Logarithms Tutorial for Chemistry Students - Class 10 PDF Download

 Page 1


Boston University CH102 - General Chemistry Spring 2012
Logarithms Tutorial for Chemistry Students
1 Logarithms
1.1 What is a logarithm?
Logarithms are the mathematical function that is used to represent the number (y) to which a base integer (a) is
raised in order to get the number x:
x =a
y
;
where y = log
a
(x). Most of you are familiar with the standard base-10 logarithm:
y = log
10
(x);
where x = 10
y
. A logarithm for which the base is not specied (y = logx) is always considered to be a base-10
logarithm.
1.2 Easy Logarithms
The simplest logarithms to evaluate, which most of you will be able to determine by inspection, are those where y
is an integer value. Take the power of 10's, for example:
log
10
(10) = 1 10
1
= 10
log
10
(100) = 2 10
2
= 100
log
10
(1000) = 3 10
3
= 1000
log
10
(10000) = 4 10
4
= 10000
log
10
(1) = 0 10
0
= 1
log
10
(0:1) =1 10
1
= 0:1
log
10
(0:01) =2 10
2
= 0:01
log
10
(0:001) =3 10
3
= 0:001
log
10
(0:0001) =4 10
4
= 0:0001
1.3 Rules of Manipulating Logarithms
There are four main algebraic rules used to manipulate logarithms:
Rule 1: Product Rule
log
a
uv = log
a
u + log
a
v
Rule 2: Quotient Rule
log
a
u
v
= log
a
u log
a
v
Rule 3: Power Rule
log
a
u
v
=v log
a
u
1
Page 2


Boston University CH102 - General Chemistry Spring 2012
Logarithms Tutorial for Chemistry Students
1 Logarithms
1.1 What is a logarithm?
Logarithms are the mathematical function that is used to represent the number (y) to which a base integer (a) is
raised in order to get the number x:
x =a
y
;
where y = log
a
(x). Most of you are familiar with the standard base-10 logarithm:
y = log
10
(x);
where x = 10
y
. A logarithm for which the base is not specied (y = logx) is always considered to be a base-10
logarithm.
1.2 Easy Logarithms
The simplest logarithms to evaluate, which most of you will be able to determine by inspection, are those where y
is an integer value. Take the power of 10's, for example:
log
10
(10) = 1 10
1
= 10
log
10
(100) = 2 10
2
= 100
log
10
(1000) = 3 10
3
= 1000
log
10
(10000) = 4 10
4
= 10000
log
10
(1) = 0 10
0
= 1
log
10
(0:1) =1 10
1
= 0:1
log
10
(0:01) =2 10
2
= 0:01
log
10
(0:001) =3 10
3
= 0:001
log
10
(0:0001) =4 10
4
= 0:0001
1.3 Rules of Manipulating Logarithms
There are four main algebraic rules used to manipulate logarithms:
Rule 1: Product Rule
log
a
uv = log
a
u + log
a
v
Rule 2: Quotient Rule
log
a
u
v
= log
a
u log
a
v
Rule 3: Power Rule
log
a
u
v
=v log
a
u
1
Boston University CH102 - General Chemistry Spring 2012
Caution! The most common errors come from students mistakenly using two completely ctitious rules (there
