## Logarithms Simple Law

#### Introduction

Recall the laws of exponents:

1. xm * xn = xm+n
2. xm/xn = xm-n
3. (xm)n = xmn
4. (a/b)m = am/bm

Let’s see how these laws can be used to state and prove the laws of logarithm.

#### Theorems on logarithms

Theorem 1:

If a ≠1 and m, n are positive real numbers then

logamn = logam + logan

Proof:

Let logam = x, logan = y

∴  m = ax  and   ∴ n = ay

mn = ax * ay

= ax + y

logamn = x + y

= logam+ logan

Corollary: loga ( m n p q . . . ) = logam + logan + logap + . . .

Theorem 2:

If a ≠1 m, n are positive real numbers then

loga(m/n) = logam – logan

Proof:

Let loga(a/b) = p       then (a/b) = xp       (1)

logxa = q                 then a  = xq

logxb = r                  then b = xr

a/b = xq / xr = xq-r

Since xm / xn = xm-n

From (1) a/b = xp       xp = xq-r

∴ p = q-r

i.e., logx (a/b) = logxa - logxb

Theorem 3

Logx am = mlogxa

Proof:

Let Logxam = p     then xp = am              ⇒    (1)

Logxa  = q      then a = xq               ⇒    (2)

From (1) and (2) am =  (xq)m = xqm     ⇒    (3)

From (1) and (3) am =  xp = xqm

∴  p = qm

i.e., Logxam = m Logxa

Examples

1. Prove that log 24 = log 6 + log 4
Solution:
From theorem 1, log ab = log a + log b
Right-hand side = log 6 + log 4
= log 6 * 4
= log 24 = left-hand side

2. Show that 3log 3 – 2log 5 = log 27/25
Solution:
3log 3 – 2log 5
= log 33 – log 52      (from the theorem 3)
= log 27 – log 25      (from the theorem 2)
= log 27/25

#### Try these questions:

In each of the following problems unless stated otherwise, assume the base a = 10

1. Prove that log xyz = log x + log y + log z
2. Prove that log 15 = log 5 + log 3
3. Show that log 10800 = 4log 2 + 3log 3 + 2log 5
4. Show that log 108/605 = 2log 2 + 3log 3 – log 5 – 2log 11
5. Show that 4log 3 + 2log 5 = log 2025

In each of the following problems unless stated otherwise, assume the base a = 10

1. Prove that log xyz = log x + log y + log z
Solution:
Let log10xyz = m  ⇒ 10m = xyz    (1)
log10x = a      ⇒ 10a = x         (2)
log10y = b      ⇒ 10b = y         (3)
log10z = c      ⇒ 10c = z         (4)
from (2), (3) and (4) xyz = 10a * 10b * 10c
⇒xyz = 10a+b+c
from (1) 10a = xyz
∴10a = 10a+b+c  ⇒ m = a + b + c
log10 xyz = log x + log y + log z

2. Prove that log 15 = log 5 + log 3
Solution:
From the theorem 1 log ab = log a + log b
Right-hand side = log 5 + log 3
= Log 5 *3
= log 15 = left-hand side

3. Show that log 10800 = 4log 2 + 3log 3 + 2log 5
Solution:
From the theorem 3 Logxam = mLogxa
4log 2 + 3log 3 + 2log 5
= log 24 + log 33 + log 52
= log 16 + log 27 + log 25    from the theorem 1
= log (16 * 27 * 25)
= log 10800 = left-hand side

4. Show that log 108/605 = 2log 2 + 3log 3 – log 5 – 2log 11
Solution:
2log 2 + 3log 3 – log 5 –2log 11
= log22 + log33 – log5 – log112
= log 4 + log 27 – log 5 – log 121
= (log 4 + log 27) – (log 5 + log 121)     from the theorem 1
= log (4 * 27) –- log(5 *121)
= log 108 – log 605                                  from the theorem 2
= log 108/605 = left-hand side

5. Show that 4log 3 + 2log 5 = log 2025
Solution:
4log 3 + 2log5
= log 34 + log 52
= log 81 + log 25
= log 81 * 25
= log 2025