Logarithmic Functions
Logarithmic Functions
For all positive real numbers x and b, where b  1, y = log b x or
f(x) = log b x , it is defined as the number such that 𝐛 log 𝐛 𝐱 = x that
is log b x is the exponent that b must be raised to produced x.
logarithm
(exponent)
exponent
number
y = log b x or 𝐟 𝐱 = 𝐥𝐨𝐠 𝐛 𝐱 means
x = by
number
base
base
KINDS OF LOGARITHM
Common Logarithm- the logarithm to the base 10 usually written
as log10 𝑥 or simply log 𝑥.
Natural Logarithm- the base e logarithm usually written as log 𝑒 𝑥
or ln 𝑥.
Transform the given exponential function to
logarithmic function(vice-versa)
Exponential Form
53 = 125
Logarithmic Form
(0.1)4 = 10,000
log101= 0
log2x= 0
8 = 23
102 = 100
  = 161
5=log232
log m = n
7a = -49
log1 2 1 = 4
16
log381 = 4
6-2 = 1
36
e2 = x
-2 = log 6 1
36
log
1
2
4
4 = (-2)2
64
=
27
4
log 3
-3
5 5=2
Seat work: Change each function into logarithmic function or
1.
2.
3.
4.
ax = y
25 = 32
200 = 1
63 = 216
1
5
1
( )4
2
4−3
5. ( )4 = 625
6.
7.
=
=
−4
1
16
1
64
8. 10 = 0.0001
9. 20.4 ≈ 1.320
10.100.5 ≈ 3.16
exponential function(vice-versa)
1. log 7 1 = 0
2. log 1000 = 3
3. log 2 M = −4
1
4. log 3 = −2
9
1
5. log 7 =-2
49
6. log 2 N = 2
7. log 5 x=-2
3
2
8. log 2 y =
9. log10 2.51 ≈ 0.4
10. log10 1.866 ≈ 0.9
Find the value of the given logarithmic function
Examples: a. log 2 16
1
28
b. log 1
d. log 2 64
1.
2.
3.
4.
log 3 81
log 2 32
log 5 125
1
log 1
16
1
log 2
8
4
5.
c. log 3
e. log 10,000
1
6. log 2
4
7. log 100
8. log 9 729
9. log 0.25 16
10. log 0.001
1
27
Graph of logarithmic function(𝑏 > 1)
f x =log 2 𝑥
𝐱
f(x)
𝟏
𝟖
-3
𝟏
𝟒
-2
𝟏
𝟐
-1
1
2
4
0
0
1
2
3
Domain:{x|x > 0}/
Range:(-∞,∞)
Graph of Logarithmic Function (0 < 𝑏 < 1)
f x = log 1 𝑥
2
𝐱
f(x)
8
4
2
1
-3
-2
-1
0
𝟏
𝟐
1
𝟏
𝟒
2
Domain:{x|x > 0}/
Range:(-∞,∞)
𝟏
𝟖
3
Graph of Logarithmic Function
𝑓 𝑥 = log 2 𝑥
𝑓 𝑥 = log 1 𝑥
2
The value of b in 𝑦 = log 𝑏 𝑥 (𝑏 > 1or 0 < 𝑏 < 1) determines whether the
graph is increasing or decreasing.
For all logarithmic functions of the form y = logbx or
f(x) = log b x, where b > 0, b  1, and x > 0,
1. The domain of the function is (0, ) .
2. The range of the function is (, ) .
3. The graph passes through the points
( 1 , 1),  1, 0  ,  b,1 .
b
Logarithmic Graphs
Example:
Graph the function f(x) = log10x.
10,1
1, 0 
 101 , -1
Notice that the
graph passes
through the
point (1,0).
Domain: {x|x > 0}
Range:
Exponential Function
Logarithmic Function
y = bx (b > 0, b  1)
y = logbx (b> 0, b  1)
Domain:
Range:
Points on
Graph:
 ,  
 0,  
 0,  
 ,  
 1, 1b 
 1b , 1
 0,1
 1, b 