are no rules that even resemble these): log
a
(u +v)6= log
a
u + log
a
v (logarithm of a sum) and log
b
(uv)6=
log
b
u log
b
v (logarithm of a dierence).
The practical implication of these rules, as we will see in the chapters dealing with thermodynamics, equilibrium,
and kinetics, is that we will be able to simplify complex algebraic expressions | easily.
1.4 Approximating Numerical Logarithms
In order to approximate the numerical values of non-trivial base-10 logarithms we will need (a) a good understanding
of the rules for manipulating logarithms and (b) the values of log 2 and log 3, which are 0:30 and 0:48, respectively.
Using these values and the rules we learned above, we can easily construct a table for the log values of integers
between 1 and 10:
x log x Justication
1 0:00 By denition
2 0:30 Given
3 0:48 Given
4 0:60 log 4 = log 2
2
= 2 log 2 = 2(0:3)
5 0:70 log 10 = log 2
 5 = log 2 + log 5
6 0:78 log 6 = log 2
 3 = log 2 + log 3
7 0:84 Estimated as 0:5(log 6 + log 8)
8 0:90 log 8 = log 2
3
= 3 log 2 = 3(0:3)
9 0:96 log 9 = log 3
2
= 2 log 3 = 2(0:48)
10 1:00 By denition
Notice that log 7 was determined using the approximation that it is the half-way point between log 6 and log 8.
In general, as numbers become larger, the distance between their logarithms becomes smaller. Consequently, this
approach should work well for large numbers. A graphical representation of this table is:
CH102 - Exam 2 Summer I, 2009
Useful information
!H " mc
s
#!T $ c
cal
#!T, N
A
" 6.0% 10
23
/mol, 1 cal = 4.184 J, 1 J = 1 kg m
2
! s
2
, 1 nm" 10
3
pm" 1% 10
&9
 m, 1 mL = 1 cm
3
, 1 m = 
100 cm, !E " mc
s
#!T, 1 cal = 4.184 J.
!E"!mc
2
, !E
light
" h', h" 6.6% 10
&34
J s, e" 1.6% 10
&19
 C, 1 eV" 1.6% 10
&19
 J, c" 3.0% 10
8
 m/s, J = kg m
2
! s
2
, 
N
n
" "1! 2#
n
#N
0
 1 Hz = 1/s, m
e
" 9.1% 10
&31
#kg.
p" mu" h!(, u"'(, KE"
p
2
)))))))))
2#m
"
1
))))
2
#mu
2
, PE"&
Ze
2
))))))))))))))))
4#*+
0
#r
"&
2#Z
)))))))))
r
#7.2% 10
&10
eV m, E
n
"&
Z
2
)))))))
n
2
#2.2% 10
&18
J"&13.6#eV#
Z
2
)))))))
n
2
, 
E
n
"
h
2
#n
2
)))))))))))))))
8#m L
2
.
N
A
" 6.0% 10
23
! mol, m
e
" 9.1% 10
&31
kg, J" kg m
2
! s
2
, nm " 10
&9
#m" 10#!" 1000#pm.
The table and figure below are of values of log
10
"x# versus x. Recall that ln"x#" 2.303#log
10
"x#.
x log
10
"x# x log
10
"x#
1 0.00 6 0.78
2 0.30 7 0.85
3 0.48 8 0.90
4 0.60 9 0.95
5 0.70 10 1.00
Values of log
10
(x) for 1,x, 10.
1 2 3 4 5 6 7 8 9 10
x
0.30
0.48
0.60
0.70
0.78
0.85
0.90
0.95
1
log
10
"x#
Values of log
10
(x) for 1,x, 10.
CH102 Exam 1, Monday, February 9, 2009 7
Copyright © 2009 Dan Dill (dan@bu.edu). All rights reserved
3
2
Page 3


