MATHS SYMBOLS - EXPONENTIALS + LOGARITHMS and THEIR PROPERTIES - DEFINITIONS - 10+ PROPERTIES - ANTILOGARITHMS - INVERSE OPERATIONS - NATURAL LOGARITHMS - NEPER - NAPIER - EULER'S NUMBER - LOG - LN - POWER - ROOT - PRODUCT - QUOTIENT - CHANGE of BASE - PROOFS, EXAMPLES and CALCULATIONS STEP by STEP

1. ENZO EXPOSYTO
MATHS
SYMBOLS
PROPERTIES of EXPONENTIALS and LOGARITHMS
Enzo Exposyto 1

2. 2X ex 2-x e-x
EXPONENTIALS
LOGARITHMS
log2(x) ln(x) log(x)
Enzo Exposyto 2

3.
Enzo Exposyto 3

4. 1 - Exponential - Definition 6
2 - Exponentials - Their Properties 8
3 - Exponentials and Logarithms 15
4 - Logarithm - Definition and Examples 25
5 - Logarithms - Their Properties 34
6 - log(y) and ln(y) - Properties 46
Enzo Exposyto 4

5. 7 - logb(bx) = x - Proofs 61
8 - ylogb(y) = y - Proofs 64
9 - log of a Power - Proofs 68
10 - log of a Root - Proofs 71
11 - log of a Product - Proofs 74
12 - log of a Quotient - Proofs 77
13 - Change of Base - Proofs 82
14 - zlogb(y) = ylogb(z) - Proof 91
15 - SitoGraphy 93
Enzo Exposyto 5

6. EXPONENTIAL
-
DEFINITION
Enzo Exposyto 6

7. EXPONENTIAL - DEFINITION
DEFINITION Examples
DEFINITION
bx = y ...
2-1 = 1
2
...
20 = 1
...
21 = 2
...
b > 0 and b < > 1 (*)
x element of R
y > 0
"b" is the fixed base
x is the variable exponent
(*) The symbol < > and ≠ have the same meaning
Enzo Exposyto 7

8. EXPONENTIALS
-
THEIR PROPERTIES
Enzo Exposyto 8

9. EXPONENTIALS - THEIR PROPERTIES - 1
[b, c > 0 and b, c < > 1]
[x, y elements of R]
Z+ = {1, 2, 3, …}
Property Exponentials Exponents Exponentials Exponents Result
1st bx · by = bx + y 23 · 22 = 23 + 2 = 32
2nd
bx
=
by
bx -y
23
=
22
23 -2 = 2
3rd (bx)y = bx *y (23)2 = 23 *2 = 64
4th
n Z+ n√bx = bx : n 2√24 = 24 : 2 = 4
5th bx · cx = (b · c)x 22 · 32 = (2 · 3)2 = 36
6th
bx
———- =
cx
(_b_)x
c
43
———- =
23
(_4_)3
=
2
8
Enzo Exposyto 9

10. EXPONENTIALS - THEIR PROPERTIES - 2
[b, c > 0 and b, c < > 1]
[x, y elements of R]
Z+ = {1, 2, 3, …}
Property Exponents Exponentials Exponents Exponentials Result
1st bx + y = bx · by
23 + 2 = 23 · 22 = 32
2nd bx -y =
bx
———-
by
23 -2 =
23
=
22
2
3rd bx *y = (bx)y 23 *2 = (23)2 = 64
4th n Z+ bx : n = n√bx
24 : 2 = 2√24 = 4
5th (b · c)x = bx · cx (2 · 3)2 = 22 · 32 = 36
6th
(_b_)x =
c
bx
———-
cx
(_4_)3
=
2
43
———- =
23
8
Enzo Exposyto 10

11. Exponentials - 6 Properties - Proofs/Examples
PROOFS / EXAMPLES
[b, c > 0 and b, c < > 1]
1st b3 · b2 = (b · b · b) · (b · b) = b · b · b · b · b = b5 = b3 + 2
2nd b3 = (b · b · b) = b = b1 = b3 - 2
b2 (b · b)
3rd (b3)2 = (b · b · b) · (b · b · b) = b · b · b · b · b · b = b6 = b3 * 2
4th 2√b4 = 2√(b · b · b · b) = b · b = b2 = b4 : 2
5th b3 · c3 = (b · b · b) · (c · c · c) = b · c · b · c · b · c = … = (b·c)3
6th
b3 (b · b · b) b b b b
——- = ————— = —— . —— . —— = … = (—)3
c3 (c · c · c) c c c c
Enzo Exposyto 11