x becomes y
y becomes x
1,0
 b,1
Exponential vs. Logarithmic Graphs
f(x)
f(x) = 10x
f(x) = log10x
f -1(x)
Notice that the two graphs are inverse functions.
Transformation of the graph of logarithmic
function
The value of b in 𝑦 = log 𝑏 𝑥 (𝑏 > 1or 0 < 𝑏 < 1) determines whether the graph
is increasing or decreasing.
1.
2.
3.
4.
𝒇 𝒙 = 𝒂 ∙ 𝒍𝒐𝒈𝒙
𝑓 𝑥 = log 𝑥
𝑓 𝑥 = 2 log 𝑥
1
𝑓 𝑥 = 2 log 𝑥
𝑓 𝑥 = − log 𝑥
The value of a determines
the stretching and shrinking
of the graph. Futher, if a is
negative, there's a reflection
of the graph at the x- axis.
𝒇 𝒙 = 𝐥𝐨𝐠(𝒙 + 𝒄)
1. 𝑓 𝑥 = log 𝑥
2. 𝑓 𝑥 = log(𝑥 + 2)
3. 𝑓 𝑥 = log(𝑥 − 1)
The value of c determines
the horizontal shift of the
graph. Futher, if c> 0 the
graph shift c units to the
left and c units to the right
when c< 0.
𝒇 𝒙 = 𝒂 ∙ 𝒍𝒐 𝒈 𝒙 + 𝒄 + 𝒅
1. 𝑓 𝑥 = 𝑙𝑜𝑔 𝑥
2. 𝑓 𝑥 = 2 𝑙𝑜𝑔(𝑥 + 2) + 3
1
3. 𝑓 𝑥 = 2 𝑙𝑜𝑔 𝑥 − 1 − 2
The value of d determines
the vertical shift of the
graph. Futher, if d> 0 the
graph shift d units to the
upward and d units to the
downward when d< 0.
Basic Properties of Logarithms
For positive real numbers x and b> 0, b  1
1. log b 1 = 0
3. log b bx = x
2. log b b = 1
4. if x > 0,then b log b x = x
Identify the property used, then give its value.
1
125
1. log 10
4. log 5
7. log 3 3
2. ln 𝑒 3
5. 5log 5 x
8. log 2 32
3. log 4 43
6. log 1
9. log 2 1
Spot the difference!!!
log 𝑥 2 = 4
Logarithmic Equation
An equation involving
logarithm
log 𝑥 2 ≤ 4
Logarithmic Inequality
An inequality involving
logarithm
f(x) = log 𝑥 2
logarithmic Function
Function of the form
𝒇 𝒙 = log 𝒃 𝒙where 𝒃 > 𝟎,
𝒃≠𝟏
Simplify each of the following logarithmic expression
10
1. log 2 2
2. log 5 5
3. ln e9
log7 5.5
4. 7
5. log100 1
6. log 3 27
1
7. log 2
8
log5 −2
8. 5
(x+1)
9. ln e
10. eln(4.5)
LAWS OF LOGARITHMS
Product Rule for Logarithms
For positive real numbers x, y, and b, b  1,
log b(xy)  log b x  log b y
Property 1
Example:
log5(4 · 7) = log54 + log57
log10(100 · 1000) = log10100 + log101000 = 2 + 3 = 5
log xy = log x + log y
log7(73 · 74 ) = log 7 73 + log 7 74
Quotient Rule for Logarithms
For positive real numbers x, y, and b, b  1,
log b x  log b x  log b y
y
Property 2
Example:
log 7 10  log 7 10  log 7 2
2
log10 1  log10 1  log10 1000  0  3  3
1000
log 7( 49 )  log 7 49  log 7 7  2  1  1
7
8
2
log 2( 3 )  log 2 28  log 2 23  8  3  5
2
Power Rule for Logarithms
If x and y are positive real numbers, b  1, and n is any
real number, then
log b x n  n log b x
Property 3
Example:
log10 102  2 log10 10  2 1  2
log 4 64  log 4 43  3log 4 4  3 1  3