Boston University CH102 - General Chemistry Spring 2012
Logarithms Tutorial for Chemistry Students
1 Logarithms
1.1 What is a logarithm?
Logarithms are the mathematical function that is used to represent the number (y) to which a base integer (a) is
raised in order to get the number x:
x =a
y
;
where y = log
a
(x). Most of you are familiar with the standard base-10 logarithm:
y = log
10
(x);
where x = 10
y
. A logarithm for which the base is not specied (y = logx) is always considered to be a base-10
logarithm.
1.2 Easy Logarithms
The simplest logarithms to evaluate, which most of you will be able to determine by inspection, are those where y
is an integer value. Take the power of 10's, for example:
log
10
(10) = 1 10
1
= 10
log
10
(100) = 2 10
2
= 100
log
10
(1000) = 3 10
3
= 1000
log
10
(10000) = 4 10
4
= 10000
log
10
(1) = 0 10
0
= 1
log
10
(0:1) =1 10
1
= 0:1
log
10
(0:01) =2 10
2
= 0:01
log
10
(0:001) =3 10
3
= 0:001
log
10
(0:0001) =4 10
4
= 0:0001
1.3 Rules of Manipulating Logarithms
There are four main algebraic rules used to manipulate logarithms:
Rule 1: Product Rule
log
a
uv = log
a
u + log
a
v
Rule 2: Quotient Rule
log
a
u
v
= log
a
u log
a
v
Rule 3: Power Rule
log
a
u
v
=v log
a
u
1
Boston University CH102 - General Chemistry Spring 2012
Caution! The most common errors come from students mistakenly using two completely ctitious rules (there
are no rules that even resemble these): log
a
(u +v)6= log
a
u + log
a
v (logarithm of a sum) and log
b
(uv)6=
log
b
u log
b
v (logarithm of a dierence).
The practical implication of these rules, as we will see in the chapters dealing with thermodynamics, equilibrium,
and kinetics, is that we will be able to simplify complex algebraic expressions | easily.
1.4 Approximating Numerical Logarithms
In order to approximate the numerical values of non-trivial base-10 logarithms we will need (a) a good understanding
of the rules for manipulating logarithms and (b) the values of log 2 and log 3, which are 0:30 and 0:48, respectively.
Using these values and the rules we learned above, we can easily construct a table for the log values of integers
between 1 and 10:
x log x Justication
1 0:00 By denition
2 0:30 Given
3 0:48 Given
4 0:60 log 4 = log 2
2
= 2 log 2 = 2(0:3)
5 0:70 log 10 = log 2
 5 = log 2 + log 5
6 0:78 log 6 = log 2
 3 = log 2 + log 3
7 0:84 Estimated as 0:5(log 6 + log 8)
8 0:90 log 8 = log 2
3
= 3 log 2 = 3(0:3)
9 0:96 log 9 = log 3
2
= 2 log 3 = 2(0:48)
10 1:00 By denition
Notice that log 7 was determined using the approximation that it is the half-way point between log 6 and log 8.
In general, as numbers become larger, the distance between their logarithms becomes smaller. Consequently, this
approach should work well for large numbers. A graphical representation of this table is:
CH102 - Exam 2 Summer I, 2009
Useful information
!H " mc
s
#!T $ c
cal
#!T, N
A
" 6.0% 10
23
/mol, 1 cal = 4.184 J, 1 J = 1 kg m
2
! s
2
, 1 nm" 10
3
pm" 1% 10
&9
 m, 1 mL = 1 cm
3
, 1 m = 
100 cm, !E " mc
s
#!T, 1 cal = 4.184 J.
!E"!mc
2
, !E
light
" h', h" 6.6% 10
&34
J s, e" 1.6% 10
&19
 C, 1 eV" 1.6% 10
&19
 J, c" 3.0% 10
8
 m/s, J = kg m
2
! s
2
, 
N
n
" "1! 2#
n
#N
0
 1 Hz = 1/s, m
e
" 9.1% 10
&31
#kg.
p" mu" h!(, u"'(, KE"
p
2
)))))))))
2#m
"
1
))))
2
#mu
2
, PE"&
Ze
2
))))))))))))))))
4#*+
0
#r
"&
2#Z
)))))))))
r
#7.2% 10
&10
eV m, E
n
"&
Z
2
)))))))
n
2
#2.2% 10
&18
J"&13.6#eV#
Z
2
)))))))
n
2
, 
E
n
"
h
2
#n
2
)))))))))))))))
8#m L
2
.
N
A
" 6.0% 10
23
! mol, m
e
" 9.1% 10
&31
kg, J" kg m
2
! s
2
, nm " 10
&9
#m" 10#!" 1000#pm.
The table and figure below are of values of log
10
"x# versus x. Recall that ln"x#" 2.303#log
10
"x#.
x log
10
"x# x log
10
"x#
1 0.00 6 0.78
2 0.30 7 0.85
3 0.48 8 0.90
4 0.60 9 0.95
5 0.70 10 1.00
Values of log
10
(x) for 1,x, 10.
1 2 3 4 5 6 7 8 9 10
x
0.30
0.48
0.60
0.70
0.78
0.85
0.90
0.95
1
log
10
"x#
Values of log
10
(x) for 1,x, 10.
CH102 Exam 1, Monday, February 9, 2009 7
Copyright © 2009 Dan Dill (dan@bu.edu). All rights reserved
3
2
Boston University CH102 - General Chemistry Spring 2012
The same approach can be used for numbers larger than ten (or smaller than one). Let's outline a general
approach while solving for log 0:0036
1. If the number is a decimal, express the number as a whole number times 10 to a power.
log 0:0036 = log

