Microsoft Word - Log & Expon
Transcript of Microsoft Word - Log & Expon
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Log and exponents (Rules and Examples) By Vinod Kum
EXPONENTIAL RULES
Rule 1: To multiply identical bases, add the exponents.
Rule 2: To divide identical bases, subtract the exponents.
Rule 3: When there are two or more exponents and only one base, multiply the exponents.
Rule 1:
E x a m p l e 1 :
means
which in turn can be written . According to Rule 1, you can get to the answer directly by adding the
exponents.
E x a m p l e 2 :
can be simplified to
Rule 2:
E x a m p l e 1 :
can be written
Which can be written as
or
The later can also be written . According to Rule 2, you can get to the answer directly by subtracting
the exponents
E x a m p l e 2 :
can be written . According to Rule 2, the expression can be simplify to
Rule 3:
E x a m p l e 1 :
can be written . According to Rule 1, we can add the
exponents. Can now be written , According to
Rule 3, we could have gone directly to the answer by multiplying the exponents
E x a m p l e 2 :
Simplify . According to Rule 3, the answer is .
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Log and exponents (Rules and Examples) By Vinod Kum
E x a m p l e 3 :
Simplify . The expression can be written
You could go directly to the answer by multiplying all the exponents.
E x a m p l e 4 :
Simplify . The solution is as follows:
Logarithm rules
Ther u
l e s o f l o g a
r i t h
m s
are1 ) P
r
o d
u c t R u
l e
Thel o g a
r i t h
m o f a p
r
o d
u c t is the
s
u
m
of the logarithms of the factors.
loga x y
= loga x
+ loga
y
2 ) Q
u
o
t i
e n
t R u
l e
Thel
o g a
r i t h
m
o f a
q
u
o
t i
e
n
t
is the logarithm of the numeratorm
i
n
u
s
the logarithm of the denominator
loga
= loga
x
– loga y
3 ) P o w
e
r R u
l e
loga
x
=
loga
x
4 ) C
h
a n g e o f B a s e
R u
l e
wherex
andy
are postive, and
> 0,
≠ 1
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Log and exponents (Rules and Examples) By Vinod Kum
C o m m e n t s o n L o g a r i t h m i c F u n c t i o n s
• The exponential equation could be written in terms of a logarithmic equation as
.
• The exponential equation can be written as the logarithmic equation
.
• Since logarithms are nothing more than exponents, you can use the rules of exponents
with logarithms.
• Logarithmic functions are the inverse of exponential functions. For example if (4, 16) is a
point on the graph of an exponential function, then (16, 4) would be the corresponding
point on the graph of the inverse logarithmic function.
• The two most common logarithms are calledc o m m o n
logarithms andn a t u r a l
logarithms. Common logarithms have a base of 10, and natural logarithms have a base
of
.
• Properties
• Graphs of Logarithms
• Change of base
• Rules of Logarithms
• Solving Equations
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Log and exponents (Rules and Examples) By Vinod Kum
P R O P E R T I E S O F L O G A R I T H M S
r o p e r t y 1 :
because . x
a m p l e 1 :
In the equation , the base is 14 and the exponent is 0. Remember
that a logarithm is an exponent, and the corresponding logarithmic equation is
where the 0 is the exponent.
x a m p
l e 2 :
In the equation , the base is and the exponent is 0. Remember
that a logarithm is an exponent, and the corresponding logarithmic equation is
. x a m
p
l e 3 :
Use the exponential equation to write a logarithmic equation. The
base x is greater than 0 and the exponent is 0. The corresponding logarithmic equation
is .
r o p e r t y 2 :
because . x
a m p l e 4 :
In the equation , the base is 3, the exponent is 1, and the answer is
3. Remember that a logarithm is an exponent, and the corresponding logarithmic
equation is . x
a m p l e 5 :
In the equation , the base is 87, the exponent is 1, and the answer
is 87. Remember that a logarithm is an exponent, and the corresponding logarithmic
equation is .
x
a m p l e 6 :
Use the exponential equation to write a logarithmic equation. If the
base p is greater than 0, then .
r o p e r t y 3 :
because . x
a m p l e 7 :
Since you know that , you can write the logarithmic equation with
base 3 as . x a m p l e 8 :
Since you know that , you can write the logarithmic equation
with base 13 as . x
a m p l e 9 :
Use the exponential equation to write a logarithmic equation with
base 4. You can convert the exponential equation
to the logarithmic equation . Since the 16 can be written as
, the equation can be written .
