T3T3 Review Solutions — Indices and Logarithms

1. Index laws. Fluency

   • (a) (2x³)⁴ = 2⁴x¹² = 16x¹².
   • (b) a⁵ × a⁻³ = a².
   • (c) 81^3/4 = (∜81)³ = 3³ = 27.
   • (d) (x²y³)² ÷ (x³y) = x⁴y⁶ ÷ x³y = xy⁵.

2. Scientific notation. Fluency

   • (a) 4.5 × 10⁻⁷.
   • (b) 306 000.
   • (c) (5 × 3) × 10⁶⁻² = 15 × 10⁴ = 1.5 × 10⁵.
   • (d) (8.4 ÷ 2.1) × 10⁷⁻³ = 4 × 10⁴ = 4.0 × 10⁴.

3. Logarithms. Fluency

   • (a) 2⁷ = 128. Answer: 7.
   • (b) log₇(49) = 2.
   • (c) log(8) + log(125) = log(8 × 125) = log(1000) = 3.
   • (d) x = 3⁻² = 1/9.

4. Solving exponential equations. Fluency

   • (a) 64 = 2⁶. So 3x = 6 ⇒ x = 2.
   • (b) x = log(20)/log(7) = 1.3010/0.8451 ≈ 1.54.
   • (c) t = log(2)/log(1.04) = 0.3010/0.01703 ≈ 17.67 years.
   • (d) n = log(0.5)/log(0.9) = −0.3010/(−0.04576) ≈ 6.58.

5. Interpreting exponential graphs. Understanding

   • (a) Curve A starts at 100; curve B starts at 160.
   • (b) Curve A is growth (increasing over time); curve B is decay (decreasing toward zero).
   • (c) B(3) = 160 × 0.7³ = 160 × 0.343 ≈ 55.
   • (d) 100 × 1.15^t = 200 ⇒ 1.15^t = 2. t = log(2)/log(1.15) = 0.3010/0.0607 ≈ 4.96 years.

6. Combined index and log skills. Understanding

   • (a) log₂(8⁴) = 4⋅log₂(8) = 4×3 = 12.
   • (b) √(a³b) = (a³b)^1/2 = a^3/2b^1/2.
   • (c) 2^x × (2²)^x+1 = (2³)² ⇒ 2^x × 2^2x+2 = 2⁶ ⇒ 2^3x+2 = 2⁶ ⇒ 3x+2 = 6 ⇒ x = 4/3.
   • (d) 2⋅log(5) + log(4) = log(25) + log(4) = log(100) = 2.

7. Real-world indices and logs. Understanding

   • (a) N(d) = 2 × 3^d.
   • (b) 2 × 3^d > 1000 ⇒ 3^d > 500. d > log(500)/log(3) = 2.699/0.4771 ≈ 5.66. So after 6 days.
   • (c) 1.67 × 10⁻²⁷ ÷ 9.11 × 10⁻³¹ = (1.67/9.11) × 10^{−27−(−31)} = 0.1833 × 10⁴ ≈ 1.83 × 10³ times heavier.
   • (d) Ratio = 10^7.5/10^5.5 = 10² = 100 times more energy.

8. Financial applications. Understanding

   • (a) 10 000 × 1.07⁸ = 10 000 × 1.7182 ≈ $17 182.
   • (b) 1.07^t = 2 ⇒ t = log(2)/log(1.07) = 0.3010/0.02938 ≈ 10.24 years.
   • (c) A = 5000 × (1 + 0.18/12)²⁴ = 5000 × 1.015²⁴ = 5000 × 1.4295 ≈ $7 147.
   • (d) 1200 × 0.8ⁿ = 200 ⇒ 0.8ⁿ = 1/6. n = log(1/6)/log(0.8) = −0.7782/(−0.09691) ≈ 8.03. After 9 years.

9. Logarithm problem solving. Problem Solving

   • (a) log₂(x(x+2)) = 3 ⇒ x(x+2) = 8 ⇒ x²+2x−8 = 0 ⇒ (x+4)(x−2) = 0. x = −4 (rejected, x > 0) or x = 2.
   • (b) log(x(x−3)) = 1 ⇒ x(x−3) = 10 ⇒ x²−3x−10 = 0 ⇒ (x−5)(x+2) = 0. x = −2 (rejected) or x = 5.
   • (c) Let y = log_b(x), so b^y = x. Take log: y⋅log(b) = log(x), so y = log(x)/log(b). Therefore log_b(x) = log(x)/log(b). ✓
   • (d) log₃(20) = log(20)/log(3) = 1.3010/0.4771 ≈ 2.727.

10. Exponential modelling. Problem Solving

    • (a) N(0) = 500 × 2⁰ = 500. N(4) = 500 × 2¹ = 1 000 downloads.
    • (b) 500 × 2^t/4 = 8000 ⇒ 2^t/4 = 16 = 2⁴ ⇒ t/4 = 4 ⇒ t = 16 weeks.
    • (c) N(t+8)/N(t) = 2^(t+8)/4/2^t/4 = 2² = 4. Downloads quadruple every 8 weeks.
    • (d) Set 500 × 2^t/4 = 2000 × 1.10^t. Testing: t=10: N=500×2^2.5≈2828, M=2000×1.10¹⁰≈5187 (M>N). t=15: N=500×2^3.75≈8409, M=2000×1.10¹⁵≈8354. Equal at approximately t ≈ 15 weeks.

11. Scientific measurement. Problem Solving

    • (a) 8.8 × 10²⁶ ÷ 1.06 × 10⁻¹⁰ = (8.8/1.06) × 10³⁶ ≈ 8.3 × 10³⁶ atoms.
    • (b) 8.3 × 10³⁶ ≈ 10³⁷. The ratio spans approximately 37 orders of magnitude.
    • (c) 8 μm = 8 × 10⁻⁶ m = 8 × 10⁻⁶⁺⁹ nm = 8 000 nm.
    • (d) 8 × 10⁻⁶ m ÷ 1.06 × 10⁻¹⁰ m = (8/1.06) × 10⁴ ≈ 75 500 times larger.

12. Connecting indices, logs, and equations. Problem Solving

    • (a) A₀ × rⁿ = A ⇒ rⁿ = A/A₀ ⇒ n⋅log(r) = log(A/A₀) ⇒ n = log(A/A₀)/log(r). ✓
    • (b) n = log(12000/3000)/log(1.06) = log(4)/log(1.06) = 0.6021/0.02531 ≈ 23.8 years.
    • (c) P × 1.05^t = 2P ⇒ 1.05^t = 2 ⇒ t = log(2)/log(1.05). The P cancels — doubling time is independent of P. ✓
    • (d) t = ln(2)/ln(1 + r/100) ≈ 0.693/(r/100) = 69.3/r ≈ 70/r. This is the Rule of 70. ✓