36 10
4


2. Apply the product and power rules to separate the power of ten term and evaluate it.
log

36 10
4


= log 36 + (4) log 10 = log 36 4
3. Express the remaining number (36) as a product of prime factors.
log 36 4 = log (4
 9) 4 = log

2
2
3
2


 4
4. Apply the product and power rules to separate all of the factors and use the table for log 1 to log 10 to evaluate
them
1
.
log

2
2
3
2


 4 = 2 log 2 + 2 log 3 4 = 2(0:3) + 2(0:48) 4 =2:44
The actual value
2
of log 0:0036 is2:4436.
Example: Approximating the value of log

2:2 10
5


Following the same steps as above:
log

2:2 10
5


= log

22 10
6


= log 22 + (6) log 10
= log (2
 11) 6
= log 2 + log 11 6
= 0:3 + 1:04

 6
= 4:66 (Exact =4:658)

Here, log 11 was computed by taking the average of log 10 (= 1:00) and log 12 (= 1:08).
Exercises:
1. Approximate, numerically, the value of the following logarithms:
(a) log 0:24 (b) log 0:0027 (c) log 0:045
(d) log 810 (e) log 6:3 (f) log 14:7
(g) log 2:8 10
2
(h) log 1:7 10
5
(i) log 7:3 10
3
(j) log 0:25
2
(k) log
p
1:8 10
5
(l) log 75
(1=3)
2. Use your scientic calculator to compute the precise value of the above logarithms. If there are any signicant
discrepancies, try them again! This exercise can be repeated, using any random numbers, until you feel
comfortable computing logarithms by hand.
1
You may run into a prime factor that is greater than 7. If that is the case, use the same approach we used to solve log 7 to solve
the logarithm of that prime factor.
2
Computed using a Texas Instruments scientic calculator
3
Page 4


