Logarithms
Mathematics Skills Guide
This is one of a series of guides designed to help you increase your confidence in
handling mathematics. This guide contains both theory and exercises which cover:1. Revision of Indices
2. What are Logarithms?
3. Rules for manipulation 4. Solving equations using logarithms
5. The straight line law
6. Graph of lnx and its inverse
There are often different ways of doing things in mathematics and the methods
suggested in the guides may not be the ones you were taught. If you are successful
and happy with the methods you use it may not be necessary for you to change
them. If you have problems or need help in any part of the work then there are a
number of ways you can get help.
For students at the University of Hull
 Ask your lecturers.
 You can contact a maths Skills Adviser from the Skills Team on the email shown
below.
 Access more maths Skills Guides and resources at the website below.
 Look at one of the many textbooks in the library.
Web: www.hull.ac.uk/skills
Email: [email protected]
1. Revision of Indices
a) Logarithms are really indices so understanding indices and how to manipulate
them will help you to deal with logarithms.
Section b) shows why the rules of indices work – basically because we define them
to work in the way we want them to! It will help you sort out logarithms if you
understand why we define the rules of indices in the way we do.
If you are really in a hurry you could skip section b) and go straight to section c)
which starts with the list of rules!
b) The rules of indices
This section is taken from the booklet Algebra 1.
If m and n are positive integers and a  0 , then
a m  a n  a m n
 n  amn
a m  a n  a mn (if m  n ) and a m
this means x 4  x3  x 43  x7 ; 54  56  546  510 ; y8  y3  y83  y5 etc.
That’s OK but what about other values of the indices (the m and n above)
Consider
x3
x
5
; expanding gives
x
If the ‘rule’ works
x3
x
Consider
x5
x
5
x3
5
If the ‘rule’ works
x5
x
x
x5

x x x
1

x x x x x
x2
 x3  x5  x35  x  2
; expanding gives
5
5
5

So we define x 2 to be
1
x2
x x x x x
 1
x x x x x
 x 5 5  x 0
So we define x 0 to be 1
1
Consider x 2 we know x  x  x (for instance 9  9  3  3  9 etc.), so
1
1
1
So we define x 2  x
If the ‘rule’ works x 2  x 2  x1  x
Note for x to exist x  0 as you can’t take the square root of a negative number.
We extend the meaning and use of indices by defining (for x  0 )
1
1
, x n as n x and x 0 as 1.
x  n as
xn
It is important to realise that these rules fit what we know for positive integers and are
defined to be true for all indices (positive, negative or fractional).
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c) Summary of the rules:
For all values of a (where a  0 )
(i) a m  a n  a m  n
(ii) a m  a n  a m n
(iii) a m
(iv) a n  n a
 n  amn
1
Combining the last two means that a
m
n
 
m
n
 am  n a
For practice on these rules see Algebra 1 exercises 4 and 5.
2. What are Logarithms?
If y  a x then we introduce the inverse function logarithm and define loga y  x
(read as log base a of y equals x ).
or
y  a x  loga y  x
Where  means “implies and is implied by” i.e. it works both ways!
Note this means that, going from exponent form to logarithmic form:
102  100  log10 (100)  2
102  0.01 log10 (0.01)  2
100  1 
1
92  3 
log10 (1)  0
25  32 
log9 (3)  1
83  4 
2
2
log2 (32)  5
log8 ( 4)  2
3
And in going from logarithmic form to exponent form:
log10 (10)  1  101  10
log10 (0.001)  3  103  0.001
log3 (81)  4  34  81
log2 (1)  0  20  1
1
log100 (10)  1 
100 2  10
2
log5 (5 5 )  3 
2
3
52  5 5
In practice we use, mainly, logs base e ( e  2.718... ) and, occasionally base 10.
On your calculator you should find a button labelled LOG with the 2nd function 10 x
and a second button labelled LN (or ln) with the second function e x .
If you put 1.45 into your calculator, find 101.45 and then find the LOG of the answer
you should get back to 1.45. This works for any number you choose.The reverse
also works (except for 0) if you find the LOG first and then 10 to the power of the
answer. You get the same results when using the LN button.
x
use 10 x button
then LOG button
1.45
28.183832931 1.45
-0.254 0.5571857489 -0.245
Try a few numbers for yourself!
use e x button
4.263114515
0.775691802
2
then LN
button
1.45
-0.254
Exercise 1
1. Write the equivalent logarithmic form for the following
(i) 104  10000
1
(ii ) 35  243 (iii ) 125 3  5 (iv ) 42  0.0625
2. Write the equivalent exponent form for the following
(i ) log10 (1000000)  6 (ii ) log4 64  3 (iii ) log4096 8  1
(iv ) log5 0.0016  4
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3. Given that 100.30105  2 write down the values of (i ) log10 2 (ii ) log10 200
4. Use your calculator to write down the values of the following
(i ) log 234 (ii ) ln 234 (iii ) log0.1245 (iv ) ln 0.1245
3. Rules for the manipulation of logs
Remember in y  a x the x is an index and as x  loga y this is why logs work like
indices.
Proofs of the rules
(i) Given p  a x and q  a y then pq  a x  a y  a x  y
By the definition
p  a x  x  loga p
q  ay