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Log and exponents (Rules and Examples) By Vinod Kum
The above rules are the same for all positive bases. The most common bases are the base 10
and the base . Logarithms with a base 10 are calledc o m m o n l o g a r i t h m s
, and logarithms with
a base aren a t u r a l l o g a r i t h m s
. On your calculator, the base 10 logarithm is noted byl o g
, and
the base
logarithm is noted byl n
.
There are an infinite number of bases and only a few buttons on your calculator. You can
convert a logarithm with a base that is not 10 or
to an equivalent logarithm with base 10 or e.
If you are interested in a discussion on how to change the bases of a logarithm, click on C h a n g
o f B
a s .
For a discussion of the relationship between the graphs of logarithmic functions and
exponential functions, click on graphs.
G R A H S O F X O N N T I A L A N D L O G A R I T H M I C F U N C T I O N S
G R A
H S O F
X O N N T I A L F U
N C
T I O N S
I n t h i s s e c t i o n w e w i l l i l l u s t r a t e , i n t e r p r e t , a n d d i s c u s s t h e g r a p h s o e
x
p o n e n t i a l a n d
l o g a r i t h m i c u n c t i o n s W e w i l l a l s o i l l u s t r a t e h o w y o u c a n u s e g r a p h s t o H
L
y o u s o l v e
e x
p o n e n t i a l a n d l o g a r i t h m i c p r o b l e m s a n d c h e c k y o u r s o l u t i o n s
r p F u c t i i e e r l :
Recall that graphs are made up of points that are plotted on rectangular coordinate systems. The points
consist of two parts: (independent variable, dependent variable). The dependent variable is so named
because it's value and behavior depend on the value and behavior of the independent variable. For
purposes of discussion, we will refer to the independent variable as x and the dependent variable as y or
f(x). In reality, you could label them anything: (p, q), (a, b), (d, c), etc.R e m i n d e r :
Remember that when we talk about the function, the function value, the value of
the function, y or f(x), we are talking about the value and behavior of the y part of the point (x,
y) in the full set of the points that form the graph.. x
p o n e n t i a l G r a p h s :
Once you know the shape of an exponential graph , you can shift it vertically or horizontally,
stretch it, shrink it, reflect it, check answers with it, and most important interpret the graph.
x
a m p l e 1 :
The function is always positive. There is simply no value of x that will
cause the value of to be negative. What does this mean in terms of a graph? It means that
the entire graph of the function is located in quadrants I and II.
G r a p h t h e u n c t i o n
. Notice that the graph never crosses the x-axis. Why is that so?
It is because there is no value of x that will cause the value of f(x) in the formula to
equal 0.
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Log and exponents (Rules and Examples) By Vinod Kum
Notice that the graph crosses the y-axis at 1. Why is that so? The value of x is always zero on
the y-axis. Substitute 0 for x in the equation : . This translates to
the point (0, 1).
Notice on the graph that, as the value of x increases, the value of f(x) also increases. This means
that the function is an increasing function. Recall that an increasing function is a one-to-one-
function, and a one-to-one function has a unique inverse.
The inverse of an exponential function is a logarithmic function and the inverse of a logarithmic
function is an exponential function.
Notice also on the graph that as x gets larger and larger, the function value of f(x) is increasing
more and more dramatically. This is why the function is called an exponential function.
If you are interested in reviewing the graphs of exponential functions, examples and problems,
click on Exponential. L o g a r i t h m i c G r a p h s :
Once you know the shape of a logarithmic graph , you can shift it vertically or horizontally,
stretch it, shrink it, reflect it, check answers with it, and most important interpret the graph. x
a m p l e 2 :
Graph the function . Notice that the graph of this function is located entirely in
quadrants I and IV. Notice also that the graph never touches the y-axis.
What does that mean? It means that the value of x (domain of the function f(x) in the equation
is always positive. Why is this so? Recall that the equation can
be rewritten as the exponential function . There is no value of f(x) that can cause the
value of x to be negative or zero.
The graph of will never cross the y-axis because x can never equal 0. The graph
will always cross the x-axis at 1.
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Log and exponents (Rules and Examples) By Vinod Kum
Notice on the graph that, as x increases, the f(x) also increases. This means that the function is
an increasing function. Recall that an increasing function is a one-to-one-function, and a one-to-
one function has a unique inverse.
Notice on the graph that the increase in the value of the function is most dramatic between 0
and 1. After x = 1, as x gets larger and larger, the increasing function values begin to slow down
(the increase get smaller and smaller as x gets larger and larger).
Notice on the graph that the u n c t i o n v a l u e s
are positive for x's that are greater than 1 and
negative for x's less than 1.