Boston University CH102 - General Chemistry Spring 2012
Logarithms Tutorial for Chemistry Students
1 Logarithms
1.1 What is a logarithm?
Logarithms are the mathematical function that is used to represent the number (y) to which a base integer (a) is
raised in order to get the number x:
x =a
y
;
where y = log
a
(x). Most of you are familiar with the standard base-10 logarithm:
y = log
10
(x);
where x = 10
y
. A logarithm for which the base is not specied (y = logx) is always considered to be a base-10
logarithm.
1.2 Easy Logarithms
The simplest logarithms to evaluate, which most of you will be able to determine by inspection, are those where y
is an integer value. Take the power of 10's, for example:
log
10
(10) = 1 10
1
= 10
log
10
(100) = 2 10
2
= 100
log
10
(1000) = 3 10
3
= 1000
log
10
(10000) = 4 10
4
= 10000
log
10
(1) = 0 10
0
= 1
log
10
(0:1) =1 10
1
= 0:1
log
10
(0:01) =2 10
2
= 0:01
log
10
(0:001) =3 10
3
= 0:001
log
10
(0:0001) =4 10
4
= 0:0001
1.3 Rules of Manipulating Logarithms
There are four main algebraic rules used to manipulate logarithms:
Rule 1: Product Rule
log
a
uv = log
a
u + log
a
v
Rule 2: Quotient Rule
log
a
u
v
= log
a
u log
a
v
Rule 3: Power Rule
log
a
u
v
=v log
a
u
1
Boston University CH102 - General Chemistry Spring 2012
Caution! The most common errors come from students mistakenly using two completely ctitious rules (there
are no rules that even resemble these): log
a
(u +v)6= log
a
u + log
a
v (logarithm of a sum) and log
b
(uv)6=
log
b
u log
b
v (logarithm of a dierence).
The practical implication of these rules, as we will see in the chapters dealing with thermodynamics, equilibrium,
and kinetics, is that we will be able to simplify complex algebraic expressions | easily.
1.4 Approximating Numerical Logarithms
In order to approximate the numerical values of non-trivial base-10 logarithms we will need (a) a good understanding
of the rules for manipulating logarithms and (b) the values of log 2 and log 3, which are 0:30 and 0:48, respectively.
Using these values and the rules we learned above, we can easily construct a table for the log values of integers
between 1 and 10:
x log x Justication
1 0:00 By denition
2 0:30 Given
3 0:48 Given
4 0:60 log 4 = log 2
2
= 2 log 2 = 2(0:3)
5 0:70 log 10 = log 2
 5 = log 2 + log 5
6 0:78 log 6 = log 2
 3 = log 2 + log 3
7 0:84 Estimated as 0:5(log 6 + log 8)
8 0:90 log 8 = log 2
3
= 3 log 2 = 3(0:3)
9 0:96 log 9 = log 3
2
= 2 log 3 = 2(0:48)
10 1:00 By denition
Notice that log 7 was determined using the approximation that it is the half-way point between log 6 and log 8.
In general, as numbers become larger, the distance between their logarithms becomes smaller. Consequently, this
approach should work well for large numbers. A graphical representation of this table is:
CH102 - Exam 2 Summer I, 2009
Useful information
!H " mc
s
#!T $ c
cal
#!T, N
A
" 6.0% 10
23
/mol, 1 cal = 4.184 J, 1 J = 1 kg m
2
! s
2
, 1 nm" 10
3
pm" 1% 10
&9
 m, 1 mL = 1 cm
3
, 1 m = 
100 cm, !E " mc
s
#!T, 1 cal = 4.184 J.
!E"!mc
2
, !E
light
" h', h" 6.6% 10
&34
J s, e" 1.6% 10
&19
 C, 1 eV" 1.6% 10
&19
 J, c" 3.0% 10
8
 m/s, J = kg m
2
! s
2
, 
N
n
" "1! 2#
n
#N
0
 1 Hz = 1/s, m
e
" 9.1% 10
&31
#kg.
p" mu" h!(, u"'(, KE"
p
2
)))))))))
2#m
"
1
))))
2
#mu
2
, PE"&
Ze
2
))))))))))))))))
4#*+
0
#r
"&
2#Z
)))))))))
r
#7.2% 10
&10
eV m, E
n
"&
Z
2
)))))))
n
2
#2.2% 10
&18
J"&13.6#eV#
Z
2
)))))))
n
2
, 
E
n
"
h
2
#n
2
)))))))))))))))
8#m L
2
.
N
A
" 6.0% 10
23
! mol, m
e
" 9.1% 10
&31
kg, J" kg m
2
! s
2
, nm " 10
&9
#m" 10#!" 1000#pm.
The table and figure below are of values of log
10
"x# versus x. Recall that ln"x#" 2.303#log
10
"x#.
x log
10
"x# x log
10
"x#
1 0.00 6 0.78
2 0.30 7 0.85
3 0.48 8 0.90
4 0.60 9 0.95
5 0.70 10 1.00
Values of log
10
(x) for 1,x, 10.
1 2 3 4 5 6 7 8 9 10
x
0.30
0.48
0.60
0.70
0.78
0.85
0.90
0.95
1
log
10
"x#
Values of log
10
(x) for 1,x, 10.
CH102 Exam 1, Monday, February 9, 2009 7
Copyright © 2009 Dan Dill (dan@bu.edu). All rights reserved
3
2
Boston University CH102 - General Chemistry Spring 2012
The same approach can be used for numbers larger than ten (or smaller than one). Let's outline a general
approach while solving for log 0:0036
1. If the number is a decimal, express the number as a whole number times 10 to a power.
log 0:0036 = log

36 10
4


2. Apply the product and power rules to separate the power of ten term and evaluate it.
log

36 10
4


= log 36 + (4) log 10 = log 36 4
3. Express the remaining number (36) as a product of prime factors.
log 36 4 = log (4
 9) 4 = log

2
2
3
2


 4
4. Apply the product and power rules to separate all of the factors and use the table for log 1 to log 10 to evaluate
them
1
.
log

2
2
3
2


 4 = 2 log 2 + 2 log 3 4 = 2(0:3) + 2(0:48) 4 =2:44
The actual value
2
of log 0:0036 is2:4436.
Example: Approximating the value of log

2:2 10
5


Following the same steps as above:
log

2:2 10
5


= log

22 10
6


= log 22 + (6) log 10
= log (2
 11) 6
= log 2 + log 11 6
= 0:3 + 1:04

 6
= 4:66 (Exact =4:658)

