#8 - LOGARITHMS CHANGE of BASE 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. CHANGE
of BASE
PROOFS
Enzo Exposyto 82

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

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

16. 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

17. SitoGraphy
Enzo Exposyto 93

18. 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