12. Exponentials - 2nd property - b superscript 0
EXPONENTIALS - THEIR PROPERTIES - 3
Reference Property Notice Proof Example
2nd
Property
b0 = 1
b > 0
b ≠ 1
x R
b0 = bx - x 20 = 23 - 3
= bx
bx
= 23
23
= 1 = 1
1 = b0 b > 0
b ≠ 1
1 = 20
Enzo Exposyto 12

13. Exponentials - 2nd property - b superscript (-x)
EXPONENTIALS - THEIR PROPERTIES - 4
Reference Property Notice Proof Example
2nd
Property
b-x = 1
bx
b > 0
b ≠ 1
x R
b-x = b0-x 2-3 = 20-3
= b0
bx
= 20
23
= 1
bx
= 1
23
1 = b-x
bx
b > 0
b ≠ 1
x R
1 = 2-3
23
Enzo Exposyto 13

14. Exponentials - 2nd property - Reciprocal of bx
EXPONENTIALS and THEIR PROPERTIES - 5
Reference Property Notice Proof Example
2nd
Property
bx = 1
b-x
b > 0
b ≠ 1
x R
23 = 1
2-3
1 = bx
b-x
b > 0
b ≠ 1
x R
1 = 23
2-3
Enzo Exposyto 14

15. EXPONENTIALS
and
LOGARITHMS
Enzo Exposyto 15

16. EXPONENTIAL BASE 2:
23 = 2 × 2 x 2 = 8
The LOGARITHM BASE 2 OF 8
goes the OTHER WAY:
Enzo Exposyto 16

17. The logarithm base 2 of 8 is 3,
BECAUSE
2 cubed is 8;
so the logarithm base 2 of 8 is 3:
Enzo Exposyto 17

18. THE LOGARITHM BASE 2 OF 8
IS THE OPERATION THAT ALLOWS US
OF GOING BACK TO THE EXPONENT 3
EXPONENTIAL 2x and LOGARITHM BASE 2
EXPONENT x 2x
1 2
2 4
3 8
4 16
Enzo Exposyto 18

19. In other words …
THE LOGARITHM BASE 2 OF 8
IS
THE EXPONENT 3
… MORE PRECISELY …
THE LOGARITHM "3"
IS THE EXPONENT
WHICH WE HAVE TO PUT ON
THE BASE “2”
TO GET “8”
Enzo Exposyto 19

20. Now, since
log2(8) = 3 and 8 = 23
then
log2(23) = 3
We can see that
THE LOGARITHM "3"
IS THE EXPONENT
WHICH WE HAVE TO PUT ON
THE BASE “2”
TO GET “23”
Enzo Exposyto 20

21. EXPONENTIAL and LOGARITHM
with the same base
CANCEL EACH OTHER.
This is true because
exponential and logarithm
with the same base
are INVERSE OPERATIONS
It is just like
Addition and Subtraction,
Multiplication and Division,
Exponentiation and Root, …
when they're
INVERSE OPERATIONS
Enzo Exposyto 21

22. Now, we can introduce
the ANTILOGARITHM BASE b:
antilogb(x) = bx
It's, simply, an EXPONENTIAL
and represents the antilogarithm
when we operate
with a logarithm …
It’s such that
logb(antilogb(x)) = x
The meaning is
logb(bx) = x
Enzo Exposyto 22

23. And, of course,
antilogb(logb(y)) = y
The meaning is
blogb(y) = y
Enzo Exposyto 23

24. These phrases - with 8 - are equivalent
log2(8) = 3 < = > 23 = 8
OR
23 = 8 < = > log2(8) = 3
These phrases - with 23 - are equivalent
log2(23) = 3 < = > 23 = 23
OR
23 = 23 < = > log2(23) = 3
Enzo Exposyto 24

25. LOGARITHM
-
DEFINITION
and
EXAMPLES
Enzo Exposyto 25

26. LOGARITHM - DEFINITION
DEFINITION Example
DEFINITION
logb(y) = x
iff (if and only if)
bx = y
log2(8) = 3
because
23 = 8
b > 0 and b < > 1
y > 0
x element of R
2 > 0 and 2 < > 1
8 > 0
3 element of R
The logarithm "x" is the exponent
which we have to put on
the base “b”
to get “y”
The logarithm
"3"
is the exponent
which we have
to put on
the base “2”
to get “8”
Enzo Exposyto 26