C H
A N
G I N
G T H B
A S O
F A L O
G A R I T H M
L e t a , b , a n d x
b e p o s i t i v e r e a l n u m b e r s s u c h t h a t a n d ( r e m e m b e r x
m u s t b e
g r e a t e r t h a n 0 ) T
h e n c a n b e c o n v e r t e d t o t h e b a s e b b y t h e o r m u l a
Let's verify this with a few examples.
x
a m p l e 1 :
Find to an accuracy of six decimals. Note that the answer will be between 1
and 2 because and , and 7 is between 3 and 9. According to the change of
logarithm rule, can be written .
When the base is 10, we can leave off the 10 in the notation. Therefore can be written
. Using your calculator,
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Log and exponents (Rules and Examples) By Vinod Kum
You will note that the answer is between 1 and 2.
Let's check the answer. If , our answer is correct. . Close
enough. Why isn't it 7 exactly? Well we rounded to six places, so our answer won't check
exactly. If we rounded to ten places, then when we checked the answer, it would be closer to 7
than this answer. x
a m p l e 2 :
We could work the same problem by converting to the base e. According to the
change of logarithm rule, can be written . When the base is e, we can leave off
the e in the notation and can be written . Using your calculator,
You will note that the answer is between 1 and 2.
x
a m p l e 3 :
Find . We know that and , and that 18 is between 16 and
32; therefore, we know that the exponent we are looking for is between 4 and 5. In fact, it is
closer to 4 than to 5 because 18 is closer to 16 than it is to 32.
Let's solve this problem by changing the base to 10. can be written . Using your
calculator,
Let's check the answer. If , our answer is correct. .
Close enough. Why isn't it 18 exactly? Since we rounded to six places, our answer won't
check exactly. If we rounded to ten places, then when we checked the answer, it would be
closer to 18 than this answer. x
a m p l e 4 :
We could work the problem in Example 3 by converting to the base e. According to
the logarithmic rule, can be written . Using your calculator,
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Log and exponents (Rules and Examples) By Vinod Kum
The answer is the same as the answer you found when you converted the base to 10.I
y o u w o u l d l i k e t o r e v i e w m o r e e x
a m p l e s o c h a n g i n g t h e b a s e o a l o g a r i t h m , c l i c k o n
x
a m p l e
Work the following problems and click on answer if you want to check your answer and review
the solution.
r o b l e m 1 :
Find .
Answer.S
o l u t i o n 1 :
Convert to base 10 logarithms.
C h e c k :
Let's check your answer. If , then the answer is correct.
is not exact but close enough for our check.
Remember why it is not exact in the check. the -11.18900388 was rounded; therefore, the
answer won't be exact.
S o l u t i o n 2 :
Convert to base e logarithms.
r o b l e m 2 :
Find .
Answer.S
o l u t i o n 1 :
Convert to the base 10.
rounded to 0.43068. If you use continuous calculation on your calculator, your answer will be
0.4306766.
C h e c k :
Let's check your answer. If , then the answer is
correct.
is not exact but close enough for our check. Remember
why it is not exact in the check. the value of log4 and log25 were rounded, divided, and the
answer rounded again.; therefore, the answer won't be exact. If you use continuous calculation
and memory on your calculator, the answer will be even closer.
S o l u t i o n 2 :
Convert to the base e.
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Log and exponents (Rules and Examples) By Vinod Kum
If you used continuous calculation and memory in your computer, your answer would be
0.4306766
r o b l e m 3 :
Find .Answer.
S o l u t i o n :
By the definition of logarithms, the base must be greater than zero.
r o b l e m 4 :
Convert to the base 2.
Answer.S
o l u t i o n :
Convert to the base 2.
The numerator can be calculated by converting the numerator to the base 10 or the base e.
. The answer can also be written
.
Note that for this problem to be valid, the base (a + b) must be greater than zero.
