Solutions — Introducing Logarithms and Logarithmic Laws

← Back to Questions›Logarithms

1. Q1 — Exponential and logarithmic form

   ▶ Show Answer

   (a) 4³ = 64 → log₄(64) = 3

   (b) 10⁻³ = 0.001 → log(0.001) = −3

   (c) log₅(125) = 3 → 5³ = 125

   (d) log₂(1/8) = −3 → 2⁻³ = 1/8

2. Q2 — Evaluating logarithms

   ▶ Show Answer

   (a) log₂(16) = log₂(2⁴) = 4

   (b) log₃(27) = log₃(3³) = 3

   (c) log₁₀(10 000) = log(10⁴) = 4

   (d) log₅(1) = 0 (since 5⁰ = 1)

   (e) log₄(1/16) = log₄(4⁻²) = −2

   (f) log₇(7) = 1 (since 7¹ = 7)

3. Q3 — Special values

   ▶ Show Answer

   (a) log₆(6⁵) = 5 (log and exp cancel)

   (b) 3^log₃(11) = 11 (exp and log cancel)

   (c) log₂(2⁻⁴) = −4

   (d) 10^log(7) = 7

4. Q4 — Expanding using log laws

   ▶ Show Answer

   (a) log₂(8x) = log₂(8) + log₂(x) = 3 + log₂(x)

   (b) log(100/x) = log(100) − log(x) = 2 − log(x)

   (c) log₃(x⁴) = 4log₃(x)

   (d) log(5x³y) = log(5) + 3log(x) + log(y) ≈ 0.699 + 3log(x) + log(y)

5. Q5 — Writing as a single log

   ▶ Show Answer

   (a) log(3) + log(4) = log(3 × 4) = log(12)

   (b) log₂(20) − log₂(5) = log₂(20/5) = log₂(4) = 2

   (c) 3log(x) + log(2) = log(x³) + log(2) = log(2x³)

   (d) 2log(x) − log(y) + log(z) = log(x²) − log(y) + log(z) = log(x²z/y)

6. Q6 — Simplifying to a number

   ▶ Show Answer

   (a) log₂(4) + log₂(8) = 2 + 3 = 5 [or: log₂(32) = 5]

   (b) log(1000) − 2log(10) = 3 − 2(1) = 1

   (c) log₃(27²) = 2log₃(27) = 2 × 3 = 6

   (d) 2log₅(25) + log₅(5) = 2(2) + 1 = 5

7. Q7 — Change of base

   ▶ Show Answer

   (a) log₂(10) = log(10)/log(2) = 1/0.30103 ≈ 3.322

   (b) log₃(50) = log(50)/log(3) = 1.699/0.4771 ≈ 3.561

   (c) log₇(200) = log(200)/log(7) = 2.301/0.8451 ≈ 2.723

8. Q8 — Full expansion

   ▶ Show Answer

   (a) log₂(8x³/y²)
   = log₂(8) + log₂(x³) − log₂(y²)
   = 3 + 3log₂(x) − 2log₂(y)

   (b) ln(√(x/(x+1))) = ln(x/(x+1))^1/2 = ½ln(x/(x+1))
   = ½[ln(x) − ln(x+1)]
   = ½ln(x) − ½ln(x+1)

9. Q9 — Solving logarithmic equations

   ▶ Show Answer

   (a) log₂(x) = 5 → x = 2⁵ = 32

   (b) log₃(x + 1) = 4 → x + 1 = 3⁴ = 81 → x = 80

   (c) log(x) + log(x − 3) = 1 → log(x(x−3)) = 1 → x(x−3) = 10
   x² − 3x − 10 = 0 → (x − 5)(x + 2) = 0
   x = 5 or x = −2. Since x > 0 and x − 3 > 0 require x > 3: x = 5

10. Q10 — Prove and apply

    ▶ Show Answer

    (a) 72 = 8 × 9 = 2³ × 3². So log_a(72) = log_a(2³) + log_a(3²) = 3log_a(2) + 2log_a(3) = 3p + 2q

    (b) 36 = 4 × 9 = 2² × 3². log₂(36) = 2log₂(2) + 2log₂(3) = 2 + 2k = 2(1 + k)

    (c) Using change of base: log_a(b) = ln(b)/ln(a) and log_b(a) = ln(a)/ln(b).
    Product: [ln(b)/ln(a)] × [ln(a)/ln(b)] = ln(b) ln(a) / (ln(a) ln(b)) = 1 ✓