log 9 34  4 log 9 3  4 1  2
2
Example1:
Write the following as the logarithm of a single
expression.
(log 2 x  2 log 2 y)  3log 2 z 
(log 2 x  log y 2)  log 2 z 3
Power Rule
log 2(xy 2)  log 2 z 3
Product Rule
 (xy 2 
log 2  3 
 z 
Quotient Rule
Example2:
Write the following as the logarithm of a single
expression.
1 log x  3log (x  1)
3
2 3
1
2
log 3 x  log 3(x  1)3
log 3
Power Rule
1
2
x
(x  1)3


x
log 3 
3
(
x

1)


Quotient Rule
Example3:
Write the following as the logarithm of a single
expression.
5log6(x  3)  [2log 6(x  4)  3log 6 x] 
log 6(x  3)5  [log 6(x  4) 2  log 6 x3] 
Power Rule
log 6(x  3)5  [log 6(x  4) 2  x3] 
Product Rule
 (x  3)5 
log 6 
2 3
(
x

4)
x 

Quotient Rule
Write the following as a single logarithm
1.log x + log z
6. log 3 x 4 − log 3 x 5
2.log y − log z
7. log x + log(x + 3)
3.log 8 + log 3
8. log 2 x + log 2 y − log 2 z
4.log 2 3 + log 2 x 3 9. log 3 (3x − 6) − log 3 3
1
3
1
5.2 log 3 x- 5 log 3 y + 4 log 3 z
10.2 log 6 −
1
log 8
2
− log 12
Example1:
Expand the logarithmic expression.
1
2
2
xz
log 6( 1 ) log 6 x  log 6 z  log 6 y 3
y3
 log 6 x  2 log 6 z  1 log 6 y
3
Example1:
Expand the logarithmic expression.
1

1
(x  1) 2 

3
2
log 3

log
(
x

1)

log
x
3
3
 x3 


1
1
3 2
2
 log 3(x  1)  log 3(x )
 
 1 log 3(x  1)  1 log 3 x 3
2
2
Expand the logarithm using the rules of logarithm
1.log 3 (xyz)
x
2.log 4 ( y)
2 3
3.log 3 (x y )
x4
4.log 5 ( z )
xy
5.log 4 ( z )
x
6. log 5 ( )
z
2
7. log xy
1
2
2
8. log 3 (x y )
3
9. log 2 (2x )
10. log 4 (
3 1
x 5 y2
1
z4
)
CHANGE-OF-BASE FORMULA
log a x
ln x
log b x 
or
log b x 
log a b
ln b
EXAMPLE: Use common logarithms and natural logarithms to
find each logarithm:
b) log 0.8 70
log 65
log 5
log 70

log 0.8
c) log 2 0.1
log 0.1

log 2
a) log 5 65

LOGARITHMIC EQUATIONS
Properties for Solving Logarithmic Equations
Properties 6a-6b
a. If x = y, then logbx = logby (x > 0, y > 0).
b. If logbx = logby, then x = y (x > 0, y > 0).
Solving Logarithmic Equations
Guidelines:
1. Isolate the logarithmic term on one side of the equation; you may first
need to combine the logarithmic terms.
2. Write the equation in exponential form.
3. Solve for the variable.
EXAMPLE 1: Solve the following:
8
a ) log x
3
27
5
b) log 4 x 
2
c) log 2
5
d ) log 3
8
4
x

