always identify the base number b first. All the following exponential expressions yield 64.


43 → 64

82 → 64

26 → 64

exp–form

The Logarithmic FunctionsWhen working with the exponential form or the logarithmic expressions, always identify the base number b first. All the following exponential expressions yield 64. Their corresponding log–form are differentiated by the bases and the different exponents required.

43 → 64

82 → 64

26 → 64

exp–form log–form


43 → 64

82 → 64

26 → 64

log4(64)

log8(64)

log2(64)

exp–form log–formTheir corresponding log–form are differentiated by the bases and the different exponents required.

The Logarithmic FunctionsWhen working with the exponential form or the logarithmic expressions, always identify the base number b first. All the following exponential expressions yield 64. Their corresponding log–form are differentiated by the bases and the different exponents required.

43 → 64

82 → 64

26 → 64

log4(64) →

log8(64) →

log2(64) →



43 → 64

82 → 64

26 → 64

log4(64) → 3

log8(64) →

log2(64) →



43 → 64

82 → 64

26 → 64

log4(64) → 3

log8(64) → 2

log2(64) →



43 → 64

82 → 64

26 → 64

log4(64) → 3

log8(64) → 2

log2(64) → 6



43 → 64

82 → 64

26 → 64

log4(64) → 3

log8(64) → 2

log2(64) → 6


Both numbers b and y appeared in the log notation “logb(y)” must be positive.


43 → 64

82 → 64

26 → 64

log4(64) → 3

log8(64) → 2

log2(64) → 6


Both numbers b and y appeared in the log notation “logb(y)” must be positive. Switch to x as the input, the domain of logb(x) is the set D = {x l x > 0 }.


43 → 64

82 → 64

26 → 64

log4(64) → 3

log8(64) → 2

log2(64) → 6


Both numbers b and y appeared in the log notation “logb(y)” must be positive. Switch to x as the input, the domain of logb(x) is the set D = {x l x > 0 }.We would get an error message if we execute log2(–1) with software.

The Logarithmic FunctionsTo convert the exp-form to the log–form:

b = yx


b = yx

logb( y ) = x→Identity the base and the correct log–function


b = yx

logb( y ) = x→insert the exponential output.


b = yx

logb( y ) = x→The log–output is the required exponent.


Example A. Rewrite the exp-form into the log-form.

a. 42 = 16 b. w = u2+v

To convert the exp-form to the log–form:

b = yx

logb( y ) = x→



a. 42 = 16 log4(16) = 2 b. w = u2+v


b = yx

logb( y ) = x→



a. 42 = 16 log4(16) = 2 b. w = u2+v


b = yx

logb( y ) = x→



a. 42 = 16 log4(16) = 2 b. w = u2+v


b = yx

logb( y ) = x→



a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→



a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→



a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→



a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

→

To convert the log–form to the exp–form:

logb( y ) = x

logb( y ) = x



a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

→


b = yx

logb( y ) = x→

logb( y ) = x



a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

→


b = yx

logb( y ) = x→

logb( y ) = x



a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

→


b = yx

logb( y ) = x→

logb( y ) = x


Example B. Rewrite the log-form into the exp-form.

a. log3(1/9) = –2b. 2w = logv(a – b)


a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→


b = yx

logb( y ) = x→



a. log3(1/9) = –2 3-2 = 1/9b. 2w = logv(a – b)


a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→


b = yx

logb( y ) = x→



a. log3(1/9) = –2 3-2 = 1/9b. 2w = logv(a – b)


a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→


b = yx

logb( y ) = x→



a. log3(1/9) = –2 3-2 = 1/9b. 2w = logv(a – b)


a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→


b = yx

logb( y ) = x→



a. log3(1/9) = –2 3-2 = 1/9b. 2w = logv(a – b) v2w = a – b


a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→


b = yx

logb( y ) = x→





a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→


b = yx

logb( y ) = x→





a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→


b = yx

logb( y ) = x→





a. 42 = 16 log4(16) = 2 b. w = u2+v logu(w) = 2+v


b = yx

logb( y ) = x→


b = yx

logb( y ) = x→

The output of logb(x), i.e. the exponent in the defined relation, may be positive or negative.

