Interpreting Exponential Parameters - Math 1

1. The company switches to monthly valuations and wants to maintain the same effective quarterly growth rate. If the new model is $V(m) = 2M \cdot r^m$ where $m$ is months, what is $r$, and how does the coefficient 2M change its meaning in the monthly model?
   Answer: $r = (1.4)^{1/3} ≈ 1.118$; coefficient represents the same initial valuation with monthly precision To maintain the same quarterly growth, after 3 months the valuation should equal what it would be after 1 quarter. So: $r^3 = 1.4$, giving $r = (1.4)^{1/3} ≈ 1.118$. The coefficient 2M still represents the initial valuation at time zero - changing the measurement frequency doesn't change what the starting value represents, only how often growth is compounded.
2. A radioactive substance follows $N(t) = N_0 \cdot 0.88^t$ where $t$ is in years. If measurements show the half-life is actually 4 years rather than the 8 years this model predicts, what should the decay factor be, and how does this change the interpretation of $N_0$?
   Answer: Decay factor 0.841; $N_0$ represents the same initial amount with corrected decay rate For half-life of 4 years, we need $N(4) = 0.5N_0$, so $N_0 \cdot b^4 = 0.5N_0$, giving $b^4 = 0.5$, thus $b = (0.5)^{1/4} ≈ 0.841$. The parameter $N_0$ still represents the initial amount - changing the decay rate doesn't change what the initial value represents, only how quickly the substance decays from that starting point.
3. The temperature of a cooling object follows $T(t) = 72 + 128 \cdot 0.85^t$ degrees Fahrenheit, where $t$ is time in minutes. What will be the temperature after a very long time, and what was the initial temperature difference from this long-term temperature?
   Answer: Long-term: 72°F, Initial difference: 128°F When you encounter an exponential decay function like this one, you're looking at a model where a quantity approaches a limiting value over time. The key is understanding what happens to each part of the equation as time increases. To find the long-term temperature, examine what happens as $t$ approaches infinity. Since $0.85 < 1$, the term $0.85^t$ gets smaller and smaller as $t$ increases, eventually approaching zero. This means $128 \cdot 0.85^t$ also approaches zero, leaving only the constant term: $T(\infty) = 72 + 0 = 72°F$. This is your long-term temperature. For the initial temperature difference, find $T(0)$ first: $T(0) = 72 + 128 \cdot 0.85^0 = 72 + 128 \cdot 1 = 200°F$. The initial difference from the long-term temperature is $200 - 72 = 128°F$. Choice A incorrectly calculates the initial difference as $128 - 72 = 56°F$, suggesting confusion about which temperature to subtract from which. Choice B mistakenly identifies the initial temperature (200°F) as the long-term temperature, reversing the concept entirely. Choice C incorrectly identifies 128°F as the long-term temperature—this is actually the coefficient in the decay term, not the limiting value. The correct answer is D: the temperature approaches 72°F in the long run, and the initial difference from this limiting temperature was 128°F. Remember: in exponential decay functions of the form $y = a + b \cdot r^t$ where $0 < r < 1$, the constant term $a$ is always your long-term limiting value.
4. If the account actually contains $5832 after 3 years, what was the bank's calculated growth factor, and what does this reveal about their error?
   Answer: 1.06, indicating they used 6% instead of the actual 4% annual rate With actual value $5832 after 3 years: $5832 = 5000(1+r)^3$, so $(1+r)^3 = 1.1664$, giving $1+r = 1.04$ or $r = 0.04$. The bank used $r + 0.02 = 0.06$, so their growth factor was $1.06$. This confirms they calculated with 6% instead of the actual 4% rate.
5. A savings account with quarterly compounding follows $A(t) = 5000(1.02)^{4t}$ where $t$ is in years. The bank changes to monthly compounding while maintaining the same annual percentage yield. What is the new monthly growth factor, and how does this change the interpretation of the coefficient 5000?
   Answer: Monthly factor $(1.0824)^{1/12} ≈ 1.0066$; coefficient still represents the original principal deposit The annual yield is $(1.02)^4 = 1.0824$ or 8.24%. To maintain this with monthly compounding: $(1 + r)^{12} = 1.0824$, so $r = (1.0824)^{1/12} - 1 ≈ 0.0066$, giving monthly factor ≈ 1.0066. The coefficient 5000 represents the initial principal and doesn't change when compounding frequency changes - it's still the starting amount.
6. A company's revenue follows $R(q) = 2400 \cdot 1.12^{q-3}$ thousand dollars, where $q$ is the number of quarters since the start of 2020. What was the company's revenue at the reference point used in this model?