25
4
9
 2x  2
64
Solving Equations
Example 1:
Solve the equation log(x  3)  log x  log 4.
log(x  3)x  log 4
Product Rule
(x  3)x  4
Property 6d.
x 2  3x  4
x 2  3x  4  0
Check:
log(4  3)  log(4)  log 4.
Stop! Logs of negative numbers
are not real numbers.
(x  4)(x  1)  0
log(1  3)  log(1)  log 4.
x  4 or x  1
log 4  0  log 4
log 4  log 4 True
Solving Equations
Example 2:
Solve the equation log(x  3)  2.
102  x  3
Change the given into exponential form
100  x  3
x  97
Check:
log(x  3)  2.
log(97  3)  2.
log(100)  2.
100  102
100  100
Substitute the value of x
True
Solving Equations
Example 3:
Solve the equation log 2( x  3)  log 2( x  3)  4.
log 2(x  3)(x  3)  4 Product Rule
(x 2  9)  24 Change the given into exponential form
x 2  9  16
x 2  25
x  5
Solve the following logarithmic equation
1.log x + log 5=1
2.log x − log 9=1
3.log 2 3 + log 2 7=log 2 x
4.log 4 18 − log 4 x=log 4 6 5
5. log 3 (x − 4) + log 3 (x + 4) = 2
6. log 3 56 − log 3 8=log 3 x
1
7. 2 log 6 4 − log 6 8=log 6 x
3
8. log 3 (x − 4) + log 3 (x + 2) = 4
Solve the following logarithmic equation
1. log x + log 10=2
2. log 2x − log 5=1
3. log 5 4 − log 5 x=log 5 36
4. log 6 (x + 6) + log 6 (x − 3) = 2
5. log 4 (8x) + log 4 (x − 1) = 2
6. 2log 7 3 + 3log 7 2=log 7 x
7. log(x 2 − 1) − log(x − 1)=log 6
8. log 2 x − 4 − log 2 3 = 1
9. log 1 x − log 1 7x − 4 = −1
3
3
10. log 7 (x − 5) + log 7 (x − 5) = 2
Find the value without using calculator, if
log 2 = 0.3010 , log 3 = 0.4771, log 5 = 0.6990,log 7 = 0.8451
Examples:
a. log 21
1. log 12
2. log 15
3. log 8
2
4. log 4 3
5. log(10.5)
b. log 3.5
c. log
9
6. log( )
14
25
7. log( )
2
8. log 0.21
9. log 90
10. log 171.5
6
5
Application of Logarithmic function
1. Earthquake Magnitude -Richter Scale
2.Sound Intensity-decibel
3. Measure of acidity -pH level
4.Growth rate
5.Half-life
Earthquake Magnitude on a Richter scale
The magnitude R of an earthquake is given by:
2
𝐸
𝑅 = log 4.40
, where 𝐸-energy in joules released by an earthquake
3
10
1. Suppose that an earthquake released an energy of 1012 joules.
a. What is the magnitude of the earthquake?
b. How much more energy does these earthquake release than by the
reference earthquake?
2
E
a. R = 3 log 104.40
R=
R=
2
3
2
3
2
3
1012
log 104.40
log 107.6
R=
7.6
R ≈ 5.1
E
b. Energy released than by the reference angle = 104.40
1012
104.40
=
≈ 39810717
Sound Intensity
Acoustic as a science of sound is measured by a decibel (dB) is given by:
𝐼
𝑤𝑎𝑡𝑡𝑠
dB= 10 log −12
, where 𝐼-sound intensity in 2
10
𝑚
Note: (the quantity 10−12is the least sound a human ear can hear)
1.
𝑤𝑎𝑡𝑡𝑠
−6
The decibel of a sound in a quiet office is10
,
𝑚2
a. What is the corresponding sounds in decibel?
b. How much more intense is this sound than the least audible sound
a human ear can hear?
10−6
10 log −12
10
10 log 106
a. dB =
dB =
dB = 10(6)
dB = 60
10−6
b. The least intensity of sound a human ear can hear = 10−12
= 106
= 1,000,000
Measure of acidity -pH level
In chemistry , the pH level measures the acidity of a water- based solution that
is measured by the concentration of hydrogen ions(𝐻 + ), in moles per liter
1
+
which is obtained by: pH = − log 𝐻 or pH = log +
𝐻
1. 1 litter of solution contains 0.0001 moles of hydrogen ions . Find its pH
level.
a.pH = − log 𝐻 +
pH = − log 10−5
pH = − −5 log 10
pH = 5 1
pH = 5
Note: 0.0001=10−5
Growth rate
1. The growth rate of a certain bacterial culture can be calculated using the
𝑡
48
formula 𝐵 = 1000(2) , where 𝐵is the number of bacteria and 𝑡is the
elapsed time in hours.
a. How many bacteria will be present after 4 hours?
b. How long will it take to the nearest hour, for there to be 12,000
bacteria in
the culture?
𝑡
𝑡
a. B = 1000 2
B = 1000 2
B ≈ 1059
48
4
48
b.
B = 1000 2
12,000 = 1000 2
𝑡
48
48
𝑡
48
12 = 2
𝑡
log 2 12 = 48
t = 48log 2 12
t=
48 log 12
log 2
t ≈ 172
Growth rate
1. In a certain town in Mindanao with 15,000 people, the spread of a rumor that
there would be ABUSAYAF coming such that 𝐭 hours after the rumors has
started, where f(t) persons heard the rumor, where experience over time has