27. iff
bx = y Enzo Exposyto 27

28. LOGARITHMS - EXAMPLES - 1
log DEFINITION because
log3(9) = 2
The logarithm "2" is the exponent
which we have to put on the base “3”
to get “9”
log3(9) = 2
because
32 = 9
log3(27) = 3
The logarithm "3" is the exponent
which we have to put on the base “3”
to get “27”
log3(27) = 3
because
33 = 27
log4(4) = 1
The logarithm "1" is the exponent
which we have to put on the base “4”
to get “(4)”
log4(4) = 1
because
41 = (4)
Enzo Exposyto 28

29. LOGARITHMS - EXAMPLES - 2
log DEFINITION because
log4(16) = 2
The logarithm "2" is the exponent
which we have to put on the base “4”
to get “16”
log4(16) = 2
because
42 = 16
log5(25) = 2
The logarithm "2" is the exponent
which we have to put on the base “5”
to get “25”
log5(25) = 2
because
52 = 25
log10(100) = 2
The logarithm "2" is the exponent
which we have to put on the base “10”
to get “100”
log10(100) = 2
because
102 = 100
Enzo Exposyto 29

30. LOGARITHMS - EXAMPLES - 3
log DEFINITION because
log4(1) = -1
4
The logarithm "-1" is the exponent
which we have to put on the base “4”
to get “1”
4
log4(1) = -1
4
because
4-1 = 1
4
log10( 1 ) = -1
10
The logarithm "-1" is the exponent
which we have to put on the base “10”
to get “ 1 ”
10
log10( 1 ) = -1
10
because
10-1 = 1
10
Enzo Exposyto 30

31. LOGARITHMS - EXAMPLES - 4
log DEFINITION because
log1/2(2) = -1
The logarithm "-1" is the exponent
which we have to put on the base “1”
2
to get “2”
log1/2(2) = -1
because
(1)-1 = 2
2
log1/4(4) = -1
The logarithm "-1" is the exponent
which we have to put on the base “1”
4
to get “4”
log1/4(4) = -1
because
(1)-1 = 4
4
Enzo Exposyto 31

32. LOGARITHMS - EXAMPLES - 5
log DEFINITION because
log1/2(4) = -2
The logarithm "-2" is the exponent
which we have to put on the base “1”
2
to get “4”
log1/2(4) = -2
because
(1)-2 = 4
2
log1/3(9) = -2
The logarithm "-2" is the exponent
which we have to put on the base “1”
3
to get “9”
log1/3(9) = -2
because
(1)-2 = 9
3
Enzo Exposyto 32

33. LOGARITHMS - EXAMPLES - 6
e = 2.718281828…
log DEFINITION because
loge(e) = 1
The logarithm "1" is the exponent
which we have to put on the base “e”
to get “(e)”
loge(e) = 1
because
e1 = (e)
loge(1) = 0
The logarithm "0" is the exponent
which we have to put on the base “e”
to get “1”
loge(1) = 0
because
e0 = 1
loge(e2) = 2
The logarithm "2" is the exponent
which we have to put on the base “e”
to get “(e2)”
loge(e2) = 2
because
e2 = (e2)
Enzo Exposyto 33

34. LOGARITHMS
-
THEIR PROPERTIES
Enzo Exposyto 34

35. LOGARITHMS - THEIR PROPERTIES - 1
Name Property Example
base = 0
log0(y) is undefined log0(2) is undefined
y > 0
For any x R-{0}, 0x ≠ 2
(really 0x = 0) and, then,
log0(2) Does Not Exist
base = 1
log1(y) is undefined log1(2) is undefined
y > 0
For any x R, 1x ≠ 2
(really 1x = 1) and, then,
log1(2) Does Not Exist
logb(0)
logb(0) is undefined log2(0) is undefined
b > 0 and b < > 1
For any x R, 2x ≠ 0
(really 2x > 0) and, then,
log2(0) Does Not Exist
Enzo Exposyto 35

36. LOGARITHMS - THEIR PROPERTIES - 2
Name Property Examples
logb(1)
logb(1) = 0 log2(1) = 0
because
20 = 1b > 0 and b < > 1
logb(b)
logb(b) = 1 log4(4) = 1
because
41 = (4)b > 0 and b < > 1
Enzo Exposyto 36

