Solutions — Exponential Functions

1. Key features from equations. Fluency

   • (a) y = 3^x: Base 3. 3 > 1 ⇒ growth. y-intercept (0, 1). Asymptote y = 0.
   • (b) y = (½)^x: Base ½. ½ < 1 ⇒ decay. y-intercept (0, 1). Asymptote y = 0.
   • (c) y = 2 ⋅ 3^x: Base 3. Growth. At x=0: y=2×1=2, so y-intercept (0, 2). Asymptote y = 0.
   • (d) y = 5^x + 2: Base 5. Growth. At x=0: y=1+2=3, so y-intercept (0, 3). As x→−∞, 5^x→0, so asymptote y = 2.

2. y = 4^x — evaluate and solve. Fluency

   • (a) Table of values: x=−2: 4⁻²=1/16. x=−1: 1/4. x=0: 1. x=1: 4. x=2: 16.
   • (b) y = 64: 4^x=64. Since 64=4³, we get x = 3.
   • (c) y = 1/64: 4^x=1/64=4⁻³. So x = −3.
   • (d) Does it cross the x-axis?: No. 4^x > 0 for all real x. The curve approaches y = 0 (asymptote) but never reaches or crosses it.

3. Growth or decay? Fluency

   • (a) y = 4^x: Base 4 > 1 ⇒ growth. y-intercept (0, 1). Asymptote y = 0.
   • (b) y = (0.5)^x: Base 0.5 < 1 ⇒ decay. y-intercept (0, 1). Asymptote y = 0.
   • (c) y = 3^−x: 3^−x = (1/3)^x. Base 1/3 < 1 ⇒ decay. y-intercept (0, 1). Asymptote y = 0.
   • (d) y = 2^x − 3: Base 2 > 1 ⇒ growth. At x=0: y=1−3=−2, so y-intercept (0, −2). Asymptote y = −3 (as x→−∞, 2^x→0).

4. Transformations — y-intercept and asymptote. Fluency

   • (a) y = 2^x + 5: At x=0: y=1+5=6. y-intercept (0, 6). Asymptote y = 5.
   • (b) y = 3^x − 1: At x=0: y=1−1=0. y-intercept (0, 0). Asymptote y = −1.
   • (c) y = 4 × 2^x: At x=0: y=4×1=4. y-intercept (0, 4). Asymptote y = 0.
   • (d) y = 2^x+1: At x=0: y=2¹=2. y-intercept (0, 2). Asymptote y = 0 (no vertical shift).

5. Find the equation y = a^x. Understanding

   • (a) Through (2, 9): a²=9 ⇒ a=3 (since a>0). y = 3^x
   • (b) Through (3, 64): a³=64=4³ ⇒ a=4. y = 4^x
   • (c) Through (2, ¼): a²=1/4 ⇒ a=1/2 (since a>0). y = (½)^x
   • (d) Through (−1, 3): a⁻¹=3 ⇒ 1/a=3 ⇒ a=1/3. y = (⅓)^x

6. Bacteria growth. Understanding

   • (a) After 3 hours: P(3)=500×2³=500×8=4000 bacteria.
   • (b) After 6 hours: P(6)=500×2⁶=500×64=32 000 bacteria.
   • (c) When P = 8000: 500×2^t=8000 ⇒ 2^t=16=2⁴ ⇒ t = 4 hours.
   • (d) Factor in first 4 hours: P(4)/P(0) = 8000/500 = 16. The population is 16 times larger (= 2⁴, because it doubles 4 times).

7. Drug clearance. Understanding

   • (a) After 4 hours: A(4)=200×(0.5)^4/4=200×0.5=100 mg.
   • (b) After 12 hours: A(12)=200×(0.5)^12/4=200×(0.5)³=200×0.125=25 mg.
   • (c) When A < 50 mg: t=4: A=100 (above). t=8: A=200×(0.5)²=50 mg (equal, not yet below). t=12: A=25 mg (below). So the amount first falls below 50 mg just after t = 8 hours.
   • (d) As t → ∞: (0.5)^t/4→0, so A→0. Asymptote A = 0. In context: the drug is theoretically never fully eliminated — it keeps halving and approaching zero without ever reaching it.

8. Read from the graph: y = 2^x and y = 3^x. Understanding

   • (a) Shared point: Both curves meet at (0, 1). Because a⁰ = 1 for any base a, every basic exponential passes through (0, 1).
   • (b) Which rises faster for x > 0?: y = 3^x (orange) rises more steeply. At x = 2: 3² = 9 vs 2² = 4. A higher base produces faster growth.
   • (c) Which is higher for x < 0?: y = 2^x (blue) sits above y = 3^x for x < 0. For example at x = −2: 2⁻² = 0.25 > 3⁻² ≈ 0.11. A larger base gives a smaller value for negative exponents.
   • (d) Shared asymptote: Both curves have horizontal asymptote y = 0. As x → −∞, both curves approach (but never reach) the x-axis.

9. Compound interest. Problem Solving

   • (a) After 1 year: A(1)=2000×1.06=$2120.00.
   • (b) After 10 years: A(10)=2000×1.06¹⁰=2000×1.79085≈$3581.70.
   • (c) When does it exceed $4000?: Need 1.06^t>2. Trial: 1.06¹¹≈1.8983 (below 2), 1.06¹²≈2.0122 (above 2). The investment first exceeds $4000 after 12 complete years.
   • (d) At 8%: New formula: A = 2000 × 1.08^t. Doubling takes fewer years — by trial, 1.08⁹≈2.00, so the investment doubles in about 9 years (compared to 12 years at 6%).

10. Coffee cooling. Problem Solving

    • (a) Initial temperature: T(0)=20+70×(0.9)⁰=20+70×1=90°C.
    • (b) After 5 minutes: T(5)=20+70×(0.9)⁵=20+70×0.5905=20+41.3≈61°C.
    • (c) As t → ∞: (0.9)^t→0, so T→20. Asymptote T = 20°C. In context: the coffee eventually reaches room temperature (20°C) — it can never cool below this.
    • (d) When T < 48°C: t=8: T=20+70×(0.9)⁸=20+70×0.4305=20+30.1=50.1°C (still above 48). t=9: T=20+70×(0.9)⁹=20+70×0.3874=20+27.1=47.1°C (below 48). The coffee drops below 48°C between 8 and 9 minutes, so approximately 9 minutes.