Using Exponential Models - Math 1

The atmospheric pressure P (in inches of mercury) at altitude h (in thousands of feet) is modeled by P(h) = $29.92e^{-0.125h}$ . At what altitude is the pressure exactly 75% of sea level pressure?
Answer: 2,300 feet When you encounter exponential decay problems involving percentages of original values, you're solving for the input that produces a specific fraction of the initial output. Here, sea level pressure occurs at h = 0, so you need to find where P(h) equals 75% of P(0). First, calculate the sea level pressure: $P(0) = 29.92e^{-0.125(0)} = 29.92e^0 = 29.92$ inches of mercury. Next, find 75% of this value: $0.75 × 29.92 = 22.44$ inches of mercury. Now solve for h when P(h) = 22.44: $22.44 = 29.92e^{-0.125h}$ Divide both sides by 29.92: $\frac{22.44}{29.92} = e^{-0.125h}$ $0.75 = e^{-0.125h}$ Take the natural logarithm of both sides: $\ln(0.75) = -0.125h$ $-0.2877 = -0.125h$ $h = \frac{0.2877}{0.125} = 2.30$ Since h is in thousands of feet, this equals 2,300 feet, which is answer D. The other options represent common calculation errors: A) 2,100 feet likely comes from rounding errors in the logarithm calculation. B) 2,800 feet might result from sign errors when handling the negative exponent. C) 2,500 feet could come from approximating ln(0.75) incorrectly. Remember: In exponential decay problems, always identify your reference point first (here, sea level), then set up your equation systematically. Double-check your logarithm calculations, as small errors compound quickly.
A radioactive substance has a half-life of 12 years. If a sample initially contains 80 grams, which expression best represents the amount remaining after 30 years?
Answer: Both A and C are correct For half-life problems, the general form is A(t) = A₀ $(0.5)^{t/h}$ where h is the half-life. This gives 80 $(0.5)^{30/12}$ . Since (0.5) = $2^{-1}$ , we can also write this as 80 $(2^{-1})^{30/12}$ = 80 $(2)^{-30/12}$ . Both A and C are mathematically equivalent. Choice B incorrectly inverts the exponent fraction.
The value of a car depreciates exponentially according to the model $V(t) = 25000 \cdot (0.82)^t$ , where $t$ is the number of years after purchase. After how many complete years will the car's value first drop below $10,000?
Answer: 5 years Set up the inequality: 25000 $(0.82)^t$ < 10000. Dividing by 25000: $(0.82)^t$ < 0.4. Taking natural log: t ln(0.82) < ln(0.4), so t > ln(0.4)/ln(0.82) ≈ 4.67. Since we need complete years, the answer is 5 years. Choice A uses t ≥ 4.67 incorrectly. Choice C rounds up incorrectly. Choice D miscalculates the logarithm.
The number of active users on a social media platform grows according to U(m) = 50,000 × $1.4^m$ , where m is months since launch. Based on this model, by what factor will the user base multiply every 6 months?
Answer: 7.5 To find the multiplication factor every 6 months, calculate U(m+6)/U(m) = [50,000 × $1.4^{m+6}$ ]/[50,000 × $1.4^m$ ] = $1.4^{m+6-m}$ = $1.4^6$ . Now $1.4^6$ = (1.4²)³ = (1.96)³ = 1.96 × 1.96 × 1.96 ≈ 7.53. So the user base multiplies by approximately 7.5 every 6 months. Choice B would correspond to a higher growth rate. Choice C would require $1.4^6$ ≈ 9.1, which is incorrect. Choice D is too high for this growth rate.
A bacterial culture initially contains 500 bacteria. The population doubles every 3 hours. If the exponential model $P(t) = 500 \cdot 2^{t/3}$ represents the population after $t$ hours, approximately how many bacteria will be present after 8 hours?
Answer: 3,175 bacteria Using the model P(t) = 500 · $2^{t/3}$ , substitute t = 8: P(8) = 500 · $2^{8/3}$ = 500 · $2^{2.67}$ ≈ 500 · 6.35 ≈ 3,175. Choice B assumes linear growth. Choice C doubles the correct answer (calculation error). Choice D assumes the population doubles every hour instead of every 3 hours.
The temperature of a cooling object follows Newton's Law of Cooling: $T(t) = 75 + 125e^{-0.1t}$ , where $T$ is temperature in °F and $t$ is time in minutes. What does the value 75 represent in this model?
Answer: The ambient room temperature that the object approaches In the model T(t) = 75 + $125e^{-0.1t}$ , as t approaches infinity, $e^{-0.1t}$ approaches 0, so T(t) approaches 75. This represents the ambient temperature. The initial temperature is T(0) = 75 + 125 = 200°F. Choice A confuses the asymptotic value with initial value. Choice C mistakes a constant for a rate. Choice D misinterprets the horizontal asymptote.
A radioactive substance has a half-life of 8 days. If a sample initially contains 240 grams, how much will remain after 20 days?
