#7 - LOGARITHMS LOG of a PRODUCT LOG of a QUOTIENT PROOFS ANTILOGARITHMS INVERSE OPERATIONS NATURAL LOGARITHMS NEPER - NAPIER - EULER'S NUMBER LOG - LN POWER - ROOT PRODUCT - QUOTIENT CHANGE of BASE PROOFS - EXAMPLES CALCULATIONS STEP by STEP MATHS SYMBOLS www.enzoexposito.it/mobile/matematica.html www.enzoexposito.it/mobile/expon_log.html

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. log of
a PRODUCT
PROOFS
Enzo Exposyto 74

7. 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

8. 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

9. log of
a QUOTIENT
PROOFS
Enzo Exposyto 77

10. 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

11. 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

12. 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

13. 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