37. LOGARITHMS - THEIR PROPERTIES - 3
Name Property Example
logb(bx)
logb(bx) = x log2(23) = 3
because
23 = (23)
b > 0 and b < > 1
x element of R
blogb(y)
blogb(y) = y 2log2(8) = 8
because
23 = 8
b > 0 and b < > 1
y > 0
Enzo Exposyto 37

38. LOGARITHMS - THEIR PROPERTIES - 4
Name Property Examples
log of
a Power
logb(yz) = z logb(y) log2(42) = 2 log2(4)
b > 0 and b < > 1
y > 0
z element of R
b = 2
y = 4
z = 2
log of
a Reciprocal
logb(1) = logb(y-1) = - logb(y)
y
log2(1) = log2(4-1) = - log2(4)
4
b > 0 and b < > 1
y > 0
b = 2
y = 4
Enzo Exposyto 38

39. LOGARITHMS - THEIR PROPERTIES - 5
Name Property Examples
log of
a Root - 1
logb(n√y) = logb(y)
n
log2(3√8) = log2(8)
3
b > 0 and b < > 1
y > 0
n Z+
Z+ = {1, 2, 3, …}
b = 2
y = 8
n = 3
log of
a Root - 2
logb(n√yz) = z logb(y)
n
log2(3√26) = 6 log2(2)
3
b > 0 and b < > 1
y > 0
z element of R
n Z+
Z+ = {1, 2, 3, …}
b = 2
y = 2
z = 6
n = 3
Enzo Exposyto 39

40. LOGARITHMS - THEIR PROPERTIES - 6
Name Property Examples
log of a
Product - 1
logb(y z) = logb(y) + logb(z) log2(4 2) = log2(4) + log2(2)
b > 0 and b < > 1
y, z > 0
b = 2
y = 4; z = 2
log of a
Product - 2
logb(yn . zp) = n.logb(y)+p.logb(z) log2(43 . 24) = 3.log2(4)+4.log2(2)
b > 0 and b < > 1
y, z > 0
n, p elements of R
b = 2
y = 4; z = 2
n = 3; p = 4
Enzo Exposyto 40

41. LOGARITHMS - THEIR PROPERTIES - 7
Name Property Examples
log of a
Quotient - 1
logb(y) = logb(y.z-1) = logb(y)-logb(z)
z
log2(8) = log2(8.4-1) = log2(8) - log2(4)
4
b > 0 and b < > 1
y, z > 0
b = 2
y = 8; z = 4
log of a
Quotient - 2
logb(y) = logb(y) - logb(z)
z
log2(4) = log2(4) - log2(2)
2
b > 0 and b < > 1
y, z > 0
b = 2
y = 4; z = 2
log of a
Quotient - 3
logb(yn) = n.logb(y)-p.logb(z)
zp
log2(43) = 3.log2(4)-4.log2(2)
24
b > 0 and b < > 1
y, z > 0
n, p elements of R
b = 2
y = 4; z = 2
n = 3; p = 4
Enzo Exposyto 41

42. LOGARITHMS - THEIR PROPERTIES - 8
Name Property Examples
Base Change - 1
logb(y) = logc(y)
logc(b) log2(16) = log4(16) = 2 = 4
log4(2) 1
2b, c > 0 and b, c < > 1
y > 0
Base Switch - 1
logb(c) = logc(c) = 1
logc(b) logc(b)
log2(4) = log4(4) = 1
log4(2) log4(2)
logc(c) = 1 log4(4) = 1
b, c > 0 and b, c < > 1 b = 2; c= 4
Base Switch - 2 logb(c) * logc(b) = 1 log2(4) * log4(2) = 1
Enzo Exposyto 42

43. LOGARITHMS - THEIR PROPERTIES - 9
Name Property Examples
Base Change - 2
logbn(y) = logb(y)
n
log2-3(8) = log2(8)
-3
b > 0 and b < > 1
n element of R, n < > 0
y > 0
b = 2
n = -3
y = 8
Base Change - 3a
n . logbn(y) = logb(y) -3 . log2-3(8) = log2(8)
n element of R, n < > 0
b > 0 and b < > 1
y > 0
n = -3
b = 2
y = 8
Base Change - 3b
logb(y) = n . logbn(y) log2(4) = 2 . log22(4)
b > 0 and b < > 1
y > 0
n element of R, n < > 0
b = 2
y = 4
n = 2
Enzo Exposyto 43