15,000
shown that f t =
.
1+7499e−0.8t
a. How many people started the rumors?
15,000
a. f t = 1+7499e−0.8t
15,000
f t = 1+7499e−0.8(0)
15,000
f t = 1+7499e0
15,000
f t = 1+7499
15,000
f t = 75000
f(t) = 2
people started the rumor
Continuation
Growth rate
1. In a certain town in Mindanao with 15,000 people, the spread of a rumor that
there would be ABUSAYAF coming such that 𝐭 hours after the rumors has
started, where f(t) persons heard the rumor, where experience over time has
15,000
shown that f t =
.
1+7499e−0.8t
b. How many people heard the rumor after 5 hours?
15,000
b. f t = 1+7499e−0.8t
15,000
f t = 1+7499e−0.8(5)
15,000
f t =
1+7499e−40
f(t) = 108
people heard the rumor
Continuation
Growth rate
1. In a certain town in Mindanao with 15,000 people, the spread of a rumor that
there would be ABUSAYAF coming such that 𝐭 hours after the rumors has
started, where f(t) persons heard the rumor, where experience over time has
15,000
shown that f t =
.
1+7499e−0.8t
c. How many hours does it take for 14,000 people to hear the rumors?
c.
15,000
1,000
f t = 1+7499e−0.8t
−0.8t = log 𝑒 104,986,000
14,000= 1+7499e−0.8𝑡
−0.8t = (ln 104,986,000)
15,000
14,000(1 + 7499e−0.8𝑡 ) = 15,000
14,000+14,000(7499e−0.8𝑡 ) = 15,000
14,000(7499e−0.8𝑡 ) = 1,000
104,986,000e−0.8𝑡 = 1,000
1,000
e−0.8𝑡 = 104,986,000
1,000
1,000
(ln104,986,000)
t=
−0.8
t ≈ 14.45 hours
14 hours and 45 minutes for
14,000 people to hear the rumor
Half-life
1. A fossil that originally contained 100 mg of carbon -14 contains 75 mg of
the isotope. Determine the approximate age of the fossil, to the nearest
100 years, if the half-life of carbon-14 is 5570years.
𝑡
A = 𝐴0
1 ℎ
2
𝑡
or
A = 𝐴0 2
−ℎ
𝑡
75 = 100 2
−5570
𝑡
−5570
0.75 = 2
𝑡
log 2 0.75 = − 5570
t =-5570log 2 0.75
−5570 log 0.75
t =
𝑙𝑜𝑔2
t = 2311.76
t = 2300 years old
Seat Work: Solve :
1. The growth rate for a particular bacterial culture can be calculated using the
𝑡
50
formula 𝐵 = 900(2) , where 𝐵is the number of bacteria and 𝑡 is the
elapsed time in hours. How many bacteria present after 5 hours?
2. Using question number 1,how many hours will it take for there to be 18,000
bacteria present?
3. A radio active isotope has a half life of 4 days . How many days for a 15 g
sample of this isotope to decay to 3 g?
4. Dan invested 5,000 pesos at 8% annual interest compounded monthly, how
many years will it take him to earn 2,000 ?
𝑤𝑎𝑡𝑡𝑠
𝐼
−4
5. The decibel of a sound in a quiet office is10
, using dB= 10 log −12
𝑚2
10
a. What is the corresponding sounds in decibel?
b. How much more intense is this sound than the least audible sound
a human ear can hear?
Application: (Exponential and Logarithmic Equations)
1.
The growth rate for a particular bacterial culture can be calculated using the formula
B = 900(2)t/50, where B is the number of bacteria and t is the elapsed time in hours. How
many bacteria will be present after 5 hours?
1.
How many hours will it take for there to be 18,000 bacteria present in the culture in
example (1)?
2.
A fossil that originally contained 100 mg of carbon-14 now contains 75 mg of the isotope.
Determine the approximate age of the fossil, to the nearest 100 years, if the half-life of
carbon-14 is 5,570 years.
A  Ao 2

t
h
where: A  present amt. of isotope
A o  orig . amt. of isotope
t  time it takes to reduce orig. amt. of isotope present
h  Half  life of the isotope
4.
In a town of 15,000 people, the spread of a rumor that the local transit company would go
on strike was such that t hours after the rumor started, f(t) persons heard the rumor,
where experience over time has shown that
15,000
f t  
1  7499e  0.8 t
a) How many people started the rumor?
b) How many people heard the rumor after 5 hours?
5. A sum of Php5,000 is invested at an interest rate of 5% per year. Find the time required
for the money to double if the interest is compounded (a) semi-annually (b) continuously.
A  A0 1  r   simple int erest for 1 year
n t
r

A  t   Ao  1  
 int erest compounded n times per year
n

A  Pe r t  int erest compounded continuously