RULES OF LOGARITHMS
L e t a b e a p o s i t i v e n u m b e r s u c h t h a t a d o e s n o t e q u a l 1 , l e t n b e a r e a l n u m b e r , a n d l e t u a n d
v b e p o s i t i v e r e a l n u m b e r s
L o g a r i t h m i c R u l e 1 :
L o g a r i t h m i c R u l e 2 :
L o g a r i t h m i c R u l e 3 :
Since logarithms are nothing more than exponents, these rules come from the rules of
exponents. Let a be greater than 0 and not equal to 1, and let n and m be real numbers. x
p o n e n t i a l R u l e 1 :
x
a m p l e :
Let a = 5, n = 2, and m = 6. and
x
p o n e n t i a l R u l e 2 :
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Log and exponents (Rules and Examples) By Vinod Kum
x
a m p l e :
Let a = 5, n = 2, and m = 6. and
x p o n e n t i a l R u l e 3 :
x
a m p l e :
Let a = 5, n = 2, and m = 6. and
E x a m p l e :
Simplify the following, expressing each as a single logarithm:
a) log 2 4 + log 2 5
b) log a 28 – log a 4
c) 2 log a 5 – 3 log a 2S o
l u t i
o n
:
a) log 2 4 + log 2 5 = log 2 (4 × 5) = log 2 20
b) log a 28 – log a 4 = log a (28 ÷ 4) = log a 7
c) 2 log a 5 – 3 log a 2 = log a 52
– log a 23
= log a E x a m p l e :
Evaluate 2 log3 5 + log3 40 – 3 log3 10S o
l u t i
o n
:
2 log3 5 + log3 40 – 3 log3 10
= log3 52
+ log3 40 – log3 103
= log3 25 + log3 40 – log3 1000
= log3
= log3 1
= 0E x a m p l e :
Given that log2 3 = 1.585 and log2 5 = 2.322, evaluate log4 15S o
l u t i
o n
:
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Log and exponents (Rules and Examples) By Vinod Kum
RULES OF LOGARITHMS - Rule 1
L e
t
a b e a p o s
i t i v
e n
u
m b e
r
s
u c h t h
a
t
a d o e s n o
t
e q
u
a l 1 , l e
t
n b e a
r
e a l n
u
m b e
r
, a n d l e
t u
a n d
v
b e
p o s
i t i v
e
r
e a l n
u
m b e
r
s .
L o g a
r i t h
m
i c R u
l e 1 :
E x a m p l e 1 :
Suppose that a base is 6 and exponents are 3 and 5. We could solve the exponential problem
by calculating and and multiplying the results. and and their
product is . You could also solve the problem by first combing the
exponents .
The same is true of logarithms. Suppose you wanted to calculate . You could
calculate the answer by first multiplying 216 by 7776, changing the base of 6 to either 10 or e and
calculating the results.
Or you could first combine the logarithms using Rule 1 and then change the bases.
Rounded to 8
E x a m p l e 2 :
Calculate .S
o l
u t i
o n :
Since a base is not indicated, we know that the base is 10. We can calculate the logarithms
directly by first multiplying the 30 and the 5. Note that can be written
. By Rule 1 we can also write as
To write this problem in turns of exponents, note that the base is 10 and the exponents are 1.4771213
and 0.69897. , rounded to 30.
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Log and exponents (Rules and Examples) By Vinod Kum
, rounded to 5. We could also combine the bases
I f y o
u
w o
u
l d l
i k
e
t
o
r
e
v i
e w a n o
t h
e
r
e x a m p l e ,
c
l
i c k
o n E x a m p l e .
W o
r k t h
e f o l l o w
i
n g p
r
o b l e m s . I f y o
u
w o
u
l d l
i k
e
t
o
r
e
v i
e w
t h
e a n s w e
r
s a n d s o l
u t i
o n s ,
c
l
i c k
o n a n s w e
r
.
P
r
o b l e m 1 :
Calculate .
Answer S
o l
u t i
o n :
The problem can be worked two ways, as is, and expanded.
can be writtenL
280=5.63478960317 .
can also be written
Since the answers are the same, then
N o
t
e :
The numbers 1.94591014906, 3.68887945411, and 5.63478960317 are the exponents. Ln means
the base is e. Let's check the answer in terms of exponents:
P
r
o b l e m 2 :
Calculate .
Answer:
The problem can be worked two ways, as is, and expanded.
can be written because
.
can also be writtenL o g
100+L o g
10+L o g
1000=2+1+3=6
Since the answers are the same, then is equivalent to
L o g 100+
L o g 10+
L o g 1000
N o
t
e :
The numbers 2, 1, 3, and 6 are the exponents. Log means the base is 10. Let's check the answer in
terms of exponents.
, , , and .
P
r
o b l e m 3 :
Calculate .
Answer S
o l
u t i
o n 1 :
The problem can be worked two ways, as is, and expanded.
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Log and exponents (Rules and Examples) By Vinod Kum
can be written . The can be written
S
o l
u t i
o n 2 :
can also be written
C
h
e
c k
:
The answers are the same. This means that and
are equivalent.N o
t
e :
The numbers -.0686215613241, 1.48535725521, and 1.41673569389 are the exponents. Let's
check the answer in terms of exponents.
P
r
o b l e m 4 :
Calculate .
