Recognizing Exponential Growth/Decay - Math 1

1. A radioactive substance has a half-life of 6 days. If you start with 80 grams, which expression represents the amount remaining after $d$ days?
   Answer: $80 \cdot (0.5)^{d/6}$ Half-life of 6 days means the substance reduces to half its amount every 6 days. This gives us $A(d) = 80 \cdot (0.5)^{d/6}$. At $d = 6$, we get $80 \cdot (0.5)^1 = 40$ grams, which is correct. Choice B decays too rapidly, Choice C represents linear decay, and Choice D has the exponent inverted.
2. An investment account grows according to the equation $A(t) = 5000(1.08)^t$, where $t$ is time in years. After how many complete years will the account first exceed $8000?
   Answer: 7 years This is an exponential growth problem where you need to find when an investment first crosses a specific threshold. When you see equations in the form $A(t) = P(1 + r)^t$, you're dealing with compound growth, and the key is setting up an inequality to find when the account exceeds the target amount. To solve this, set up the inequality $5000(1.08)^t > 8000$. Divide both sides by 5000 to get $(1.08)^t > 1.6$. Taking the natural logarithm of both sides: $t \ln(1.08) > \ln(1.6)$. This gives us $t > \frac{\ln(1.6)}{\ln(1.08)} \approx \frac{0.4700}{0.0770} \approx 6.1$ years. Since the question asks for complete years when the account first exceeds $8000, you need 7 complete years. Let's verify: $A(6) = 5000(1.08)^6 \approx 7934$ (still under $8000), while $A(7) = 5000(1.08)^7 \approx 8569$ (exceeds $8000). Therefore, D) 7 years is correct. Choice A) 6 years is wrong because after 6 years, the account is still below $8000. Choice C) 8 years represents calculating correctly but adding an extra year unnecessarily. Choice B) 9 years might result from computational errors or misunderstanding the exponential formula. Remember that "first exceeds" problems require you to round up to the next whole number, even if your calculated value is just slightly above a whole number. Always verify your answer by plugging back into the original equation.
3. A population of rabbits grows exponentially. If the population doubles every 6 months and starts at 50 rabbits, which function models the population after $m$ months?
   Answer: $P(m) = 50 \cdot 2^{m/6}$ When you encounter exponential growth problems, focus on identifying three key components: the initial value, the growth factor, and the time period for that growth. Here, rabbits start at 50, double (growth factor of 2), and do so every 6 months. The general form for exponential growth is $P(t) = P_0 \cdot b^{t/k}$, where $P_0$ is initial population, $b$ is the growth factor, and $k$ is the time period for one complete growth cycle. Since the population doubles every 6 months, we need $P(m) = 50 \cdot 2^{m/6}$. This ensures that when $m = 6$, the exponent becomes $6/6 = 1$, giving us $50 \cdot 2^1 = 100$ rabbits—exactly double the initial population. Choice A uses 6 as the base instead of 2, which would mean the population multiplies by 6 rather than doubles. Choice B has the time relationship backwards with $2^{6m}$, causing explosive growth—after just one month, you'd have $50 \cdot 2^6 = 3200$ rabbits instead of the gradual doubling pattern. Choice C represents linear growth (adding a constant amount each month) rather than exponential growth, completely missing the doubling behavior. To verify: at $m = 0$, we get 50 rabbits; at $m = 6$, we get $50 \cdot 2^1 = 100$; at $m = 12$, we get $50 \cdot 2^2 = 200$—perfect doubling every 6 months. Strategy tip: In exponential growth problems, always check your function by plugging in the given time period. Your growth factor should appear exactly once as the result.
4. A radioactive substance has a half-life of 6 years. If a sample initially contains 80 grams of the substance, approximately how many grams will remain after 15 years?
   Answer: 14.1 grams After 15 years, the substance undergoes 15/6 = 2.5 half-life periods. The amount remaining is 80 × $(1/2)^{2.5}$ = 80 × $(1/2)^2$ × $(1/2)^{0.5}$ = 80 × 0.25 × 0.707 ≈ 14.1 grams. Choice B uses only 2 half-lives. Choice C uses 3 half-lives. Choice D uses an incorrect calculation method.
