CHAPTER 3

INDICES AND LOGARITHMS

Law 1: a^m x aⁿ = a^m+n

Example 5³ x 5⁴ = 5³⁺⁴ = 5⁷

Law 2: a^m ÷ aⁿ = a^m-n

Example 5⁷ ÷ 5³ = 5^7-3 = 5⁴

Law 3: (aⁿ)^m = a^mn

Example (3²)³ = 3² x 3² x 3² = 3²⁺²⁺² = 3⁶

Law 4: a⁰ = 1

Example 7⁰ = 7^3-3 = 7³ ÷ 7³ = 1

Law 5: a^-n =   1      .

                    aⁿ

Example 5^–2 = 5^{2- 4} = 5² ÷ 5⁴ =         5 x 5       =    1     =   1

                                                    5 x 5 x 5 x 5     5 x 5       5²

Law 6: a^1/n = ⁿ√a

Example 27^1/3 = ³√27 = 3

SIMPLE EXPONENTIAL EQUATIONS

Example 1: Solve the equation 3^x = 81

Solution: 3^x = 3⁴

Hence x = 4

Example 2: Solve 8^x = 0.25

Solution: 2^3x = 1/4 = 1/2² = 2^-2

Hence 3x = -2

Therefore x = - 2/3

Example: Solve 2^2x + 4(2^x) – 32 = 0

Solution: (2^x)² + 4(2^x) – 32 = 0

Let 2^x = p, hence

(2^x)² + 4(2^x) – 32 =  p² + 4p – 32 = 0

Solving for p, (p + 8) (p – 4) = 0

Hence p = - 8 or 4

2^x = -8 or 2^x = 4, but 2^x = -8 has no solution

Hence, 2^x = 4 = 2²

x = 2

Example: Solve the equation 3^{2(x – 1)} – 8 (3^{x – 2} ) = 1

Solution: 3^{2(x – 1)} – 8 (3^{x – 2} ) – 1 = 0

3^{2x – 2} – 8 (3^x x 3^–2 ) - 1 = 0

(3^x)² x 3^–2 – 8 (3^x x 3^–2 ) - 1 = 0

Let 3^x = P

 P² – 8( P ) – 1 = 0

 9          9

Multiply through by 9

P² – 8P – 9 = 0

(P – 9) (P + 1) = 0

P = 9 or – 1

P = - 1 has no solution, but P = 9, since 3^x = P

3^x = 3²

x = 2

LOGARITHMS

Law 1: If y = a^x, then x = log_a y

Example: 9 = 3², therefore log₃ 9 = 2     

Law 2: log_a PQ = log_a P  +  log_a Q 

Example: log_a 6 = log_a (3 x 2) =  log_a 3 + log_a 2

Law 3: log_a (P/Q) = log_a P – log_a Q

Example: log_a 5 = log_a (10/2) = log_a 10 – loga 2 

Law 4: log_a Pⁿ =  nlog_a P

Example: log_a 9 = log_a 3² = 2 log_a 3

Law 5: log_a 1 = 0 (The log of 1 to any base is zero)

Example: log₅ 1 = 0,  log₇ 1 = 0

Law 6: log_p P = 1 (The logarithm of the same base always equal to 1)

Example: log₂ 2 = 1, log₆ 6 = 1

Law 7: log_p 1/P  = log_p P^-1 = -1 (The logarithm of any fraction of the same base is equal to the exponent of that fraction)

Example: log₂ 1/8 = log₂ 8^-1 = log₂ 2^-3 = - 3

Law 8: log_b^a b = 1/a

Example: log₉ 3 = log₃² 3 = ½.

Examination Questions

1. Simplify log₄ 9 + log4 21 – log4 7

Solution:

Log₄ 9 + log₄ 21 – log₄ 7 = log₄ (9 x 21 ÷ 7)

Log₄ 27 = log₄ 3³

3 log₄ 3

2. If log₇ 2 = 0.356 and log₇ 3 = 0.566 find the value of 2 log₇ (7/15) +  log₇ (25/12) – 2 log₇ (7/3)

Solution:

= 2 log₇ (7/15) + log₇ (25/12) – 2 log₇ (7/3)

= log₇ (7/15)² + log₇ (25/12) – (7/3)²

= log₇ {(7/15)² x (25/12)} = log₇ { 7  x  7  x 25 x  3  x  3 }

                    (7/3)²                               15    15    12     7      7

= log₇ (1/12) = log₇ 1 – log₇ 12 = - log₇ 12 (since log₇ 1 = 0 Law 5)