44. LOGARITHMS - THEIR PROPERTIES - 10
Name Property Examples
Base Change - 4
log1/b(y) = - logb(y) log1/8(8) = - log8(8)
b > 0 and b < > 1
y > 0
b = 8
y = 8
Base Change - 5
logb(1) = log1/b(y)
y
log2(1) = log1/2(4)
4
b > 0 and b < > 1
y > 0
b = 2
y = 4
Enzo Exposyto 44

45. LOGARITHMS - THEIR PROPERTIES - 11
Name Property Examples
zlogb(y)
zlogb(y) = ylogb(z) 2log2(8) = 8log2(2)
z > 0
b > 0 and b < > 1
y > 0
z = 2
b = 2
y = 8
Enzo Exposyto 45

46. log(y) AND ln(y)
-
THEIR
PROPERTIES
Enzo Exposyto 46

47. log(y) AND ln(y) - THEIR PROPERTIES
REMARKS:
• log(y) always refers to log base 10,
i. e.,
log(y) = log10(y)
Therefore,
log(y) = x
if and only if
10x = y
Enzo Exposyto 47

48. • ln(y) is called the natural logarithm
and is used to represent loge(y),
where the irrational number e 2.718281828:
ln(y) = loge(y)
Therefore,
ln(y) = x
if and only if
ex = y
Enzo Exposyto 48

49. • Most calculators can directly compute
logs base 10
and/or
the natural log.
For any other base
it is necessary to use
the change of the base formula:
logb(y) = log10(y) = log(y) log2(8) = log(8)
log10(b) log(b). log(2)
or
logb(y) = ln(y) log2(8) = ln(8)
ln(b) ln(2)
Enzo Exposyto 49

50. log(y) AND ln(y) - THEIR PROPERTIES - 1
Property log ln
base = 0
base = 10 base = e
base = 1
base = 10 base = e
logb(0)
log(0) is undefined ln(0) is undefined
For any x R, 10x ≠ 0
(really 10x > 0) and, then,
log(0) Does Not Exist
For any x R, ex ≠ 0
(really ex > 0) and, then,
ln(0) Does Not Exist
Enzo Exposyto 50

51. log(x) AND ln(x) - THEIR PROPERTIES - 2
Property log ln
logb(1)
log(1) = 0 ln(1) = 0
because
100 = 1
because
e0 = 1
logb(b)
log(10) = 1 ln(e) = 1
because
101 = 10
because
e1 = e
Enzo Exposyto 51

52. log(y) AND ln(y) - THEIR PROPERTIES - 3
Property log ln
logb(bx)
log(10x) = x ln(ex) = x
x element of R x R
blogb(y)
10log(y) = y eln(y) = y
y > 0 y > 0
Enzo Exposyto 52

53. log(y) AND ln(y) - THEIR PROPERTIES - 4
Property log ln
log of a
Power
log(yz) = z log(y) ln(yz) = z ln(y)
y > 0
z element of R
y > 0
z element of R
log of
a Reciprocal
log(1) = log(y-1) = - log(y)
y
ln(1) = ln(y-1) = - ln(y)
y
y > 0 y > 0
Enzo Exposyto 53

54. log(y) AND ln(y) - THEIR PROPERTIES - 5
Property log ln
log of
a Root - 1
log(n√y) = log(y)
n
ln(n√y) = ln(y)
n
n Z+
Z+ = {1, 2, 3, …}
y > 0
n Z+
Z+ = {1, 2, 3, …}
y > 0
log of
a Root - 2
log(n√yz) = z log(y)
n
ln(n√yz) = z ln(y)
n
n element of Z+
Z+ = {1, 2, 3, …}
y > 0
z element of R
n element of Z+
Z+ = {1, 2, 3, …}
y > 0
z element of R
Enzo Exposyto 54

55. log(y) AND ln(y) - THEIR PROPERTIES - 6
Property log ln
log of a
Product - 1
log(y z) = log(y) + log(z) ln(y z) = ln(y) + ln(z)
y, z > 0 y, z > 0
log of a
Product - 2
log(yn . zp) = n.log(y)+p.log(z) ln(yn . zp) = n.ln(y)+p.ln(z)
y, z > 0
n, p elements of R
y, z > 0
n, p elements of R
Enzo Exposyto 55