The Logarithmic FunctionsExample C. a. Rewrite the exp-form into the log-form.

4–3 = 1/64

8–2 = 1/64

log4(1/64) = –3

log8(1/64) = –2


b. Rewrite the log-form into the exp-form.

(1/2)–2 = 4log1/2(4) = –2

log1/2(8) = –3

exp–formlog–form

(1/2)–3 = 8


The Common Log and the Natural Log

Example C. a. Rewrite the exp-form into the log-form.

4–3 = 1/64

8–2 = 1/64

log4(1/64) = –3

log8(1/64) = –2



(1/2)–2 = 4log1/2(4) = –2

log1/2(8) = –3


(1/2)–3 = 8


Base 10 is called the common base.The Common Log and the Natural Log


4–3 = 1/64

8–2 = 1/64

log4(1/64) = –3

log8(1/64) = –2



(1/2)–2 = 4log1/2(4) = –2

log1/2(8) = –3


(1/2)–3 = 8


Base 10 is called the common base. Log with base10,. written as log(x) without the base number b, is called the common log,



4–3 = 1/64

8–2 = 1/64

log4(1/64) = –3

log8(1/64) = –2



(1/2)–2 = 4log1/2(4) = –2

log1/2(8) = –3


(1/2)–3 = 8


Base 10 is called the common base. Log with base10,. written as log(x) without the base number b, is called the common log, i.e. log(x) is log10(x).



4–3 = 1/64

8–2 = 1/64

log4(1/64) = –3

log8(1/64) = –2



(1/2)–2 = 4log1/2(4) = –2

log1/2(8) = –3


(1/2)–3 = 8

Base e is called the natural base. The Common Log and the Natural Log

Base e is called the natural base. Log with base e is written as ln(x) and it’s called the natural log,


Base e is called the natural base. Log with base e is written as ln(x) and it’s called the natural log, i.e. In(x) is loge(x).




Example D. Convert to the other form. exp-form log-form

103 = 1000

ln(1/e2) = -2

ert =

log(1) = 0

AP




103 = 1000 log(1000) = 3

ln(1/e2) = -2

ert =

log(1) = 0

AP




103 = 1000 log(1000) = 3

e-2 = 1/e2 ln(1/e2) = -2

ert =

log(1) = 0

AP




103 = 1000 log(1000) = 3

e-2 = 1/e2 ln(1/e2) = -2

ert = ln( ) = rt

log(1) = 0

AP

AP




103 = 1000 log(1000) = 3

e-2 = 1/e2 ln(1/e2) = -2

ert = ln( ) = rt

100 = 1 log(1) = 0

AP

AP




103 = 1000 log(1000) = 3

e-2 = 1/e2 ln(1/e2) = -2

ert = ln( ) = rt

100 = 1 log(1) = 0

AP

AP

Most log and powers can’t be computed efficiently by hand.




103 = 1000 log(1000) = 3

e-2 = 1/e2 ln(1/e2) = -2

ert = ln( ) = rt

100 = 1 log(1) = 0

AP

AP

Most log and powers can’t be computed efficiently by hand. We need a calculation device to extract numerical solutions.

The Common Log and the Natural LogExample E. Calculate each of the following logs using a calculator. Then convert the relation into the exp–form and confirm the exp–form with a calculator. a. log(50) =

The Common Log and the Natural LogExample E. Calculate each of the following logs using a calculator. Then convert the relation into the exp–form and confirm the exp–form with a calculator. a. log(50) = 1.69897...