y  loga q
So x  y  loga p  loga q
But pq  a x  y

x  y  loga  pq 
This shows that loga  pq   x  y  loga p  loga q
p
(ii) In the same way we can show that loga    loga p  loga q
q
 
(iii) loga  p n  n loga p
By the definition
Also
Hence
p  ax

x  loga p
 n  anx  nx  loga p n 
loga p n   nx  n loga p
pn  a x
The rules are:
loga ( pq)  loga p  loga q
p
loga    loga p  loga q
q
 
 
loga p n  n loga p
You can check some of these on your calculator. Take various values of a and b
and, using either the log button or the ln button, compare the values of
log a  logb with log(ab) , log a  logb with log( a ) l, log(a b ) with blog a .
b
3
Using logs base 10
a
ab
b
logab
loga
logb
log a  log b
5
7.2
36
1.5563
0.6990
0.8573
1.5563
3.2
1.9
6.08
0.7839
0.5051
0.2788
0.7839
a
b
a
b
loga
logb
log a  log b
5
7.2
0.6940
-0.1584
0.6990
0.8573
-0.1584
3.2
1.9
1.6842
0.2264
0.5051
0.2788
0.2264
log

a
b
Very Important Notes Using p  a x  x  loga p ;
1. Putting x  0 , p  a 0  1  0  loga 1 i.e. the log (any base) of 1 is zero.
2. Putting x  1 , p  a1  a  1  loga a hence lne  1, log10 10  1 etc
 
3. Taking logs base a of both sides gives loga p  loga a x but as x  loga p
 
so we have loga a x  x for any base.
4. Putting x  loga p into p  a x gives p  a loga p or a loga p  p .
5. As a  0 then for all values of x , a x is positive and, as p  a x then p  0
for all x , then we can only take the log of a positive number.
Examples (Where the question says log, this implies loga where a  0
1. Write as single logarithmic terms
(i) log 4  log 7 :
log 4  log7  log(4  7)  log 28
3log5  log(53 )  log125
(ii) 3log5 :
(iii) 1 log 25 :
1 log 25
2
2
(iv) –log6 :
 
2. Simplify log 3bd :
 
3. Simplify ln ce kx :
1
 log 25 2  log 25  log5
 log 6   1  log 6  log 6 1  log 1
6
3
d
log
 log 3d  logb  log3  log d  logb
b
 
 
 
 
ln ce kx  ln c  ln e kx  ln c  kx ln e  ln c  kx
4. Simplify 3 ln 4  2 ln 3  5 ln 2  ln 6 :
3 ln 4  2 ln 3  5 ln 2  ln 6  ln 4 3  ln 3 2  ln 2 5  ln 6
2 
 43  32 
 6
  ln 2  3   ln 3
 ln
 25  6 
 25  2  3 