56. log(y) AND ln(y) - THEIR PROPERTIES - 7
Property log ln
log of a
Quotient - 1
log(y) = log(y.z-1) = log(y)-log(z)
z
ln(y) = ln(y.z-1) = ln(y)-ln(z)
z
y, z > 0 y, z > 0
log of a
Quotient - 2
log(y) = log(y) - log(z)
z
ln(y) = ln(y) - ln(z)
z
y, z > 0 y, z > 0
log of a
Quotient - 3
log(yn) = n.log(y)-p.log(z)
zp
ln(yn) = n.ln(y)-p.ln(z)
zp
y, z > 0
n, p elements of R
y, z > 0
n, p elements of R
Enzo Exposyto 56

57. log(y) AND ln(y) - THEIR PROPERTIES - 8
Property log ln
Base Change - 1
log(y) = logc(y)
logc(10)
ln(y) = logc(y)
logc(e)
y > 0
c > 0 and c < > 1
y > 0
c > 0 and c < > 1
Base Switch - 1
log(c) = logc(c) = 1
logc(10) logc(10)
ln(c) = logc(c) = 1
logc(e) logc(e)
logc(c) = 1 logc(c) = 1
c > 0 and c < > 1 c > 0 and c < > 1
Base Switch - 2
log(c) * logc(10) = 1 ln(c) * logc(e) = 1
c > 0 and c < > 1 c > 0 and c < > 1
Enzo Exposyto 57

58. log10(y) AND loge(y) - THEIR PROPERTIES - 9
Property log10 loge
Base Change - 2
log10n(y) = log10(y)
n
logen(y) = loge(y)
n
n element of R, n < > 0
y > 0
n element of R, n < > 0
y > 0
Base Change - 3a
n . log10n(y) = log10(y) n . logen(y) = loge(y)
n element of R, n < > 0
y > 0
n element of R, n < > 0
y > 0
Base Change - 3b
log10(y) = n . log10n(y) loge(y) = n . logen(y)
y > 0
n element of R, n < > 0
y > 0
n element of R, n < > 0
Enzo Exposyto 58

59. log10(y) AND loge(y) - THEIR PROPERTIES - 10
Property log10 loge
Base Change - 4
log1/10(y) = - log10(y) log1/e(y) = - loge(y)
y > 0 y > 0
Base Change - 5
log10(1) = log1/10(y)
y
loge(1) = log1/e(y)
y
y > 0 y > 0
Enzo Exposyto 59

60. log10(y) AND loge(y) - THEIR PROPERTIES - 11
Property log10 loge
zlogb(y)
zlog10(y) = ylog10(z) zloge(y) = yloge(z)
z > 0
y > 0
z > 0
y > 0
Enzo Exposyto 60

61. logb(bx) = x
PROOFS
Enzo Exposyto 61

62. logb(bx) = x
a) The log of bx is the exponent which we have to put on the
base b to get bx itself and, therefore, it’s “x”
b) If,
from x, we ‘go’ to bx
and, then,
from bx, we ‘go’ to logb(bx),
since
logb(bx) is the inverse operation of bx,
we go back to x …
therefore,
logb(bx) = x
In other words
(remembering that bx = antilogb(x)):
logb(antilogb(x)) = x
Enzo Exposyto 62

63. logb(bx) = x
c) Let's set
bx = y
and, then, we ‘do’ the logarithms in base b of both sides;
we get
logb(bx) = logb(y)
Since
bx = y <=> logb(y) = x
we can write
logb(bx) = logb(y)
= x
Therefore,
logb(bx) = x
Q.E.D.
Enzo Exposyto 63

64. blogb(y) = y
PROOFS
Enzo Exposyto 64

65. blogb(y) = y
a) If
from y, we ‘go’ to logb(y)
and, then,
from logb(y), we ‘go’ to blogb(y),
since
blogb(y) is the inverse operation of logb(y),
we go back to y …
therefore,
blogb(y) = y
In other words
(remembering that blogb(y) = antilogb(logb(y))):
antilogb(logb(y)) = y
Enzo Exposyto 65

66. blogb(y) = y
b) Let's set
bx = y
and, then, we ‘do’ the log in base b of both sides; we get
logb(bx) = logb(y)
Remembering that (pages 62-63)
logb(bx) = x
we can write
logb(bx) = logb(y)
x = logb(y)
or
logb(y) = x
Now, we ‘do’ the exponentials in base b of both sides and we get
blogb(y) = bx
Since
bx = y
we get
blogb(y) = y
Q.E.D.
Enzo Exposyto 66

67. blogb(y) = y
c) Let's set
logb(y) = x
and, then,
on the left hand side of the equation,
we get
blogb(y) = bx
Since
logb(y) = x <=> bx = y
it's
blogb(y) = bx = y
and, therefore,
blogb(y) = y
Q.E.D.
Enzo Exposyto 67