   Answer: $2,400,000 during quarter 3 (Q3 2020) When you encounter exponential functions with expressions like $(q-3)$ in the exponent, you're looking at a horizontally shifted exponential model. The key insight is identifying the reference point where the model takes its base value. To find the reference point, look for where the exponent equals zero, since any number raised to the power of zero equals 1. Setting the exponent $q-3 = 0$, we get $q = 3$. At this point, $R(3) = 2400 \cdot 1.12^{3-3} = 2400 \cdot 1.12^0 = 2400 \cdot 1 = 2400$ thousand dollars, or $2,400,000. Since $q$ represents quarters since the start of 2020, quarter 3 corresponds to Q3 2020. Looking at the wrong answers: Choice B incorrectly assumes the reference point is at $q = 0$ (before 2020), but at $q = 0$, we'd have $R(0) = 2400 \cdot 1.12^{-3}$, which is much less than $2,400,000. Choice C mistakenly thinks Q1 2020 is the reference point, but that's $q = 1$, giving $R(1) = 2400 \cdot 1.12^{-2}$, again much smaller than the base value. Choice D suggests an extrapolation to the company's founding, which isn't what the mathematical reference point represents. Remember: In exponential models of the form $a \cdot b^{(x-h)}$, the reference point is always at $x = h$, where the function equals $a$. Don't confuse the mathematical reference point with the beginning of the time period or other contextual starting points.
7. A population of bacteria follows the model $P(t) = 850 \cdot 1.15^t$, where $t$ is time in hours. If the growth rate decreases by 40% due to limited nutrients, what would be the new growth factor in the modified exponential model?
   Answer: 1.09 The original growth factor is 1.15, which represents a 15% growth rate. If the growth rate decreases by 40%, the new growth rate is 15% × (1 - 0.40) = 15% × 0.60 = 9%. Therefore, the new growth factor is 1 + 0.09 = 1.09. Choice A incorrectly multiplies the entire growth factor by 0.60. Choice B subtracts 0.40 from the growth factor. Choice D incorrectly adds 40% to the original growth factor.
8. A radioactive substance decays according to $N(t) = 450 \cdot (0.5)^{t/12}$, where $t$ is time in hours. What does the parameter 12 represent in this context?
   Answer: The number of hours for the substance to decay to 50% of its original amount When t = 12, we have N(12) = 450 × $(0.5)^{12/12}$ = 450 × 0.5 = 225, which is exactly half the initial amount of 450. So 12 represents the half-life. Choice A describes when the amount reaches 25% (which occurs at t = 24). Choice C describes when it reaches 12.5% (which occurs at t = 36). Choice D misinterprets the parameter as a rate rather than a time period.
9. The temperature of a cooling object follows $T(t) = 75 + 125 \cdot 0.85^t$ where $T(t)$ is temperature in °F after $t$ minutes. A student claims that 75 represents the initial temperature and 125 represents the cooling rate. Which part of this interpretation needs correction, and what do these parameters actually represent?
   Answer: Both are wrong - 75 is ambient temperature, 125 is initial temperature difference above ambient This is the form $T(t) = T_{ambient} + (T_0 - T_{ambient}) \cdot b^t$. Here 75°F is the ambient/room temperature that the object approaches as $t → ∞$. The value 125°F represents the initial temperature difference above ambient (so initial temp was 75 + 125 = 200°F). The student incorrectly identified both parameters.
10. An investment account grows according to $A(t) = 5200 \cdot 1.045^t$ dollars after $t$ years. If the account owner makes an additional one-time deposit of $800 at the start, but the annual growth rate drops to 3.8%, what is the ratio of the new initial value to the new growth factor?
    Answer: 5780.3 The new initial value is 5200 + 800 = 6000. The new growth factor is 1 + 0.038 = 1.038. The ratio is 6000/1.038 ≈ 5780.3. Choice A uses the original growth factor 1.045. Choice C gives just the new initial value without division. Choice D incorrectly adds the deposit after applying the original growth factor once.
11. Two population models are given: City A follows $P_A(t) = 80000 \cdot 1.025^t$ and City B follows $P_B(t) = 95000 \cdot 0.98^t$, where $t$ is years after 2020. If both cities implement policies that double their respective annual percentage changes, which statement correctly interprets the new parameters?
    Answer: City A: growth factor 1.05, meaning 5% annual growth; City B: decay factor 0.96, meaning 4% annual decline City A originally grows at 2.5% annually (factor 1.025). Doubling this rate gives 5% growth (factor 1.05). City B originally declines at 2% annually (factor 0.98 = 1 - 0.02). Doubling the decline rate gives 4% annual decline (factor 0.96 = 1 - 0.04). The initial populations remain unchanged as policy affects growth rates, not starting values.
12. An exponential model $y = ab^x$ has the property that when $x$ increases by 3, $y$ increases by 125%. If $b > 1$ and the model passes through $(0, 20)$, what are the values of $a$ and $b$, and what does $b$ represent in terms of unit growth?
    Answer: $a = 20$, $b = (2.25)^{1/3}$; represents growth rate per unit yielding 125% per 3 units From $(0,20)$: $a = 20$. A 125% increase means the new value is 225% of original, so $y(x+3) = 2.25y(x)$. This gives us $ab^{x+3} = 2.25ab^x$, so $b^3 = 2.25$, thus $b = (2.25)^{1/3} ≈ 1.31$. This $b$ represents the growth factor per unit of $x$ that compounds to give 125% total growth over 3 units.

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