5. A savings account earns compound interest according to $A = P(1 + r)^t$. If an account grows from $2000 to $2420 in 2 years, what is the approximate annual interest rate?
   Answer: 10.0% Using A = $P(1 + r)^t$: 2420 = 2000$(1 + r)^2$. Solving: $(1 + r)^2$ = 1.21, so 1 + r = 1.1, giving r = 0.1 or 10%. Choice A is slightly too low. Choice C is slightly too high. Choice D incorrectly uses the total 2-year growth rate as the annual rate.
6. An investment account grows according to $A(t) = 5000(1.08)^t$ where $t$ is years. After how many years will the account value exceed $8000 for the first time?
   Answer: 7 years We need $5000(1.08)^t > 8000$, so $(1.08)^t > 1.6$. Testing: $t=6: (1.08)^6 ≈ 1.586 < 1.6$; $t=7: (1.08)^7 ≈ 1.714 > 1.6$. Therefore, it takes 7 years to first exceed $8000. Choices A and B are too early, while Choice D is unnecessarily late.
7. A population of rabbits in a park follows the model $P(t) = 150(0.85)^t$, where $t$ is time in months. What does the value 0.85 represent in this context?
   Answer: The population retains 85% of its previous value each month, indicating a 15% monthly decline When you encounter exponential models like $P(t) = 150(0.85)^t$, focus on interpreting the base of the exponent. This number reveals how the population changes from one time period to the next. In this model, each month the population equals 85% of what it was the previous month. If you start with 150 rabbits, after one month you'd have $150 \times 0.85 = 127.5$ rabbits. After two months: $150 \times (0.85)^2 = 108.4$ rabbits. The factor 0.85 means the population retains 85% of its value each month, which corresponds to losing 15% monthly (since $1 - 0.85 = 0.15 = 15\%$). Choice A incorrectly states the population decreases by 85% each month. If that were true, only 15% would remain each month, and the model would be $P(t) = 150(0.15)^t$. Choice B suggests an 85% increase each month, which would require a base greater than 1, like $P(t) = 150(1.85)^t$. Since our base is 0.85 (less than 1), the population is declining, not growing. Choice D misinterprets 0.85 as a long-term stabilization value, but 0.85 is the monthly retention factor, not an equilibrium point. Remember: in exponential models $A \cdot b^t$, when the base $b$ is between 0 and 1, it represents the fraction retained each time period. The percentage decrease is always $1 - b$.
8. A scientist observes that a bacterial culture triples in size every 4 hours. If the culture starts with 200 bacteria, which expression best represents the number of bacteria after $t$ hours?
   Answer: $200 \cdot 3^{t/4}$ Since the culture triples every 4 hours, after t hours it will have undergone t/4 complete cycles of tripling. The correct expression is 200 · $3^{t/4}$. Choice B incorrectly uses 4t as the exponent. Choice C confuses the base and time relationship. Choice D uses an incorrect growth factor.
9. The value of a car decreases according to the function $V(t) = 25000(0.85)^t$, where $t$ is the number of years after purchase. After how many complete years will the car's value first drop below $15,000?
   Answer: 4 years We need to find when $25000(0.85)^t < 15000$, so $(0.85)^t < 0.6$. Testing values: $t=3: (0.85)^3 ≈ 0.614 > 0.6$; $t=4: (0.85)^4 ≈ 0.522 < 0.6$. So after 4 complete years, the value drops below $15,000. Choice A is too early, while Choices C and D are too late.
10. A scientist observes that a bacterial culture triples in size every 4 hours. If the culture starts with 200 bacteria at time $t = 0$, which function best models the population $P(t)$ after $t$ hours?
    Answer: $P(t) = 200 \cdot 3^{t/4}$ Since the culture triples every 4 hours, we need $P(t) = 200 \cdot 3^{t/4}$. When $t = 4$, this gives $P(4) = 200 \cdot 3^1 = 600$, which is triple the initial amount. Choice B grows too rapidly (triples every hour), Choice C is linear growth, and Choice D uses incorrect base and exponent relationship.

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