68. log of
a POWER
PROOFS
Enzo Exposyto 68

69. logb(yz) = z . logb(y)
1) Let's set
logb(y) = l
and, then, the right hand side of the equation becomes
z . logb(y) = z . l
2) Since
logb(y) = l <=> bl = y
or y = bl
the left hand side of the equation becomes
logb(yz) = logb((bl)z)
= logb(blz)
Remembering that (pages 62-63)
logb(bx) = x
it's
logb(blz) = lz = z . l
and we can write
logb(yz) = logb(blz) = z . l
3) Since the left hand side and the right hand side are equal to z . l, they are equal:
logb(yz) = z . logb(y)
Q.E.D.
Enzo Exposyto 69

70. logb(1) = - logb(y)
y
OR
- logb(y) = logb(1)
y
Remembering that
logb(yz) = z . logb(y) [log of a Power, previous page]
we get
logb(1) = logb(y-1)
y
= (-1) . logb(y)
= - logb(y)
Q.E.D.
Enzo Exposyto 70

71. log of
a ROOT
PROOFS
Enzo Exposyto 71

72. logb(n√y) = logb(y)
n
Remembering that
logb(yz) = z . logb(y) [log of a Power, page 69]
we get
logb(n√y) = logb(y1/n)
= 1 logb(y)
n
= logb(y)
n
Q.E.D.
Enzo Exposyto 72

73. logb(n√yz) = z . logb(y)
n
Remembering that
logb(yz) = z . logb(y) [log of a Power, page 69]
we get
logb(n√yz) = logb(yz/n)
= z logb(y)
n
= z . logb(y)
n
Q.E.D.
Enzo Exposyto 73

74. log of
a PRODUCT
PROOFS
Enzo Exposyto 74

75. logb(y . z) = logb(y) + logb(z)
1) Let's set
logb(y) = l
logb(z) = m
and, then, the right hand side of the equation becomes
logb(y) + logb(z) = l + m
2) Since
logb(y) = l <=> bl = y or y = bl
logb(z) = m <=> bm = z or z = bm
the left hand side of the equation becomes
logb(y . z) = logb(bl . bm) = logb(bl+m)
Remembering that (pages 62-63)
logb(bx) = x
it's
logb(bl+m) = l + m
and we can write
logb(y . z) = logb(bl+m) = l + m
3) Since the left hand side and the right hand side are equal to l + m, they are equal:
logb(y . z) = logb(y) + logb(z)
Q.E.D.
Enzo Exposyto 75

76. logb(yn . zp) = n . logb(y) + p . logb(z)
Remembering that
logb(y . z) = logb(y) + logb(z) [previous page]
and
logb(yz) = z . logb(y) [log of a Power, page 69]
we get
logb(yn . zp) = logb(yn) + logb(zp)
= n . logb(y) + p . logb(z)
Q.E.D.
Enzo Exposyto 76

77. log of
a QUOTIENT
PROOFS
Enzo Exposyto 77

78. logb(y) = logb(y) - logb(z)
z
Remembering that
logb(y . z) = logb(y) + logb(z) [page 75]
and
logb(yz) = z . logb(y) [log of a Power, page 69]
we get
logb(y) = logb(y . 1) = logb(y . z-1)
z z
= logb(y) + logb(z-1)
= logb(y) + (-1) . logb(z)
= logb(y) - logb(z)
Q.E.D.
Enzo Exposyto 78

79. logb(y) = logb(y) - logb(z)
z
1) Let's set
logb(y) = l
logb(z) = m
and, then, the right hand side of the equation becomes
logb(y) - logb(z) = l - m
2) Since
logb(y) = l <=> bl = y or y = bl
logb(z) = m <=> bm = z or z = bm
the left hand side of the equation becomes
logb(y) = logb(bl ) = logb(bl-m)
z bm
Remembering that (pages 62-63)
logb(bx) = x
it's
logb(bl-m) = l - m
and we can write
logb(y) = logb(bl-m) = l - m
z
Enzo Exposyto 79

80. logb(y) = logb(y) - logb(z)
z
3) Since the left hand side and the right hand side are equal to l - m, they are equal:
logb(y) = logb(y) - logb(z)
z
Q.E.D.
Enzo Exposyto 80