Answer
S
o l
u t i
o n 1 :
The problem can be worked two ways, as is, and expanded. can be
written . The can be written
S
o l
u t i
o n 2 :
can also be written
The answers are the same. This means that and
are equivalent.C
h
e
c k
:
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Log and exponents (Rules and Examples) By Vinod Kum
P
r
o b l e m 5 :
Simplify and write the answer in terms of a base 10. What
assumptions must be made about a, b, d, and d before you can work this problem?Answer
S
o l
u t i
o n :
if you want the answer in terms of the base 3.
if you want the answer in terms of the base 10.
if you want the answer in terms of the base e.
if you want the answer in terms of the base 5.N o
t
e :
None of the above is valid unless
> 0,b
> 0,c
> 0, andd
> 0.C
h
e
c k
:
Suppose that a = 2, b = 3, c = 4, and d = 5. Then
and
RULES OF LOGARITHMS - Rule 2L e
t
a b e a p o s
i t i v
e n
u
m b e
r
s
u c h t h
a
t
a d o e s n o
t
e q
u
a l 1 , l e
t
n b e a
r
e a l n
u
m b e
r
, a n d l e
t u
a n d
v
b e
p o s
i t i v
e
r
e a l n
u
m b e
r
s .
L
o g a
r i t h
m
i c R u
l e 2 :
E x a m p l e 1 :
Suppose that a base is 6 and exponents are 10 and 3. We could solve the exponential
problem by calculating and
and dividing the results. , and their quotient is
. You could also solve the problem by first combining the exponents
The same is true of logarithms. Suppose you wanted to calculate
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Log and exponents (Rules and Examples) By Vinod Kum
You could calculate the answer by first dividing 60,466,176 by 216, changing the base of 6 to either 10 or
e and calculating the results.
Or you could first combine the logarithms using Rule 2 and then change the bases.
E x a m p l e 2 :
Calculate .S
o l
u t i
o n :
Note that
Using Rule 2 you could also work the problem by separating .
I f y o
u
w o
u
l d l
i k
e
t
o
r
e
v i
e w a n o
t h
e
r
e x a m p l e ,
c
l
i c k
o n E x a m p l e .
W o
r k t h
e
f o l l
o w
i
n g p
r
o b l e m
s .
I f y o
u
w o
u
l
d l
i k
e
t
o
r
e
v i
e
w
t h
e a
n s w e
r
s a
n d s o l
u t i
o n s ,
c
l
i c k
o n a
n s w e
r
.
P
r
o b l e m 1 :
Calculate .
Answer
S
o l
u t i
o n :
Since , you can calculate this problem in two ways:
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Log and exponents (Rules and Examples) By Vinod Kum
First find rounded to -1.743.C
h
e
c k
:
.
Another way to calculate this problem is to find
P
r
o b l e m 2 :
Calculate .
Answer B
e
f o
r
e
w e b e
g
i
n :
Since many individuals still have problems with how to simplify the expression
, let's walk through the process. The expression can be written
which in turn can be written .
Also the expression can be written
S
o l
u t i
o n :
can be written which can be expanded to
This problem can be solved in several ways:1
Find
rounded to 0.301.C
h
e
c k
:
.2
Another way to calculate this problem is to find
rounded to 0.301.3
Still another way to calculate this problem is to convert
to
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Log and exponents (Rules and Examples) By Vinod Kum
P
r
o b l e m 3 :
Calculate .
Answer S
o l
u t i
o n :
Since,
you can calculate this problem in two ways:
Find
rounded to 1.417.
A second way to work the problem is to write as
rounded to 1.417.
C
h
e
c k
:
.
P
r
o b l e m 4 :
Calculate .
Answer S
o l
u t i
o n :
There are no rules about simplifying logarithms of sums or differences; therefore to work this
problem, you must simplify 5 + 80 - 400 Since can be writtenL o g
(-315) , we
know that there is no solution.
You cannot take the logarithm of a negative numbers. This is like saying that if you raise 50 to any
power, there is no way you are going to get a negative answer.
P
r
o b l e m 5 :
Simplify and write the answer in terms of a base 10. What assumptions
must be made about a, b, d, and d before you can work this problem?
Answer
S
o l
u t i
o n :
To work the problem as it is, the value of the expression must be greater than zero.
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Log and exponents (Rules and Examples) By Vinod Kum
The expression can be simplified to
as long as the value of the expressions abc and 2de are both positive.
The expression can be written as
as long as a, b, c, d, and e are all positive.
If the values of a, b, c, d, and e are all positive, then we can expand the original expression to the last
expression. If the values are not all positive, then we cannot expand the original expression.
can be written in terms of base 10 as
C
h
e
c k
:
Since is equivalent to
let us choose values for a, b, c, d, and e and substitute them into the original expression and in our final
expression, the answers should be equal. Let us try it.