In the exp–form, it’s101.69897 =


In the exp–form, it’s101.69897 = 49.9999995...≈50



b. ln(9) =



b. ln(9) = 2.1972245..



b. ln(9) = 2.1972245..In the exp–form, it’s e2.1972245 =



b. ln(9) = 2.1972245..In the exp–form, it’s e2.1972245 = 8.9999993...≈ 9



b. ln(9) = 2.1972245..

c. Calculate the power using a calculator. Then convert the relation into the log–form and confirm the log–form by the calculator.

e4.3 =

In the exp–form, it’s e2.1972245 = 8.9999993...≈ 9



b. ln(9) = 2.1972245..


e4.3 = 73.699793..




b. ln(9) = 2.1972245..


e4.3 = 73.699793..→ In(73.699793) =




b. ln(9) = 2.1972245..


e4.3 = 73.699793..→ In(73.699793) = 4.299999..≈ 4.3




b. ln(9) = 2.1972245..


e4.3 = 73.699793..→ In(73.699793) = 4.299999..≈ 4.3

Your turn. Follow the instructions in part c for 10π.


Equation may be formed with log–notation. The Common Log and the Natural Log

Equation may be formed with log–notation. Often we need to restate them in the exp–form.


Equation may be formed with log–notation. Often we need to restate them in the exp–form. We say we "drop the log" when this step is taken.


Example F. Solve for x

a. log9(x) = –1




a. log9(x) = –1Drop the log and get x = 9–1.




a. log9(x) = –1Drop the log and get x = 9–1. So x = 1/9





b. logx(9) = –2





b. logx(9) = –2

Drop the log and get 9 = x–2,





b. logx(9) = –2

Drop the log and get 9 = x–2, i.e. 9 = 1x2





b. logx(9) = –2

Drop the log and get 9 = x–2, i.e. 9 = So 9x2 = 1

1x2





b. logx(9) = –2

Drop the log and get 9 = x–2, i.e. 9 = So 9x2 = 1 x2 = 1/9 x = 1/3 or x= –1/3

1x2





b. logx(9) = –2

Drop the log and get 9 = x–2, i.e. 9 = So 9x2 = 1 x2 = 1/9 x = 1/3 or x= –1/3Since the base b > 0, so x = 1/3 is the only solution.

1x2




Graphs of the Logarithmic Functions

Recall that the domain of logb(x) is the set of all x > 0.



1/4

1/2

1

2

4

8

x y=log2(x)

Recall that the domain of logb(x) is the set of all x > 0.Hence to make a table to plot the graph of y = log2(x), we only select positive x’s.



2

4

8

x y=log2(x)

Recall that the domain of logb(x) is the set of all x > 0.Hence to make a table to plot the graph of y = log2(x), we only select positive x’s. In particular we select x’s related to base 2 for easy computation of the y’s.



1/4

1/2

1

2

4

8

x y=log2(x)




1/4 -21/2

1

2

4

8

x y=log2(x)




1/4 -21/2 -1

1

2

4

8

x y=log2(x)




1/4 -21/2 -1

1 0

2

4

8

x y=log2(x)




1/4 -21/2 -1

1 0

2 1

4

8

x y=log2(x)




1/4 -21/2 -1

1 0

2 1

4 2

8 3

x y=log2(x)



(1, 0)

(2, 1)(4, 2)

(8, 3)(16, 4)

(1/2, -1)

(1/4, -2)y=log2(x)


1/4 -21/2 -1

1 0

2 1

4 2

8 3

x y=log2(x)


No G

raph Zone

x

y

The Logarithmic FunctionsTo graph log with base b = ½, we havelog1/2(4) = –2, log1/2(8) = –3, log1/2(16) = –4


x

y

(1, 0)

(8, -3)

To graph log with base b = ½, we havelog1/2(4) = –2, log1/2(8) = –3, log1/2(16) = –4

(4, -2)

(16, -4)

y = log1/2(x)

No G

raph Zone


x

y

(1, 0)

(8, -3)

To graph log with base b = ½, we havelog1/2(4) = –2, log1/2(8) = –3, log1/2(16) = –4

(4, -2)

(16, -4)

y = log1/2(x)

x x

y

(1, 0)(1, 0)

y = logb(x), b > 1

y = logb(x), 1 > b

Here are the general shapes of log–functions. yN

o Graph Z

one

No G

raph Zone

No G

raph Zone

(b, 1)(b, 1)

Documents

64 introduction to logarithm