= - log₇ (2² x 3) = - 2 log₇ 2 + log₇ 3

= - (2 x 0.356) + 0.566

= -1.278

3. Find x if a^2-x = T

Solution: log_a a^2-x = log_a T

Then 2 – x = log_a T

x = 2 – log_a T

x = 2 log_a a – log_a T

x = log_a a² – log_a T

x = log_a (a²/T)

LOGARITHMIC EQUATIONS

1. Solve the equations (a) log₁₀ (2x² + 5x + 97) = 2

(b) log₁₀ (3x² + 8) = 1 + log₁₀ {(x/2) + 1}

Solutions:

(a) log₁₀ (2x² + 5x + 97) = log₁₀ 100

Taking away log₁₀

Hence 2x² + 5x + 97 = 100

2x² + 5x – 3 = 0

Therefore (2x – 1) (x + 3) = 0, giving x = ½ or 3

(b) log₁₀ (3x² + 8) = log₁₀ 10 + log₁₀ {(x/2) + 1}

log₁₀ (3x² + 8) = log₁₀ {10[(x/2) + 1]}

log₁₀ (3x² + 8) = log₁₀ (5x + 10)

3x² + 8 = 5x + 10 or 3x² + 5x – 2 = 0

(3x + 1) (x – 2) = 0, giving x = - ⅓ or 2

2. If log₉ (a + 3) – log₉ b = c + ½ ……… (1)

log₃ (a – 3) + log₃ b = c – 1 ………….….(2)

Show that a² = 9 + 27^c and find the possible values of a and b when c = 1.

Solution:

From equation (1)

log₉ (a + 3) – log₉ b = c + log₉ 3 (since 9^½ = 3)

log₉ (a + 3) – log₉ b - log₉ 3 = c

log₉ (a + 3) – log₉ (3b) = c

log₉  a + 3 = c

         3b

 a + 3  = 9^c ……………………………...(3)

   3b

From equation (2)

log₃ (a – 3) + log₃ b = c – log₃ 3 (since log₃ 3 = 1)

log₃ (a – 3) + log₃ b + log₃ 3 = c

log₃ 3b(a – 3) = c

3b(a – 3) = 3^c ……………………………..(4)

(3) x (4) = (a + 3)(a – 3) = 9^c  x 3^c

= a² – 9 = 27^c

Hence a² = 9 + 27^c

If c = 1, then a² = 36

a = +6 or -6

But a cannot equal – 6 as then neither log₉ (a + 3) or log₃ (a – 3) would exist.

When a = 6, equation (1) gives 6 + 3 = 9

                                                  3b

9/3b= 9 or b = ⅓.  Hence a = 6, b = ⅓

COMMON LOGARITHMS

Express 3200 in standard form; hence we have 3.2 x 10³. To find the logarithm of 3200 ie log₁₀ 3200 = log₁₀ 3.2 x 10³ = log₁₀ 3.2 + log₁₀ 10³ = 0.505 + 3 log₁₀ 10 = 3.505. 

For any number expressed in standard form such as 3.2 x 10ⁿ, log₁₀ 3.2 x 10ⁿ = n + 0.505. The part 0.505 is called the Mantissa while n is called the Characteristic.

Example: Find log₇ 12

Solution:

x = log₇ 12

7^x = 12

Taking logarithms of both sides

log 7^x = log 12

x log 7 = log 12

x = log 12   = 1.0792   = 1.278

       log 7        0.8451

Using logarithm table to solve 1.0792

                                             0.8451

No.                  Log

1.0792             0.0331

0.8451            -1.9269

1.278               0.1062

QUESTIONS

1. Find the value of:

(a) log₃ 5,  (b) log₂ 7,  (c) log₅ 20  (d) log₇ 35  (e) log₈ 10  (f) log₃ (1/8)  (g) log₁₂ 94

(h) log₁₂ 9.4  (i) log₂ 0.47

2. Solve the equations:

(a) 2^x = 5  (b) 3^x = 5  (c) 1.8^x = 2.7  (d) 5^x = 10  (e) 3^x+2 = 7  (f) 2^x-1 = 5^x  (g) 3^x+2 = 5^x-1

(h) (2.5)^x-3 = (0.4)^x   (i) (0.37)^2x-1 = 9   (j) (0.4)^x+3 = (0.5)2^x   (k) 3^x+2 = 5 x 2^x