Suppose a = 2, b = 3, c = 4, d = 5, and e = 6, the original expression has a value
Now let's substitute these same values in the final expression.
Both answers are the same, therefore the original expression is equivalent to the final expression as long
as a, b, c, d, and e are all positive numbers.
A last check using our numbers:
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Log and exponents (Rules and Examples) By Vinod Kum
RULES OF LOGARITHMS - RULE 3 L e t a b e a p o s i t i v e n u m b e r s u c h t h a t a d o e s n o t e q u a l 1 , l e t n b e a r e a l n u m b e r , a n d l e t u a n d
v b e p o s i t i v e r e a l n u m b e r s
L o g a r i t h m i c R u l e 3 :
x
a m p l e 1 :
Find two ways.
S o l u t i o n :
Since can be written , the expression can be written
which in turn can be written
We have
and
x
a m p l e 2 :
Find .
S o l u t i o n :
The expression can be written which in turn
can be written . This last expression can be rewritten using Rule 1
as
This represents 6 identical terms and we can write the sum of the six terms as .
C h e c k :
The original expression can be written
The last expression can be written
I y o u w o u l d l i k e t o r e v i e w a n o t h e r e
x
a m p l e , c l i c k o n x
a m p l e
W o r k t h e o l l o w i n g p r o b l e m s a n d i y o u w a n t t o c h e c k y o u r a n s w e r , c l i c k o n a n s w e r
r o b l e m 1 :
Find
Answer S
o l u t i o n :
The can be written
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Log and exponents (Rules and Examples) By Vinod Kum
rounded to 5.02109. We could also do the calculation without first simplifying
rounded to 5.02109.
r o b l e m 2 :
(
a )
Find(
b )
Find
Answer S
o l u t i o n :
( a )
Since a negative number raised to an odd power is negative, we cannot
work this problem. You cannot take the log of a negative number.S
o l u t i o n :
( b )
Since a negative number raised to an even power is positive, we can find the value
of this problem
r o b l e m 3 :
Simplify
AnswerS
o l u t i o n 1 :
The expression can be written
rounded to 30.866
S o l u t i o n 2 :
We can also simplify the expression first before we calculate it. can
be written
rounded to - 30.866.S
o l u t i o n 3 :
We have
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Log and exponents (Rules and Examples) By Vinod Kum
r o b l e m 4 :
Simplify
AnswerS
o l u t i o n :
There are many ways to work this problem. We have included a few of the
ways.S
o l u t i o n 1 :
You can find the value directly by writing
S o l u t i o n 2 :
You can also find the value by simplifying the expression first and then calculating
the value. can be written
r o b l e m 5 :
Simplify
AnswerS
o l u t i o n :
There are many ways to work the above problem. We have included two of
the ways.S
o l u t i o n 1 :
S t e
p 1 :
First note that you are taking the log of an expression raised to a power; therefore, you
use Rule 3. can be written .S
t e p 2 :
You will now be taking the log of a quotient; therefore, use Rule 2.
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Log and exponents (Rules and Examples) By Vinod Kum
S t e p 3 :
You will now be taking logs of products; therefore, use Rule 1.
S t e p 4 :
You will now be taking logs of expressions raised to a power, use Rule 3.
S t e p 5 :
This last expression can be written
rounded to 3.487.S
t e p 6 :
You can check your answer by calculating directly from the initial expression.
Since both answers are the same, you have correctly worked the problem.S
o l u t i o n 2 :
S
t e p 1 :
First note that you are taking the log of an expression raised to a power; therefore, you
use Rule 3.
S t e p 2 :
Simplify the numerator and simplify the denominator: can be
written
S t e p 3 :
Calculate the logarithm:
r o b l e m 6 :
Simplify . What assumptions must you make before you can
begin work on this problem?
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Log and exponents (Rules and Examples) By Vinod Kum
Answer S
o l
u t i
o n :
The only requirement about the values of a, b, and c in the original expression is that
.
is equivalent to 8L
(
+b
+c
) only when
+b
+c
>0 .
That is far as you can take it. There are no logarithm rules dealing with logs of sums.
r o b l e m 7 :
Simplify the following term completely
State the domain that makes your final answer equal to the original expression.