Answer: 42.4 grams The exponential decay model is A(t) = 240 × $(1/2)^{t/8}$ . After 20 days: A(20) = 240 × $(1/2)^{20/8}$ = 240 × $(1/2)^{2.5}$ = 240 × $(1/2)^2$ × $(1/2)^{0.5}$ = 240 × 1/4 × 1/√2 = 60/√2 ≈ 42.4 grams. Choice A incorrectly uses linear decay. Choice C uses the wrong time calculation. Choice D forgets the fractional half-life portion.
Based on this depreciation model, what will be the approximate value of the car after 5 years?
Answer: $32,768 First find the decay rate: 64,000 = 80,000 × r². Solving: r² = 0.8, so r = √0.8 ≈ 0.8944. The model is V(t) = 80,000 × $(0.8944)^t$ . After 5 years: V(5) = 80,000 × (0.8944)⁵ = 80,000 × 0.4096 = $32,768. Choice A assumes linear depreciation. Choice B uses an incorrect base calculation. Choice D uses the wrong time period in the exponential calculation.
A scientist observes that a chemical reaction rate follows the model $R(T) = 2.5e^{0.08T}$ , where $R$ is the reaction rate and $T$ is temperature in Celsius. If the temperature increases from 25°C to 45°C, by what factor does the reaction rate increase?
Answer: Approximately 4.9 R(25) = $2.5e^{0.08×25}$ = $2.5e^2$ ≈ 2.5(7.39) ≈ 18.48. R(45) = $2.5e^{0.08×45}$ = $2.5e^{3.6}$ ≈ 2.5(36.6) ≈ 91.5. Factor increase = 91.5/18.48 ≈ 4.95 ≈ 4.9. Choice A uses incorrect exponential calculation. Choice B calculates difference instead of ratio. Choice D uses wrong temperature values in calculation.
What is the half-life of this medication, and what will be the concentration after 3 hours?
Answer: Half-life: 1 hour; Concentration: 1.875 mg/L For half-life, set C(t) = 7.5 (half of initial 15 mg/L): 7.5 = $15e^{-0.693t}$ , so 0.5 = $e^{-0.693t}$ . Taking ln: ln(0.5) = -0.693t, so t = ln(0.5)/(-0.693) = 0.693/0.693 = 1 hour. For concentration after 3 hours: C(3) = $15e^{-0.693×3}$ = $15e^{-2.079}$ = 15 × 0.125 = 1.875 mg/L. Choice B has correct half-life but wrong concentration calculation. Choices C and D incorrectly calculate half-life as ln(2)/0.693 instead of recognizing that 0.693 ≈ ln(2).
If $5,000 grows to $6,500 in 4 years, what would be the value of a $12,000 investment at the same interest rate after 6 years?
Answer: $19,344 First find the interest rate: 6,500 = 5,000(1 + r)⁴, so (1 + r)⁴ = 1.3, giving 1 + r = $(1.3)^{1/4}$ ≈ 1.0678, so r ≈ 0.0678. For the $12,000 investment after 6 years: A = 12,000(1.0678)⁶ ≈ 12,000 × 1.612 ≈ $19,344. Choice A uses simple interest calculation. Choice C miscalculates the compound factor. Choice D uses the wrong time period in the rate calculation.
A coffee shop's daily profit follows an exponential decline after a competitor opens nearby. The profit P (in dollars) after t weeks is P(t) = 2400 × $0.85^t$ . In how many weeks will the profit drop to approximately $1,200?
Answer: 4.3 weeks When you encounter exponential decay problems, you're solving for when a decreasing quantity reaches a specific value. Here, you need to find when the coffee shop's profit drops from its initial $2,400 to $1,200. Set up the equation by substituting the target profit into the given formula: $1200 = 2400 \times 0.85^t$ First, isolate the exponential term by dividing both sides by 2,400: $\frac{1200}{2400} = 0.85^t$ , which simplifies to $0.5 = 0.85^t$ To solve for t, take the natural logarithm of both sides: $\ln(0.5) = \ln(0.85^t) = t \times \ln(0.85)$ Therefore: $t = \frac{\ln(0.5)}{\ln(0.85)} = \frac{-0.693}{-0.163} \approx 4.25$ Rounding to one decimal place gives approximately 4.3 weeks, making D correct. Looking at the wrong answers: A (4.1 weeks) likely comes from calculation errors or using the wrong logarithm base. B (4.9 weeks) might result from incorrectly setting up the fraction as $\frac{\ln(0.85)}{\ln(0.5)}$ instead of the correct order. C (4.6 weeks) could stem from rounding errors during intermediate steps or using approximations too early in the calculation. For exponential decay problems, always set up your equation carefully, isolate the exponential term first, then use logarithms to solve for the variable in the exponent. Double-check that your logarithm fraction has the target ratio in the numerator and the base's logarithm in the denominator.

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