Q1: Which of the following represents an exponential growth function? (a) \(y = 3x + 5\) (b) \(y = 2(0.5)^x\) (c) \(y = 4(1.2)^x\) (d) \(y = x^2 + 3\)
Solution:
Ans: (c) Explanation: An exponential growth function has the form \(y = ab^x\) where \(a > 0\) and \(b > 1\). Option (c) has \(b = 1.2 > 1\), so it represents growth. Option (a) is linear, option (b) has \(b = 0.5 < 1\)="" so="" it="" represents="" decay,="" and="" option="" (d)="" is="">
Q2: A bacteria population doubles every 3 hours. If the initial population is 500, what is the growth factor per hour? (a) 2 (b) \(2^{1/3}\) (c) 3 (d) 6
Solution:
Ans: (b) Explanation: The population doubles in 3 hours, so the growth factor for 3 hours is 2. To find the hourly growth factor, we need \(b^3 = 2\), which gives \(b = 2^{1/3}\). This is the per-hour growth factor.
Q3: The equation \(A = 5000(1.06)^t\) models an investment. What does the value 1.06 represent? (a) The initial investment amount (b) The annual growth rate of 6% (c) The final amount after t years (d) The time in years
Solution:
Ans: (b) Explanation: In the exponential model \(A = P(1 + r)^t\), the base \(1 + r\) represents the growth factor. Here, \(1.06 = 1 + 0.06\), indicating an annual growth rate of 6% (or 0.06 as a decimal).
Q5: A car depreciates at a rate of 15% per year. If the initial value is $20,000, which equation models the car's value after t years? (a) \(V = 20000(1.15)^t\) (b) \(V = 20000(0.15)^t\) (c) \(V = 20000(0.85)^t\) (d) \(V = 20000 - 0.15t\)
Solution:
Ans: (c) Explanation: A depreciation rate of 15% means the car retains 85% of its value each year. The decay factor is \(1 - 0.15 = 0.85\). The correct model is \(V = 20000(0.85)^t\). Option (a) represents growth, option (b) uses the wrong base, and option (d) is a linear model.
Q6: The half-life of a radioactive substance is 8 years. What fraction of the original amount remains after 24 years? (a) \(\frac{1}{2}\) (b) \(\frac{1}{3}\) (c) \(\frac{1}{4}\) (d) \(\frac{1}{8}\)
Solution:
Ans: (d) Explanation:Half-life is the time for half the substance to decay. After 24 years, there are \(\frac{24}{8} = 3\) half-lives. The remaining fraction is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\) of the original amount.
Q7: Which value of b in \(y = ab^x\) would make the function increase most rapidly? (a) \(b = 0.5\) (b) \(b = 1.1\) (c) \(b = 2.5\) (d) \(b = 0.9\)
Solution:
Ans: (c) Explanation: For exponential growth, the larger the base \(b > 1\), the more rapidly the function increases. Among the options, \(b = 2.5\) is the largest value greater than 1, so it produces the most rapid growth. Options (a) and (d) represent decay since they are less than 1.
Q8: An investment of $1,000 grows according to the formula \(A = 1000e^{0.05t}\), where t is in years. What type of growth model is this? (a) Linear growth (b) Discrete exponential growth (c) Continuous exponential growth (d) Quadratic growth
Solution:
Ans: (c) Explanation: The presence of the natural base e in the formula \(A = Pe^{rt}\) indicates continuous exponential growth. This model assumes continuous compounding, unlike discrete models that use \((1 + r)^t\).
Section B: Fill in the Blanks
Q9: The general form of an exponential function is \(y = ab^x\), where a is the __________ and b is the __________.
Solution:
Ans: initial value (or y-intercept); base (or growth/decay factor) Explanation: In the exponential function \(y = ab^x\), the parameter \(a\) represents the initial value (the value when \(x = 0\)), and \(b\) is the base or growth/decay factor.
Q10: If an exponential function has a base between 0 and 1, it represents exponential __________.
Q11: The formula for continuous compound interest is \(A = Pe^{rt}\), where e is approximately equal to __________.
Solution:
Ans: 2.718 (or 2.71828) Explanation: The mathematical constant e is the natural base and is approximately equal to 2.718. It is used in continuous exponential models.
Q12: The time required for a quantity to reduce to half its initial amount in an exponential decay model is called the __________.
Solution:
Ans: half-life Explanation: The half-life is the characteristic time in an exponential decay process for a quantity to decrease to 50% of its original value.
Q13: If a population grows from 200 to 800 in 2 hours, and the growth is exponential, the population was multiplied by a factor of __________.