AnswerS o l u t i o n
:
S t e p 1 :
The expression is of the form which can be simplified to the form
or
S t e p 2 :
The expression can be simplified to the form or
which can be written as
S t e p 3 :
This last expression is of the form which can be simplified
to
or in term of our problem
can be written as
S t e p 4 :
Both of these last terms are in the form which can be
simplified to the form
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Log and exponents (Rules and Examples) By Vinod Kum
or in terms of our problem
can be written as
S t e p 5 :
These terms can be simplified to
and again to
S t e p 6 :
The first term will work if x > 0, the second term will work if x > -5, the third term will
work if x > 8, the fourth term will always work, the fifth term will work if x > -9, and the last
term will work if x > 4. If we choose the domain as x > 8, then the original expression
is equal to the final expression
S t e p 7 :
You can check answer by graphing both expression over the domain x >8. If you see just
one graph, you have worked the problem correctly.S
t e p 8 :
You can also check the answer by assigning x a value greater than 8 and substituting
that value in both expressions. If the answers come out the same, you are correct. Let's let x =
10 in the original equation. If x = 10, the original equation has a value equal to
Let's let x = 10 in the final equation. If x = 10, the final equation has a value equal to
S t e p 9 :
Pat yourself on the back because you have correctly worked the problem.
SOLVING EXPONENTIAL EQUATIONST
o s o l v e a n e x
p o n e n t i a l e q u a t i o n , t a k e t h e l o g o b o t h s i d e s , a n d s o l v e o r t h e v a r i a b l e
x
a m p l e 1 :
Solve for x in the equation .S
o l u t i o n :
S
t e p 1 :
Take the natural log of both sides:
S t e p 2 :
Simplify the left side of the above equation using Logarithmic Rule 3:
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Log and exponents (Rules and Examples) By Vinod Kum
S t e p 3 :
Simplify the left side of the above equation: Since L n ( )=1, the equation reads
Ln(80) is the exact answer and x =4.38202663467 is an approximate answer because we
have rounded the value of Ln(80)..C h e c k
:
Check your answer in the original equation.
x a m p l e 2 :
Solve for x in the equationS
o l u t i o n :
S
t e p 1 :
Isolate the exponential term before you take the common log of both sides. Therefore,
add 8 to both sides:S
t e p 2 :
Take the common log of both sides:
S t e p 3 :
Simplify the left side of the above equation using Logarithmic Rule 3:
S t e p 4 :
Simplify the left side of the above equation: Since Log(10) = 1, the above equation can
be written
S t e p 5 :
Subtract 5 from both sides of the above equation:
is the exact answer. x = -3.16749108729 is an approximate answer..C h e c k :
Check your answer in the original equation. Does
Yes it does. x
a m p l e 3 :
Solve for x in the equation
S o l u t i o n :
S
t e p 1 :
When you graph the left side of the equation, you will note that the graph crosses the x-
axis in two places. This means the equation has two real solutions.S
t e p 2 :
Rewrite the equation in quadratic form:
S t e p 3 :
Factor the left side of the equation:
can now be written
S t e p 4 :
Solve for x. Note: The product of two terms can only equal zero if one or both of the
two terms is zero.
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Log and exponents (Rules and Examples) By Vinod Kum
S t e p 5 :
Set the first factor equal to zero and solve for x: If , then and
andx
=L n
(2) is the exact answer or is an
approximate answer.
S t e p 6 :
Set the second factor equal to zero and solve for x: If , then and
and x =L n (3) is the exact answer or is an
approximate answer. The exact answers are Ln(3) and Ln(2) and the approximate answers are
0.69314718056 and 1.09861228867.C h e c k :
These two numbers should be the same numbers where the graph crosses the x-axis.R e m a r k :
Why did we choose the L n in Example 3? Because we know that L n ( ) = 1.I
y o u w o u l d l i k e t o r e v i e w a n o t h e r e x
a m p l e , c l i c k o n x
a m p l e
W o r k t h e o l l o w i n g p r o b l e m s
I y o u w a n t t o r e v i e w t h e a n s w e r a n d t h e s o l u t i o n , c l i c k o n
a n s w e r
r o b l e m 1 :
Solve for x in the equation .
AnswerS
o l u t i o n :
S t e
p 1 :
Isolate the exponential term in the equation
using steps 2 through 5.S
t e p 2 :
Subtract 8 from both sides of the above equation:S
t e p 3 :
Since the base is 5, take the log to base 5 of both sides:
S t e p 4 :
Simplify the left side of the equation using Logarithmic Rule 3:
S t e p 5 :
Simplify the left side: We know that
(that's why we choose log with a base 5). Therefore, the left side of the equation can be
simplified to
S t e p 6 :
Subtract 3 from both sides of the above equation:
S t e p 7 :
Divide both sides of the above equation by 2:
is the exact answer.
is an approximate answer.