81. logb(yn) = n . logb(y) - p . logb(z)
zp
Remembering that
logb(y) = logb(y) - logb(z) [previous page]
z
and
logb(yz) = z . logb(y) [log of a Power, page 69]
we get
logb(yn) = logb(yn) - logb(zp)
zp
= n . logb(y) - p . logb(z)
Q.E.D.
Enzo Exposyto 81

82. CHANGE
of BASE
PROOFS
Enzo Exposyto 82

83. logb(y) = logc(y)
logc(b)
1) On the left hand side of the equation,
let's set
logb(y) = x
2) On the right hand side of the equation,
since
logb(y) = x <=> bx = y or y = bx
we substitute y by bx
logc(y) = logc(bx)
= x . logc(b) [log of a Power, page 69]
and we get
logc(y) = x . logc(b) = x
logc(b) logc(b)
3) Since the left hand side and the right hand side are equal to x, they are equal:
logb(y) = logc(y)
logc(b)
Q.E.D.
Enzo Exposyto 83

84. logb(c) = 1
logc(b)
It's
logb(c) = logc(c)
logc(b)
Since
logc(c) = 1
we get
logb(c) = logc(c)
logc(b)
= 1
logc(b)
and, then,
logb(c) = 1
logc(b)
Q.E.D.
Enzo Exposyto 84

85. logb(c) . logc(b) = 1
From
logb(c) = 1
logc(b)
multiplying both sides by logc(b), it’s
logb(c) logc(b) = 1 logc(b)
logc(b)
and, then,
logb(c) logc(b) = 1
Q.E.D.
Enzo Exposyto 85

86. logbn(y) = logb(y)
n
Let's change the base bn by b:
logbn(y) = logb(y)
logb(bn) [log of a Power, page 69]
= logb(y)
n logb(b) [remember: logb(b) = 1]
= logb(y)
n
Therefore:
logbn(y) = logb(y)
n
Q.E.D.
Enzo Exposyto 86

87. n . logbn(y) = logb(y)
OR
logb(y) = n . logbn(y)
From
logbn(y) = logb(y) [previous page]
n
multiplying both sides by n, it’s
n . logbn(y) = logb(y) . n
n
and, then,
n . logbn(y) = logb(y)
Q.E.D.
Enzo Exposyto 87

88. log1/b(y) = - logb(y)
a) From
logb(y) = n . logbn(y) [previous page]
if n = -1 we get
logb(y) = (-1) . logb-1(y) [remember: b-1 = 1]
b
logb(y) = - log1/b(y) [multiplying both sides by (-1)]
- logb(y) = log1/b(y)
and, then,
log1/b(y) = - logb(y)
Q.E.D.
Enzo Exposyto 88

89. log1/b(y) = - logb(y)
b) Let's change the base 1 by b:
b
log1/b(y) = logb(y)
logb(1) [remember: 1 = b-1]
b b
= logb(y)
logb(b-1) [log of a Power, page 69]
= logb(y)
(-1) logb(b) [remember: logb(b) = 1]
= logb(y)
(-1)
= - logb(y)
Q.E.D.
Enzo Exposyto 89

90. logb(1) = log1/b(y)
y
OR
log1/b(y) = logb(1)
y
1) From page 70:
logb(1) = - logb(y)
y
2) From previous page:
log1/b(y) = - logb(y)
3) Since the first and the second left hand side are equal to - logb(y), they are equal:
logb(1) = log1/b(y)
y
Q.E.D.
Enzo Exposyto 90

91. zlogb(y) = ylogb(z)
PROOF
Enzo Exposyto 91

92. zlogb(y) = ylogb(z)
1) On the left hand side of the equation,
let's set
logb(y) = x
and we get
zlogb(y) = zx
2) On the right hand side of the equation,
since
logb(y) = x <=> bx = y or y = bx
we substitute y by bx and we get
ylogb(z) = (bx)logb(z)
= bxlogb(z)
= blogb(zx) [log of a Power, page 69]
= zx [blogb(zx) = zx See blogb(y) = y (pages 65-67)]
3) Since the left hand side and the right hand side are equal to zx, they are equal:
zlogb(y) = ylogb(z)
Q.E.D.
Enzo Exposyto 92

93. SitoGraphy
Enzo Exposyto 93

94. https://en.m.wikipedia.org/wiki/List_of_logarithmic_identities
https://www.andrews.edu/~calkins/math/webtexts/numb17.htm
http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/eandl/logprop/logprop.html#sec2
Enzo Exposyto 94