Q1: A population of bacteria doubles every 3 hours. If there are initially 500 bacteria, which function models the population \(P(t)\) after \(t\) hours? (a) \(P(t) = 500(2)^{3t}\) (b) \(P(t) = 500(2)^{t/3}\) (c) \(P(t) = 500(3)^{2t}\) (d) \(P(t) = 500(2)^t\)
Solution:
Ans: (b) Explanation: Since the bacteria doubles every 3 hours, we need to divide the time by 3 to find how many doubling periods have occurred. The correct exponential growth function is \(P(t) = 500(2)^{t/3}\). Option (a) incorrectly multiplies \(t\) by 3, option (c) uses the wrong base, and option (d) assumes doubling every hour.
Q2: Which of the following represents exponential decay? (a) \(y = 100(1.5)^x\) (b) \(y = 50(0.8)^x\) (c) \(y = 200(2)^x\) (d) \(y = 75(1)^x\)
Solution:
Ans: (b) Explanation:Exponential decay occurs when the base of the exponential function is between 0 and 1. In option (b), the base is 0.8, which is less than 1, indicating decay. Options (a) and (c) show growth since their bases are greater than 1, and option (d) represents a constant function.
Q3: A car purchased for $25,000 depreciates at a rate of 12% per year. What is the value of the car after 4 years? (a) $15,200 (b) $14,641.28 (c) $13,500 (d) $16,000
Solution:
Ans: (b) Explanation: Use the exponential decay formula \(V(t) = V_0(1 - r)^t\) where \(V_0 = 25000\), \(r = 0.12\), and \(t = 4\). \(V(4) = 25000(0.88)^4\) \(V(4) = 25000(0.59969536)\) \(V(4) ≈ 14641.28\) The car's value after 4 years is approximately $14,641.28.
Q4: The function \(f(x) = 800(1.25)^x\) models exponential growth. What is the growth factor? (a) 800 (b) 1.25 (c) 25 (d) 0.25
Solution:
Ans: (b) Explanation: In an exponential function of the form \(f(x) = a(b)^x\), the growth factor is \(b\). In this case, \(b = 1.25\). The value 800 is the initial amount, not the growth factor. Option (c) represents the percent increase (25%), and option (d) is incorrect.
Q5: If a substance decays according to the function \(A(t) = 200(0.5)^{t/5}\), what is the half-life of the substance? (a) 5 years (b) 0.5 years (c) 10 years (d) 2.5 years
Solution:
Ans: (a) Explanation: The half-life is the time it takes for the substance to decay to half its original amount. In the function \(A(t) = 200(0.5)^{t/5}\), the exponent \(t/5\) indicates that the substance halves when \(t/5 = 1\), which means \(t = 5\). Therefore, the half-life is 5 years.
Q6: An investment of $5,000 grows at a rate of 6% per year compounded annually. Which equation represents the value \(V\) after \(t\) years? (a) \(V = 5000(0.06)^t\) (b) \(V = 5000(1.6)^t\) (c) \(V = 5000(1.06)^t\) (d) \(V = 5000(0.94)^t\)
Solution:
Ans: (c) Explanation: For exponential growth with a rate of 6%, the growth factor is \(1 + r = 1 + 0.06 = 1.06\). The correct formula is \(V = 5000(1.06)^t\). Option (a) uses only the rate, option (b) incorrectly calculates the growth factor, and option (d) represents decay.
Q7: A city's population decreases from 80,000 to 64,000 over 5 years. Assuming exponential decay, what is the annual decay rate? (a) 5% (b) 20% (c) 4.37% (d) 3.5%
Solution:
Ans: (c) Explanation: Use the formula \(P(t) = P_0(1 - r)^t\) where \(P_0 = 80000\), \(P(5) = 64000\), and \(t = 5\). \(64000 = 80000(1 - r)^5\) \(0.8 = (1 - r)^5\) \(1 - r = 0.8^{1/5} ≈ 0.9563\) \(r ≈ 0.0437\) or 4.37% Options (a) and (d) are incorrect calculations, and option (b) represents the total decrease, not the annual rate.
Q8: Which characteristic distinguishes exponential functions from linear functions? (a) Exponential functions have a constant rate of change (b) Exponential functions have a variable rate of change (c) Exponential functions always pass through the origin (d) Exponential functions form straight lines when graphed
Solution:
Ans: (b) Explanation:Exponential functions have a variable rate of change that increases or decreases multiplicatively, while linear functions have a constant rate of change. Exponential functions typically pass through \((0, a)\) where \(a\) is the initial value, not necessarily the origin, and they form curves, not straight lines.
## Section B: Fill in the Blanks
Q9: In the exponential function \(y = ab^x\), the value \(a\) is called the __________ and represents the value of \(y\) when \(x = 0\).