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Log and exponents (Rules and Examples) By Vinod Kum
C h e c k :
Let's check the approximate answer with the original problem. When we substitute the
above value of x in the left side of the equation, we get
r o b l e m 2 :
Solve for x in the equation .
AnswerS
o l u t i o n :
S
t e p 1 :
Isolate the exponential term using steps 2 through 4.
S t e p 2 :
Multiply both sides of the original equation by :
S t e p 3 :
Divide both sides of the above equation by 5:
S t e p 4 :
Subtract 2 from both sides:
S t e p 5 :
Since the base is 7, take of both sides:
S t e p 6 :
Simplify the left side of the above equation using Logarithmic Rule 3:
S t e p 7 :
We know that (that's why we choose ). Therefore, the left side of
the equation can be simplified to
S t e p 8 :
Divide both sides of the above equation by 3:
is the exact answer.
is an approximate answer.C h e c k :
Let's check the approximate answer with the original problem. When we substitute the
above value of x in the left side of the equation, we get
Close enough to 5. Remember it will not check directly because we rounded the answer. If you
choose to round to only 2 or 3 decimals, the difference between the check answer and 5 would
be greater.
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Log and exponents (Rules and Examples) By Vinod Kum
r o b l e m 3 :
Solve for x in the equation .
AnswerS
o l u t i o n :
S
t e p 1 :
If you graph the left side of the above equation, you will note that the graph crosses the
x-axis in two places, once to the left of the y-axis and once to the right of the y-axis. This means
that there will be one negative real solution and one positive real solution.S
t e p 2 :
Write the equation in quadratic form and factor:
S t e p 3 :
The only way a product of two factors is zero is when one or both of the factors is equal
to zero.
S t e p 4 :
If and . Take the natural log of both sides.
is the exact answer andx
=0.916290731874 is an approximate answer.
S t e p 5 :
If and
Take the natural log of both sides. and
is the exact answer and
is an approximate answer.C h e c k :
Let check both answers with the original problem. If when the value of x is substituted in
the left side of the equation, the value of the left side of the equation equals the right side of
the equation (in this case 0), you have found the correct answer. You could also check the
values of x with the x-intercepts on your graph. They should be the same.
They do and you have worked the problem correctly.
r o b l e m 4 :
Solve for x in the equation .
AnswerS
o l u t i o n :
S
t e p 1 :
Graph the function
Note that it crosses the x-axis only once between 6 and 7. This means there is just one real
solution and that solution is between 6 and 7.S
t e p 2 :
Take the Ln of both sides of the original equation:
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Log and exponents (Rules and Examples) By Vinod Kum
S t e p 3 :
Simplify the left side of the above equation using Logarithmic Rule 3:
S t e p 4 :
Divide both sides of the above equation by
to get
is the exact answer and
is the approximate answer.C h e c k :
Let's check the approximate answer with the original problem. When we substitute the
above value of x in the left side of the equation, we get
You can also check your answer by graphing the function
and note where it crosses the x-axis. If you have worked the problem correctly, it should be the
same values of x
r o b l e m 5 :
Solve for x in the equation .
Answer S
o l u t i o n :
S
t e p 1 :
Isolate the exponential term using steps 2 through 6.S
t e p 2 :
Divide both sides of the original equation by 5000:
S t e p 3 :
Subtract 1 from both sides of the above equation:
or
S t e p 4 :
Multiply both sides of the above equation by :
S t e p 5 :
:Divide both sides of the above equation by 0.92:
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Log and exponents (Rules and Examples) By Vinod Kum
S t e p 6 :
Subtract 4 from both sides of the above equation:
S t e p 7 :
Take the natural log of both sides of the above equation:
S t e p 8 :
Simplify the left side of the above equation using Logarithmic Rule 3:
S t e p 9 :
Since Ln(e) = 1, the above equation is simplified to
S t e p 1 0 :
Divide both sides of the above equation by -0.002:
rounded to 528.C h e c k :
Check the graph, it should cross in one place very close to x = 528. You can alsosubstitute the number in the original equation and check to see if the left side of the equation
then equals the right side of the equation.
r o b l e m 6 :
Solve for x in the equation .
Answer S o l u t i o n :
S
t e p 1 :
Graph the function
Note that the graph crosses the x-axis at the origin. This means that there is only one realsolution and that solution is 0.
S t e p 2 :
Divide by sides of the original equation by 5:
S t e p 3 :
Take the cube root of both sides of the above equation:
S t e p 4 :
Add 3 to both sides of the above equation:S
t e p 5 :
Divide both sides of the above equation by 8:S
t e p 6 :
Take the natural logs of both sides of the above equation:
S t e p 7 :
Simplify the left side of the above equation: