L36 Solutions — Solving Exponential Equations

1. Same-base method. Fluency

   • (a) 128 = 2⁷. So x = 7.
   • (b) 1/9 = 3⁻². So x = −2.
   • (c) 125 = 5³. So 2x = 3 ⇒ x = 3/2.
   • (d) 4^x−1 = (2²)^x−1 = 2^2x−2. And 8 = 2³. So 2x − 2 = 3 ⇒ x = 5/2.

2. Logarithm method. Fluency

   • (a) x = log(10)/log(2) = 1/0.3010 ≈ 3.32.
   • (b) x = log(50)/log(3) = 1.6990/0.4771 ≈ 3.56.
   • (c) x = log(10)/log(1.5) = 1/0.1761 ≈ 5.68.
   • (d) x = log(0.1)/log(0.8) = −1/(−0.0969) ≈ 10.32.

3. Growth and decay. Fluency

   • (a) V(t) = 2000 × 1.05^t.
   • (b) V(10) = 2000 × 1.05¹⁰ = 2000 × 1.6289 ≈ $3 257.79.
   • (c) V(t) = 25 000 × 0.88^t.
   • (d) V(5) = 25 000 × 0.88⁵ = 25 000 × 0.5277 ≈ $13 181.

4. Solving growth/decay for time. Fluency

   • (a) 3000 × 1.06^t = 6000 ⇒ 1.06^t = 2. t = log(2)/log(1.06) = 0.3010/0.02531 ≈ 11.9 years.
   • (b) 200 × 0.95^t = 100 ⇒ 0.95^t = 0.5. t = log(0.5)/log(0.95) ≈ −0.3010/(−0.02228) ≈ 13.5 years.
   • (c) 500 × 2^t/3 > 10 000 ⇒ 2^t/3 > 20. t/3 = log(20)/log(2) = 4.322. t > 12.97 hours ⇒ after 13 hours.
   • (d) Rule of 72: 72 ÷ 9 = 8 years.

5. Graph of exponential growth vs decay. Understanding

   • (a) Both graphs pass through (0, 1). This is their only intersection.
   • (b) For x > 0: 2^x > 1 while (1/2)^x < 1. So 2^x > (1/2)^x for all x > 0.
   • (c) Both equal 1 at x = 0. By the zero index law: 2⁰ = 1 and (1/2)⁰ = 1.
   • (d) As x → +∞: 2^x → ∞ (grows without bound); (1/2)^x → 0 (approaches but never reaches zero).

6. Mixed exponential equations. Understanding

   • (a) log(2^x+1) = log(3^x) ⇒ (x+1)log(2) = x⋅log(3) ⇒ x⋅log(2) + log(2) = x⋅log(3) ⇒ x(log(3)−log(2)) = log(2) ⇒ x = log(2)/log(3/2) ≈ 0.3010/0.1761 ≈ 1.71.
   • (b) 4^x = 6 ⇒ (2²)^x = 6 ⇒ 2^2x = 6 ⇒ 2x = log₂(6) ⇒ x = log₂(6)/2 = log₂(6)/2 ≈ 1.29.
   • (c) Let u = 3^x. Then 9^x = u². Equation: u² − 10u + 9 = 0 ⇒ (u−1)(u−9) = 0 ⇒ u = 1 or u = 9. 3^x = 1 ⇒ x = 0; 3^x = 9 ⇒ x = 2.
   • (d) Let u = 2^x. u² = 5u − 4 ⇒ u² − 5u + 4 = 0 ⇒ (u−1)(u−4) = 0 ⇒ u = 1 or u = 4. 2^x = 1 ⇒ x = 0; 2^x = 4 ⇒ x = 2.

7. Compound interest. Understanding

   • (a) A = 4000 × (1 + 0.08/4)^4t = 4000 × 1.02^4t.
   • (b) A = 4000 × 1.02²⁰ = 4000 × 1.4859 ≈ $5 943.73.
   • (c) 1.02^4t = 2.5 ⇒ 4t = log(2.5)/log(1.02) = 0.3979/0.008600 ≈ 46.27 ⇒ t ≈ 11.6 years.
   • (d) Annual compounding: A = 4000 × 1.08⁵ = 4000 × 1.4693 ≈ $5 877.31. Quarterly compounding gives $5 943.73 — slightly more.

8. Applications of decay. Understanding

   • (a) A(t) = 500 × (1/2)^t/10.
   • (b) A(25) = 500 × (1/2)^2.5 = 500 × 0.1768 ≈ 88.4 mg.
   • (c) 500 × (1/2)^t/10 = 10 ⇒ (1/2)^t/10 = 0.02. t/10 = log(0.02)/log(0.5) = −1.699/(−0.3010) = 5.644. t ≈ 56.4 years.
   • (d) The curve starts at A₀ and decreases toward zero but never reaches it. Each half-life period reduces the amount by exactly half.

9. Newton’s Law of Cooling. Problem Solving

   • (a) T(0) = 20 + 70 × 0.95⁰ = 20 + 70 = 90°C.
   • (b) The “20” is the ambient (room) temperature — the temperature the drink approaches but never goes below.
   • (c) T(10) = 20 + 70 × 0.95¹⁰ = 20 + 70 × 0.5987 ≈ 20 + 41.9 = 61.9°C.
   • (d) 20 + 70 × 0.95^t = 40 ⇒ 0.95^t = 20/70 = 2/7. t = log(2/7)/log(0.95) = −0.5441/(−0.02228) ≈ 24.4 minutes.

10. Carbon-14 dating. Problem Solving

    • (a) A(t) = A₀ × (0.5)^t/5730.
    • (b) (0.5)^t/5730 = 0.30. t/5730 = log(0.30)/log(0.5) = −0.5229/(−0.3010) = 1.7374. t = 1.7374 × 5730 ≈ 9955 years.
    • (c) Rounded: 9 900 years (or approximately 10 000 years).
    • (d) 70% remaining means less time has elapsed — the fossil is younger. t/5730 = log(0.70)/log(0.5) = −0.1549/(−0.3010) = 0.5148. t ≈ 0.5148 × 5730 ≈ 2950 years.