Here, log 11 was computed by taking the average of log 10 (= 1:00) and log 12 (= 1:08).
Exercises:
1. Approximate, numerically, the value of the following logarithms:
(a) log 0:24 (b) log 0:0027 (c) log 0:045
(d) log 810 (e) log 6:3 (f) log 14:7
(g) log 2:8 10
2
(h) log 1:7 10
5
(i) log 7:3 10
3
(j) log 0:25
2
(k) log
p
1:8 10
5
(l) log 75
(1=3)
2. Use your scientic calculator to compute the precise value of the above logarithms. If there are any signicant
discrepancies, try them again! This exercise can be repeated, using any random numbers, until you feel
comfortable computing logarithms by hand.
1
You may run into a prime factor that is greater than 7. If that is the case, use the same approach we used to solve log 7 to solve
the logarithm of that prime factor.
2
Computed using a Texas Instruments scientic calculator
3
Boston University CH102 - General Chemistry Spring 2012
1.5 Natural Logarithms
Natural logarithms are a specic subset of the general logarithm (x =a
log
a
(x)
), where the base (a) is the number
e (= 2:718:::). The natural logarithm is formally dened by:
x =e
ln(x)
;
where ln(x) (= log
e
x) is the `natural log of x'.
To compute natural logarithms we can employ the following simple identity: ln(x) = 2:303 log(x).
Example: Approximating the value of ln

2:2 10
5


Following the same steps as above:
ln

2:2 10
5


= 2:303 log

2:2 10
5


= 2:303 (4:66)
= 10:73 (Exact =10:72)

Here, log

2:2 10
5


was taken from the exercise in the previous problem.
Exercises:
1. Approximate, numerically, the value of the following logarithms:
(a) ln 0:12 (b) ln 0:00625 (c) ln 0:064
(d) ln 210 (e) ln 5:5 (f) ln 12:4
(g) ln 3:6 10
3
(h) ln 2:5 10
7
(i) ln 8:3 10
2
(j) ln 0:18
3
(k) ln
p
4:9 10
4
(l) ln 25
(1=4)
2. Use your scientic calculator to compute the precise value of the above logarithms. If there are any signicant
discrepancies, try them again! This exercise can be repeated, using any random numbers, until you feel
comfortable computing logarithms by hand.
2 Antilogarithms
The antilogarithm, or power, function eectively undoes a logarithm. The best example of this in Chemistry is to
compute the hydronium ion concentration from the pH. In this case,
pH = log[H
3
O
+
];
and the hydronium ion concentration can be found from the pH using:
[H
3
O
+
] = 10
pH
:
2.1 Approximating Base-10 Antilogs
Consider the antilog of2:16. The procedure for computing the power, 10
2:16
, is as follows:
1. Rewrite the power as ten to the power of the dierence of two numbers: a number between 0 and 1, and an
integer.
10
2:15
= 10
0:853
4
Page 5


