Introduction

Recall the laws of exponents:

1. x^m * xⁿ = x^m+n
2. x^m/xⁿ = x^m-n
3. (x^m)ⁿ = x^mn
4. (a/b)^m = a^m/b^m

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

     log_amn = log_am + log_an

     Proof:

     Let log_am = x, log_an = y

            ∴  m = a^x  and   ∴ n = a^y

              mn = a^x * a^y

                    = a^x + y

         log_amn = x + y

                     = log_am+ log_an

Corollary: log_a ( m n p q . . . ) = log_am + log_an + log_ap + . . .

Theorem 2:

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

     log_a(m/n) = log_am – log_an

 Proof:

  Let log_a(a/b) = p       then (a/b) = x^p       (1)

  log_xa = q                 then a  = x^q

  log_xb = r                  then b = x^r

  a/b = x^q / x^r = x^q-r

  Since x^m / xⁿ = x^m-n

  From (1) a/b = x^p       x^p = x^q-r

     ∴ p = q^-r

  i.e., log_x (a/b) = log_xa - log_xb

Theorem 3

Log_x a^m = mlog_xa

     Proof:

     Let Log_xa^m = p     then xp = a^m              ⇒    (1)

          Log_xa  = q      then a = x^q               ⇒    (2)

     From (1) and (2) a^m =  (x^q)m = x^qm     ⇒    (3)

     From (1) and (3) a^m =  xp = x^qm

                           ∴  p = qm

     i.e., Log_xa^m = m Log_xa

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

Answers to 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
   Solution:
   Let log₁₀xyz = m  ⇒ 10m = xyz    (1)
         log₁₀x = a      ⇒ 10^a = x         (2)
         log₁₀y = b      ⇒ 10^b = y         (3)
         log₁₀z = c      ⇒ 10^c = z         (4)
   from (2), (3) and (4) xyz = 10^a * 10^b * 10^c
   ⇒xyz = 10^a+b+c
   from (1) 10^a = xyz
   ∴10^a = 10^a+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 Log_xa^m = mLog_xa
                              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