Solution:
Ans: initial value (or initial amount/y-intercept) Explanation: The initial value is the starting amount in an exponential function. When \(x = 0\), \(y = ab^0 = a(1) = a\), confirming that \(a\) is the value when \(x = 0\).
Q10: If the growth factor in an exponential growth function is 1.08, then the percent rate of increase is __________.
Solution:
Ans: 8% (or 8) Explanation: The growth factor equals \(1 + r\) where \(r\) is the rate of increase. If the growth factor is 1.08, then \(r = 1.08 - 1 = 0.08 = 8\%\).
Q11: An exponential decay function has a base \(b\) where \(0 < b=""><>
Solution:
Ans: 1 Explanation: For exponential decay, the base must be between 0 and 1. This ensures that as \(x\) increases, the function value decreases.
Q12: The time required for a quantity to reduce to half its initial value in exponential decay is called the __________.
Solution:
Ans: half-life Explanation: The half-life is a key characteristic of exponential decay, representing the constant time period needed for the quantity to decrease by 50%.
Q13: If a population triples every 10 years, the growth factor per 10-year period is __________.
Solution:
Ans: 3 Explanation: The growth factor represents the multiplier applied to the population. Since the population triples (multiplies by 3) every 10 years, the growth factor is 3.
Q14: In the function \(N(t) = 1000(0.85)^t\), the quantity is experiencing exponential __________ at a rate of 15% per time period.
Solution:
Ans: decay Explanation: Since the base 0.85 is less than 1, this represents exponential decay. The decay rate is \(1 - 0.85 = 0.15\) or 15%.
## Section C: Word Problems
Q15: A radioactive isotope has a half-life of 8 days. If a sample contains 400 grams initially, how much of the isotope will remain after 24 days?
Solution:
Ans: Step-by-step solution: Use the formula \(A(t) = A_0(0.5)^{t/h}\) where \(A_0 = 400\) grams, \(t = 24\) days, and \(h = 8\) days (half-life). \(A(24) = 400(0.5)^{24/8}\) \(A(24) = 400(0.5)^3\) \(A(24) = 400(0.125)\) \(A(24) = 50\) Final Answer: 50 grams
Q16: Sarah invests $3,000 in an account that earns 5% interest compounded annually. How much money will be in the account after 6 years? Round your answer to the nearest cent.
Solution:
Ans: Step-by-step solution: Use the compound interest formula \(A = P(1 + r)^t\) where \(P = 3000\), \(r = 0.05\), and \(t = 6\). \(A = 3000(1.05)^6\) \(A = 3000(1.3401)\) \(A = 4020.30\) Final Answer: $4,020.30
Q17: The population of a town is decreasing exponentially at a rate of 2.5% per year. If the current population is 18,000, what will the population be in 10 years? Round to the nearest whole number.
Solution:
Ans: Step-by-step solution: Use the exponential decay formula \(P(t) = P_0(1 - r)^t\) where \(P_0 = 18000\), \(r = 0.025\), and \(t = 10\). \(P(10) = 18000(0.975)^{10}\) \(P(10) = 18000(0.7764)\) \(P(10) ≈ 13975.2\) Final Answer: 13,975 people
Q18: A virus spreads through a network of computers, doubling the number of infected computers every 2 hours. If 5 computers are initially infected, how many computers will be infected after 12 hours?
Solution:
Ans: Step-by-step solution: Use the formula \(N(t) = N_0(2)^{t/d}\) where \(N_0 = 5\), \(t = 12\) hours, and \(d = 2\) hours (doubling time). \(N(12) = 5(2)^{12/2}\) \(N(12) = 5(2)^6\) \(N(12) = 5(64)\) \(N(12) = 320\) Final Answer: 320 computers
Q19: The value of a piece of machinery depreciates at 18% per year. If the machinery was purchased for $45,000, what will its value be after 5 years? Round to the nearest dollar.
Solution:
Ans: Step-by-step solution: Use the exponential decay formula \(V(t) = V_0(1 - r)^t\) where \(V_0 = 45000\), \(r = 0.18\), and \(t = 5\). \(V(5) = 45000(0.82)^5\) \(V(5) = 45000(0.3707)\) \(V(5) ≈ 16681.5\) Final Answer: $16,682
Q20: A biologist observes that a population of 200 rabbits is growing at a rate of 15% per month. Assuming exponential growth continues, how many rabbits will there be after 8 months? Round to the nearest whole number.
Solution:
Ans: Step-by-step solution: Use the exponential growth formula \(P(t) = P_0(1 + r)^t\) where \(P_0 = 200\), \(r = 0.15\), and \(t = 8\). \(P(8) = 200(1.15)^8\) \(P(8) = 200(3.0590)\) \(P(8) ≈ 611.8\) Final Answer: 612 rabbits
The document Worksheet (with Solutions): Exponential Growth & Decay is a part of the Grade 9 Course Integrated Math 1.
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