Boston University CH102 - General Chemistry Spring 2012
Logarithms Tutorial for Chemistry Students
1 Logarithms
1.1 What is a logarithm?
Logarithms are the mathematical function that is used to represent the number (y) to which a base integer (a) is
raised in order to get the number x:
x =a
y
;
where y = log
a
(x). Most of you are familiar with the standard base-10 logarithm:
y = log
10
(x);
where x = 10
y
. A logarithm for which the base is not specied (y = logx) is always considered to be a base-10
logarithm.
1.2 Easy Logarithms
The simplest logarithms to evaluate, which most of you will be able to determine by inspection, are those where y
is an integer value. Take the power of 10's, for example:
log
10
(10) = 1 10
1
= 10
log
10
(100) = 2 10
2
= 100
log
10
(1000) = 3 10
3
= 1000
log
10
(10000) = 4 10
4
= 10000
log
10
(1) = 0 10
0
= 1
log
10
(0:1) =1 10
1
= 0:1
log
10
(0:01) =2 10
2
= 0:01
log
10
(0:001) =3 10
3
= 0:001
log
10
(0:0001) =4 10
4
= 0:0001
1.3 Rules of Manipulating Logarithms
There are four main algebraic rules used to manipulate logarithms:
Rule 1: Product Rule
log
a
uv = log
a
u + log
a
v
Rule 2: Quotient Rule
log
a
u
v
= log
a
u log
a
v
Rule 3: Power Rule
log
a
u
v
=v log
a
u
1
Boston University CH102 - General Chemistry Spring 2012
Caution! The most common errors come from students mistakenly using two completely ctitious rules (there
are no rules that even resemble these): log
a
(u +v)6= log
a
u + log
a
v (logarithm of a sum) and log
b
(uv)6=
log
b
u log
b
v (logarithm of a dierence).
The practical implication of these rules, as we will see in the chapters dealing with thermodynamics, equilibrium,
and kinetics, is that we will be able to simplify complex algebraic expressions | easily.
1.4 Approximating Numerical Logarithms
In order to approximate the numerical values of non-trivial base-10 logarithms we will need (a) a good understanding
of the rules for manipulating logarithms and (b) the values of log 2 and log 3, which are 0:30 and 0:48, respectively.
Using these values and the rules we learned above, we can easily construct a table for the log values of integers
between 1 and 10:
x log x Justication
1 0:00 By denition
2 0:30 Given
3 0:48 Given
4 0:60 log 4 = log 2
2
= 2 log 2 = 2(0:3)
5 0:70 log 10 = log 2
 5 = log 2 + log 5
6 0:78 log 6 = log 2
 3 = log 2 + log 3
7 0:84 Estimated as 0:5(log 6 + log 8)
8 0:90 log 8 = log 2
3
= 3 log 2 = 3(0:3)
9 0:96 log 9 = log 3
2
= 2 log 3 = 2(0:48)
10 1:00 By denition
Notice that log 7 was determined using the approximation that it is the half-way point between log 6 and log 8.
In general, as numbers become larger, the distance between their logarithms becomes smaller. Consequently, this
approach should work well for large numbers. A graphical representation of this table is:
CH102 - Exam 2 Summer I, 2009
Useful information
!H " mc
s
#!T $ c
cal
#!T, N
A
" 6.0% 10
23
/mol, 1 cal = 4.184 J, 1 J = 1 kg m
2
! s
2
, 1 nm" 10
3
pm" 1% 10
&9
 m, 1 mL = 1 cm
3
, 1 m = 
100 cm, !E " mc
s
#!T, 1 cal = 4.184 J.
!E"!mc
2
, !E
light
" h', h" 6.6% 10
&34
J s, e" 1.6% 10
&19
 C, 1 eV" 1.6% 10
&19
 J, c" 3.0% 10
8
 m/s, J = kg m
2
! s
2
, 
N
n
" "1! 2#
n
#N
0
 1 Hz = 1/s, m
e
" 9.1% 10
&31
#kg.
p" mu" h!(, u"'(, KE"
p
2
)))))))))
2#m
"
1
))))
2
#mu
2
, PE"&
Ze
2
))))))))))))))))
4#*+
0
#r
"&
2#Z
)))))))))
r
#7.2% 10
&10
eV m, E
n
"&
Z
2
)))))))
n
2
#2.2% 10
&18
J"&13.6#eV#
Z
2
)))))))
n
2
, 
E
n
"
h
2
#n
2
)))))))))))))))
8#m L
2
.
N
A
" 6.0% 10
23
! mol, m
e
" 9.1% 10
&31
kg, J" kg m
2
! s
2
, nm " 10
&9
#m" 10#!" 1000#pm.
The table and figure below are of values of log
10
"x# versus x. Recall that ln"x#" 2.303#log
10
"x#.
x log
10
"x# x log
10
"x#
1 0.00 6 0.78
2 0.30 7 0.85
3 0.48 8 0.90
4 0.60 9 0.95
5 0.70 10 1.00
Values of log
10
(x) for 1,x, 10.
1 2 3 4 5 6 7 8 9 10
x
0.30
0.48
0.60
0.70
0.78
0.85
0.90
0.95
1
log
10
"x#
Values of log
10
(x) for 1,x, 10.
CH102 Exam 1, Monday, February 9, 2009 7
Copyright © 2009 Dan Dill (dan@bu.edu). All rights reserved
3
2
Boston University CH102 - General Chemistry Spring 2012
The same approach can be used for numbers larger than ten (or smaller than one). Let's outline a general
approach while solving for log 0:0036
1. If the number is a decimal, express the number as a whole number times 10 to a power.
log 0:0036 = log

36 10
4


2. Apply the product and power rules to separate the power of ten term and evaluate it.
log

36 10
4


= log 36 + (4) log 10 = log 36 4
3. Express the remaining number (36) as a product of prime factors.
log 36 4 = log (4
 9) 4 = log

2
2
3
2


 4
4. Apply the product and power rules to separate all of the factors and use the table for log 1 to log 10 to evaluate
them
1
.
log

2
2
3
2


 4 = 2 log 2 + 2 log 3 4 = 2(0:3) + 2(0:48) 4 =2:44
The actual value
2
of log 0:0036 is2:4436.
Example: Approximating the value of log

