L35 Solutions — Introduction to Logarithms

1. Log and index form. Fluency

   • (a) 2⁵ = 32.
   • (b) log₄(64) = 3.
   • (c) 10⁴ = 10 000.
   • (d) log₅(0.04) = −2.

2. Evaluating logarithms. Fluency

   • (a) 2^? = 64 = 2⁶. Answer: 6.
   • (b) 10^? = 100 000 = 10⁵. Answer: 5.
   • (c) 3^? = 1/9 = 3⁻². Answer: −2.
   • (d) 4^? = 8. Write 4 = 2² and 8 = 2³. So (2²)^? = 2³ ⇒ 2? = 3 ⇒ ? = 3/2.

3. Logarithm laws. Fluency

   • (a) log(6) + log(5) = log(6 × 5) = log(30).
   • (b) log(200) − log(20) = log(200/20) = log(10) = 1.
   • (c) 3⋅log(4) = log(4³) = log(64).
   • (d) log₂(4) + log₂(8) = 2 + 3 = 5. (Or: log₂(4×8) = log₂(32) = 5.)

4. Solving log equations. Fluency

   • (a) x = 2⁵ = 32.
   • (b) x⁴ = 81 ⇒ x = 81^1/4 = 3^4×1/4 = 3.
   • (c) x = 10⁻³ = 0.001.
   • (d) x + 1 = 3³ = 27 ⇒ x = 26.

5. Reading a logarithm graph. Understanding

   • (a) The x-intercept is (1, 0). This means log₂(1) = 0 — which follows because 2⁰ = 1.
   • (b) From the graph, log₂(8) = 3. Algebraically: 2³ = 8. ✓
   • (c) log₂(x) = −1 ⇒ x = 2⁻¹ = 1/2. Check: log₂(0.5) = −1. ✓
   • (d) log_b(x) is defined only when there exists y such that b^y = x. Since b^y > 0 for all y, x must be positive. You cannot take the log of zero or a negative number.

6. Combining log laws. Understanding

   • (a) log(12) = log(4 × 3) = log(2²) + log(3) = 2⋅log(2) + log(3).
   • (b) log₂(32) − log₂(4) = 5 − 2 = 3. (Or: log₂(32/4) = log₂(8) = 3.)
   • (c) log(25) = log(100/4) = log(10²/2²) = 2⋅log(10) − 2⋅log(2) = 2 − 2⋅log(2). ✓
   • (d) 2⋅log(3) + log(4) − log(36) = log(9) + log(4) − log(36) = log(9×4/36) = log(1) = 0.

7. Logarithms in real-world formulas. Understanding

   • (a) pH = −log(10⁻⁴) = −(−4) = 4.
   • (b) 3 = −log[H⁺] ⇒ log[H⁺] = −3 ⇒ [H⁺] = 10⁻³ mol/L.
   • (c) Difference = 6 − 4 = 2. So intensity ratio = 10² = 100 times more intense.
   • (d) dB = 10⋅log(1000I₀/I₀) = 10⋅log(1000) = 10⋅3 = 30 dB.

8. Inverse relationship with exponentials. Understanding

   • (a) f⁻¹(x) = log₂(x). The inverse of 2^x is the base-2 logarithm.
   • (b) Let y = log_b(b^x). By definition, b^y = b^x, so y = x. Therefore log_b(b^x) = x. ✓
   • (c) log₅(5⁷) = 7.
   • (d) 10^log(3.7) = 3.7 — exponential and logarithm (same base) are inverses.

9. Population doubling time. Problem Solving

   • (a) 200 000 × 1.08^t = 400 000 ⇒ 1.08^t = 2. ✓
   • (b) Taking log₁₀: log(1.08^t) = log(2) ⇒ t⋅log(1.08) = log(2). ✓
   • (c) t = log(2)/log(1.08) = 0.301/0.0334 ≈ 9.01 ≈ 9 years.
   • (d) Rule of 72: 72 ÷ 8 = 9 years. This matches the exact answer closely.

10. Logarithmic scale. Problem Solving

    • (a) 10⁻¹² to 10¹ spans exponents −12 to 1 = 13 orders of magnitude.
    • (b) dB = 10⋅log(10/10⁻¹²) = 10⋅log(10¹³) = 10×13 = 130 dB.
    • (c) 60 dB: I₁/I₀ = 10⁶. 80 dB: I₂/I₀ = 10⁸. Ratio: I₂/I₁ = 10⁸/10⁶ = 100.
    • (d) A linear scale would require a ruler 10 000 000 000 000 times longer for the high end vs. the low end — impossible to display. A logarithmic scale compresses this enormous range into a manageable 0–130 dB scale.