Solution:
Ans: 4 Explanation: The growth factor is found by dividing the final amount by the initial amount: \(\frac{800}{200} = 4\). The population increased by a factor of 4.
Q14: To convert an annual growth rate of 8% to the form \(1 + r\) for use in an exponential model, we write __________.
Solution:
Ans: 1.08 Explanation: An annual growth rate of 8% means \(r = 0.08\). The growth factor is \(1 + r = 1 + 0.08 = 1.08\), which is used as the base in the exponential model.
Section C: Word Problems
Q15: A city's population is currently 50,000 and is growing at a rate of 3% per year. Write an exponential function to model the population after t years, and find the population after 10 years. Round to the nearest whole number.
Solution:
Ans:
The exponential growth model is \(P(t) = P_0(1 + r)^t\)
Given: \(P_0 = 50000\) and \(r = 0.03\)
The function is: \(P(t) = 50000(1.03)^t\)
For \(t = 10\) years:
\(P(10) = 50000(1.03)^{10}\)
\(P(10) = 50000(1.3439)\)
\(P(10) = 67195\) Final Answer: The population after 10 years is approximately 67,195 people.
Q16: A laptop costs $1,200 and depreciates at a rate of 20% per year. What will be its value after 3 years? Round to the nearest dollar.
Solution:
Ans:
The exponential decay model is \(V(t) = V_0(1 - r)^t\)
Given: \(V_0 = 1200\) and \(r = 0.20\)
The decay factor is \(1 - 0.20 = 0.80\)
The function is: \(V(t) = 1200(0.80)^t\)
For \(t = 3\) years:
\(V(3) = 1200(0.80)^3\)
\(V(3) = 1200(0.512)\)
\(V(3) = 614.40\) Final Answer: The laptop's value after 3 years is approximately $614.
Q17: A radioactive isotope has a half-life of 5 years. If you start with 80 grams, how much will remain after 15 years?
Solution:
Ans:
Using the half-life formula: \(A(t) = A_0 \left(\frac{1}{2}\right)^{t/h}\)
where \(A_0 = 80\) grams, \(t = 15\) years, and \(h = 5\) years
\(A(15) = 80 \left(\frac{1}{2}\right)^{15/5}\)
\(A(15) = 80 \left(\frac{1}{2}\right)^3\)
\(A(15) = 80 \times \frac{1}{8}\)
\(A(15) = 10\) Final Answer: 10 grams will remain after 15 years.
Q18: An investment of $2,500 earns interest compounded continuously at an annual rate of 4.5%. How much will the investment be worth after 6 years? Use the formula \(A = Pe^{rt}\) and round to the nearest cent.
Solution:
Ans:
Using the continuous compound interest formula: \(A = Pe^{rt}\)
Given: \(P = 2500\), \(r = 0.045\), \(t = 6\)
\(A = 2500e^{0.045 \times 6}\)
\(A = 2500e^{0.27}\)
\(A = 2500(1.3100)\)
\(A = 3275.00\) Final Answer: The investment will be worth $3,275.00 after 6 years.
Q19: A bacterial culture starts with 300 bacteria and triples every 4 hours. Write an exponential function to model the number of bacteria after t hours, and determine how many bacteria will be present after 12 hours.
Solution:
Ans:
The bacteria triple every 4 hours, so the growth factor per 4 hours is 3
The number of 4-hour periods in t hours is \(\frac{t}{4}\)
The exponential model is: \(N(t) = 300(3)^{t/4}\)
For \(t = 12\) hours:
\(N(12) = 300(3)^{12/4}\)
\(N(12) = 300(3)^3\)
\(N(12) = 300(27)\)
\(N(12) = 8100\) Final Answer: There will be 8,100 bacteria after 12 hours.
Q20: The value of a rare coin appreciates at 12% per year. If the coin is currently worth $500, how many years will it take for the coin to be worth at least $1,000? Use logarithms and round up to the nearest whole year.
Solution:
Ans:
Using the exponential growth model: \(A = P(1 + r)^t\)
Given: \(P = 500\), \(A = 1000\), \(r = 0.12\)
\(1000 = 500(1.12)^t\)
Divide both sides by 500:
\(2 = (1.12)^t\)
Take the logarithm of both sides:
\(\log(2) = t \log(1.12)\)
\(t = \frac{\log(2)}{\log(1.12)}\)
\(t = \frac{0.3010}{0.0492}\)
\(t = 6.12\)
Rounding up to the nearest whole year: \(t = 7\) Final Answer: It will take 7 years for the coin to be worth at least $1,000.
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