2:2 10
5


Following the same steps as above:
log

2:2 10
5


= log

22 10
6


= log 22 + (6) log 10
= log (2
 11) 6
= log 2 + log 11 6
= 0:3 + 1:04

 6
= 4:66 (Exact =4:658)

Here, log 11 was computed by taking the average of log 10 (= 1:00) and log 12 (= 1:08).
Exercises:
1. Approximate, numerically, the value of the following logarithms:
(a) log 0:24 (b) log 0:0027 (c) log 0:045
(d) log 810 (e) log 6:3 (f) log 14:7
(g) log 2:8 10
2
(h) log 1:7 10
5
(i) log 7:3 10
3
(j) log 0:25
2
(k) log
p
1:8 10
5
(l) log 75
(1=3)
2. Use your scientic calculator to compute the precise value of the above logarithms. If there are any signicant
discrepancies, try them again! This exercise can be repeated, using any random numbers, until you feel
comfortable computing logarithms by hand.
1
You may run into a prime factor that is greater than 7. If that is the case, use the same approach we used to solve log 7 to solve
the logarithm of that prime factor.
2
Computed using a Texas Instruments scientic calculator
3
Boston University CH102 - General Chemistry Spring 2012
1.5 Natural Logarithms
Natural logarithms are a specic subset of the general logarithm (x =a
log
a
(x)
), where the base (a) is the number
e (= 2:718:::). The natural logarithm is formally dened by:
x =e
ln(x)
;
where ln(x) (= log
e
x) is the `natural log of x'.
To compute natural logarithms we can employ the following simple identity: ln(x) = 2:303 log(x).
Example: Approximating the value of ln

2:2 10
5


Following the same steps as above:
ln

2:2 10
5


= 2:303 log

2:2 10
5


= 2:303 (4:66)
= 10:73 (Exact =10:72)

Here, log

2:2 10
5


was taken from the exercise in the previous problem.
Exercises:
1. Approximate, numerically, the value of the following logarithms:
(a) ln 0:12 (b) ln 0:00625 (c) ln 0:064
(d) ln 210 (e) ln 5:5 (f) ln 12:4
(g) ln 3:6 10
3
(h) ln 2:5 10
7
(i) ln 8:3 10
2
(j) ln 0:18
3
(k) ln
p
4:9 10
4
(l) ln 25
(1=4)
2. Use your scientic calculator to compute the precise value of the above logarithms. If there are any signicant
discrepancies, try them again! This exercise can be repeated, using any random numbers, until you feel
comfortable computing logarithms by hand.
2 Antilogarithms
The antilogarithm, or power, function eectively undoes a logarithm. The best example of this in Chemistry is to
compute the hydronium ion concentration from the pH. In this case,
pH = log[H
3
O
+
];
and the hydronium ion concentration can be found from the pH using:
[H
3
O
+
] = 10
pH
:
2.1 Approximating Base-10 Antilogs
Consider the antilog of2:16. The procedure for computing the power, 10
2:16
, is as follows:
1. Rewrite the power as ten to the power of the dierence of two numbers: a number between 0 and 1, and an
integer.
10
2:15
= 10
0:853
4
Boston University CH102 - General Chemistry Spring 2012
2. Separate the terms using the identity 10
a+b
= 10
a
10
b
.
10
0:853
= 10
0:85
 10
3
3. Use the denition of a base-10 logarithm (x = 10
y
) to determine the value of x. The easiest way to do this is
to use the logarithm graph (or table) from section 5.4. In this case, 10
0:85
 7.
10
0:85
 10
3
 7 10
3
Example: Approximating the value of 10
4:74
Following the same steps as above:
10
4:74
= 10
0:265
= 10
0:26
10
5
 2 10
5
(Exact = 1:8 10
5
)
Here, 10
0:26
is approximated as 2 from the graph in section 1.4.
Exercises:
1. Approximate, numerically, the value of the following antilogs to 1 signicant gure:
(a) 10
1:5
(b) 10
12:3
(c) 10
4:91
(d) 10
2:28
(e) 10
5:71
(f) 10
17:44
2. Use your scientic calculator to compute the precise value of the above logarithms. If there are any signicant
discrepancies, try them again! This exercise can be repeated, using any random numbers, until you feel
comfortable computing logarithms by hand.
5