The answer ln 3 may be more suitable than the 1.0896 that the calculator gives!
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Exercise 2
1. Express the following in terms of log p , logq and logr

p
r
(i) log( pqr ) (ii) log( p2 ) (iii) log

(iv) log 1
q
(v) log r (vi) log qr
2. Write as single logarithmic terms
(i) log5  log3 (ii) log26  log2 (iii) 4log3 (iv) 1 log 36 (v) log43  5log2
2
(vi) log4  log12  log8 (vii) 3log6  log4  2log3 .
3. Simplify (some answers have a log term)

(i) log10 3  10 4

(ii) ln(e6 )
(iii) log 1  2 log 2  3 log 2
4
 
3
(iv) ln(8 x)  ln(3 x3 )  ln(6 x2 ) (v) 1 log16 x 2  2 log3x   3 log4 x 
4
2
4. Solving equations using logs
Examples
(i) Solve the equation 10 x  3.79
The definition of logs says if y  a x then loga y  x or y  a x  x  loga y
Hence 10 x  3.79  x  log10 3.79  0.57864 (to 5 decimal places)
Check 100.57864  3.79000 (to 5 decimal places)
In practice from 10 x  3.79 we take logs to base 10 giving
 
log10 10 x  log3.79
x log10 10  log3.79
x  0.57864
(ii) Solve the equation 32 x  56
 
log10 3 2 x  log10 56
2 x log10 3  log10 56
2x 
log10 56
 3.66403...
log10 3
x  1.83201....
Check 33  27 , 34  81, we want 32 x so the value of 2x lies between 3 and 4 or
3  2x  4 which means x lies between 1.5 and 2.
This tells us that x  1.83201... is roughly correct.
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(iii) Solve the equation 4 x  3 x1
4 x  3 x 1
x log10 4  x  1 log10 3
 x log10 3  log10 3
x log10 4  x log10 3  log10 3
xlog10 4  log10 3  log10 3
x
log10 3
 3.8188..
log10 4  log10 3

4 x  43.8188..  44  256

Check  x 1
 very close!
4.8188..
5


3

3

3

243


Note you could combine terms, giving,
log10 3
log10 3
x

 3.8188..
log10 4  log10 3 log10 4
3 
(iv) Solve the equation 4 x  6  35 2 x
4 x  6  35  2 x
x  6log 4  5  2 x log 3
Take logs of both sides
x log 4  6 log 4  5 log 3  2 x log 3
Expand brackets
Collect terms x log 4  2 x log 3  5 log 3  6 log 4
Factorise the left hand side xlog 4  2 log 3  5 log 3  6 log 4
5 log 3  6 log 4
divide
x
 -0.78825
log 4  2 log 3
(Note you get the same answer by using the ln button on your calculator.)
Check 4 x 6  40.788256  45.21175  1373.368
and 35 2 x  35 2(.78825)  36.576498  1373.368
Notice that you could combine the log-terms in
5 log 3  6 log 4
log 35  4 6
x
to give x 
log 4  2 log 3
log 4  3 2




It does not really simplify things here but, in some cases, it can.
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  
(v) Solve the equation 7 3x 1  2 52 x 1

  

7 3 x 1  2 5 2 x  1
log 7  x  1log 3  log 2  2 x  1log 5
Take logs of both sides
Expand brackets log 7  x log 3  log 3  log 2  2 x log 5  log 5
Collect terms
x log 3  2 x log 5  log 2  log 5  log 7  log 3
 7 
x log 3   log30 
25
7
log30 
0.632023
7
x

 0.686371
3
log   0.920819
25
xlog 3  2 log 5  log 2  5 3
Factorise left hand side
simplify
divide
Check (not easy!!)
7 7
 

32 9
2
RHS = 252 x 1   2  50.4 

0.4
5
LHS = 7 3x 1  7  31.7 
(taking 31.7  32  9 )
2
(taking 50.4  50.5  5  2.2... )
1
5
The values of LHS and RHS are roughly the same. A more exact check could be
made using a calculator.
Exercise 3
Solve the equations
1. 3x  6
5. 42a 1  3
2. 5 x  4
6. 62 p 1  513 p
9. 53 2s  33s  4
4. 3 y 1  7
3. 22 x  5

7.
 
23 s  5
10. 5 95 x 2  7 453x
 

8. 2r 1 23r 2  128

5. The straight line law
There are many uses of logs. They are frequently required in Calculus and also
occur when analysing data from experiments.
In an experiment the following results were obtained.
x
2
4
5
6.5
y
4
10
13
19.2
8
23
11
37
Drawing a graph of values of y against x gives a curve but it is not possible to give
its equation. I.e. it is not possible to find the function which will give other values of
y for given values of x . It is possible to estimate values from the graph.
It is suggested that the graph of the function may be of the form y  ax n .
If we take logs (base e , though you could use base 10) of both sides.
 
ln y  ln ax n  ln a  n ln x
7
If we put Y  ln y, X  ln x and A  ln a this becomes Y  A  nX which is the usual
equation of a straight line.
x
y
X  ln x
Y  ln y
2
4
0.69
1.39
4
10
1.39
2.30
5
13
1.61
2.56
6.5
19.2
1.87
2.95
By drawing a straight line which fits the points
as closely as possible we can get approximate
values for A and n .
8
23
2.08
3.14
11
37
2.40
3.61
Y


3
The equation is Y  A  nX
The intercept is at A  0.7 which gives
ln a  0.7  a  2.014
The gradient is found by taking two points and
using
Gradient =
Y2  Y1
X 2  X1

A


2

1
B
Take A(2.5, 3.7), B(0, 0.7)
Gradient AB 
3.7  0.7
 1.2
2.5  0
11
2
The gradient n  1.2 .
The formula is (approximately) y  2x1.2 .
From the graph the point (0.69, 1.39) could be an error in a reading!
Exercise 4
Find the values of the unknowns given the following sets of readings
1. It is assumed that the connection is of the form y  ax n
x
0.5
2
3
5
5.6
y
1.1
8.5
15.6
33.5
39.8
8
67.9
2. It is assumed that the connection is of the form y  ax n
x
1.3
1.7
2
3.6
4.3
y
81
54
42
17.6
13.5
6
8.2
8
3
X
6 The graph of y = lnx and its inverse.
Drawing the graphs of
y = ex
y
y  e x and y  ln x
using the same scale on the axes shows that
one is the reflection of the other in the line
y  x . This fits with the definition of one
being the inverse of the other.
y=x
1
y = lnx
Take y  e x , interchanging x and y (which
gives the inverse function) gives x  e y ,
take logs base e ln x  y
Answers
Exercise 1
1) (i ) log10 (10000)  4
2) (i ) 10 6  1000000
(ii ) log3 243  5
(iv ) log4 0.0625  2
3) (i ) 0.30105
(ii ) 2.30105
(ii ) 5.4553 (iii )  0.9048 (iv)  2.0834
Exercise 2
1) (i ) log p  log q  log r
2) (i ) log15
3
1
5
(iv )  log q
(iii ) log125 5  1
(ii ) 4 3  64 (iii ) 4096 4  8 or 4 4096  8
(iv ) 5  4  0.0016 or 14  0.0016
4) (i) 2.36921
x
1
(v ) 1 log r
2
(ii ) 2 log p
(iii ) log p  log r
(vi) 1 log q  1 log r
2
2
(ii ) log13 (iii ) log81 (iv ) log6
(v) log 2
3) (i ) 4  log10 3 (ii ) 6 (iii )  log 72 (iv ) ln2 x 2
Exercise 3
1) 1.6309 2) 0.8614
6) 0.1465 7) -3.9694
3) 1.1610
8)2
4) 2.7712
9) 1.4157
(vi) log6
9 
(vii) log6
(v) log 16
5) –0.1038
10) 0.7700
Exercise 4
1) 3, 1.5
2) 120, -1.5
We would appreciate your comments on this worksheet, especially if
you’ve found any errors, so that we can improve it for future use. Please
contact the Maths Skills Adviser by email at [email protected]
The information in this leaflet can be made available in an alternative format on
request using the email above.
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