Exponential & Logarithmic Models - IB Mathematics Applications and Interpretation

1. A cup of coffee with an initial temperature of 95°C is left to cool in a room with a constant temperature of 20°C. The temperature, $T$, of the coffee after $t$ minutes is modelled by $T(t) = 20 + 75e^{-kt}$. After 5 minutes, the temperature is 60°C. Find the time it takes for the coffee to cool to 30°C.
   Answer: 16.0 minutes First, find $k$ using the given information: $60 = 20 + 75e^{-5k}$. This gives $40 = 75e^{-5k}$, so $e^{-5k} = 40/75 = 8/15$. Taking natural logs, $-5k = \ln(8/15)$, so $k = -\frac{\ln(8/15)}{5} \approx 0.1256$. Next, find $t$ when $T(t)=30$: $30 = 20 + 75e^{-0.1256t}$. This gives $10 = 75e^{-0.1256t}$, so $e^{-0.1256t} = 10/75 = 2/15$. Taking natural logs, $-0.1256t = \ln(2/15)$, so $t = \frac{\ln(2/15)}{-0.1256} \approx 16.04$ minutes. The closest answer is 16.0 minutes.
2. The loudness of a sound, $L$, in decibels (dB), is given by $L = 10 \log_{10}(\frac{I}{I_0})$, where $I$ is the intensity of the sound and $I_0 = 10^{-12} \text{ W/m}^2$. A rock concert has a loudness of 115 dB. Find the intensity, $I$, of the sound at the concert.
   Answer: $0.316 \text{ W/m}^2$ Substitute $L=115$ into the formula: $115 = 10 \log_{10}(\frac{I}{10^{-12}})$. First, divide by 10: $11.5 = \log_{10}(\frac{I}{10^{-12}})$. By the definition of logarithms, this means $10^{11.5} = \frac{I}{10^{-12}}$. To solve for $I$, multiply both sides by $10^{-12}$: $I = 10^{11.5} \times 10^{-12} = 10^{11.5 - 12} = 10^{-0.5}$. Using a calculator, $I = 10^{-0.5} \approx 0.316$ W/m$^2$.
3. The concentration of a drug in a patient's bloodstream, $C$, in mg/L, $t$ hours after administration is modelled by $C(t) = 15e^{-0.2t}$. The drug is effective as long as the concentration is above 2 mg/L. For how many hours is the drug effective? Give your answer to the nearest hour.
   Answer: 10 hours We need to find the time $t$ when the concentration $C(t)$ drops to 2 mg/L. We solve the equation $2 = 15e^{-0.2t}$. Divide by 15: $\frac{2}{15} = e^{-0.2t}$. Take the natural logarithm of both sides: $\ln(\frac{2}{15}) = -0.2t$. Solve for $t$: $t = \frac{\ln(2/15)}{-0.2} \approx \frac{-2.0149}{-0.2} \approx 10.07$ hours. The drug is effective from $t=0$ until this time. The duration is 10.07 hours. To the nearest hour, this is 10 hours.
4. The magnitude of an earthquake, $M$, on the Richter scale is given by $M = \frac{2}{3}\log_{10}(\frac{E}{E_0})$, where $E$ is the energy released and $E_0$ is a constant. An earthquake with magnitude 6.0 is followed by an aftershock with magnitude 4.0. The energy released by the first earthquake is how many times the energy released by the aftershock?
   Answer: 1000 Let $E_1$ be the energy of the first earthquake and $E_2$ be the energy of the aftershock. We have two equations: $6 = \frac{2}{3}\log_{10}(\frac{E_1}{E_0})$ and $4 = \frac{2}{3}\log_{10}(\frac{E_2}{E_0})$. From the first equation, $9 = \log_{10}(\frac{E_1}{E_0})$, so $E_1 = E_0 \times 10^9$. From the second equation, $6 = \log_{10}(\frac{E_2}{E_0})$, so $E_2 = E_0 \times 10^6$. The ratio of the energies is $\frac{E_1}{E_2} = \frac{E_0 \times 10^9}{E_0 \times 10^6} = 10^{9-6} = 10^3 = 1000$. The first earthquake released 1000 times more energy.
5. The population of a town, $P$, is modelled by the function $P(t) = 5000 \times (1.02)^{2t}$, where $t$ is the number of years since 2020. Which of the following statements correctly describes the annual percentage growth rate of the population?
   Answer: The annual percentage growth rate is 4.04%. The model is $P(t) = 5000 \times (1.02)^{2t}$. Using exponent rules, this can be rewritten as $P(t) = 5000 \times ((1.02)^2)^t$. Calculating $(1.02)^2$ gives 1.0404. So the model is $P(t) = 5000 \times (1.0404)^t$. This is in the form $P(t) = P_0(1+r)^t$, where $1+r$ is the annual growth factor. Here, the annual growth factor is 1.0404. The annual growth rate $r$ is $1.0404 - 1 = 0.0404$. As a percentage, this is 4.04%.
6. The temperature $T$ in degrees Celsius inside a building $t$ hours after a power failure is modelled by $T(t) = 5 + 17e^{-0.08t}$. How long does it take for the temperature to drop by 10°C from its initial temperature at $t=0$?
   Answer: 11.1 hours This question tests exponential decay models, which are common in real-world applications like temperature changes, radioactive decay, and population decline. When you see a model like $T(t) = 5 + 17e^{-0.08t}$, recognize that the constant term (5) represents the limiting value, while the exponential term (17) represents the initial difference from that limit. First, find the initial temperature at $t = 0$: $T(0) = 5 + 17e^{0} = 5 + 17(1) = 22°C$. After dropping 10°C, the target temperature is $22 - 10 = 12°C$. Now solve for when $T(t) = 12$: $12 = 5 + 17e^{-0.08t}$ $7 = 17e^{-0.08t}$ $\frac{7}{17} = e^{-0.08t}$ Taking the natural logarithm of both sides: $\ln\left(\frac{7}{17}\right) = -0.08t$ $t = \frac{\ln(7/17)}{-0.08} = \frac{-0.916}{-0.08} = 11.1 \text{ hours}$ Answer A (8.7 hours) likely comes from calculation errors in the logarithm step. Answer B (9.9 hours) might result from using the wrong base for the logarithm or arithmetic mistakes. Answer C (12.5 hours) could come from setting up the equation incorrectly or misunderstanding what "drop by 10°C" means. Remember: in exponential decay problems, always identify your initial value first, then your target value, before setting up the equation. Double-check that you're using natural logarithms when solving equations with base $e$.
7. A person invests $5000 in an account that pays 6% annual interest, compounded monthly. How many full years will it take for the investment to be worth at least $8000?
   Answer: 8 years The formula for compound interest is $FV = PV(1 + \frac{r}{k})^{kt}$. Here, $PV=5000$, $FV=8000$, $r=0.06$, and $k=12$. We need to solve for $t$: $8000 = 5000(1 + \frac{0.06}{12})^{12t}$. $1.6 = (1.005)^{12t}$. Take the natural logarithm of both sides: $\ln(1.6) = 12t \ln(1.005)$. $t = \frac{\ln(1.6)}{12 \ln(1.005)} \approx 7.85$ years. Since the question asks for the number of full years for the value to be at least $8000, we must round up to the next integer. After 7 years, the amount would be less than $8000. So, it will take 8 full years.
8. A biological culture contains 3000 bacteria. The population is observed to triple every 4 hours. The population $P$ after $t$ hours is modelled by $P(t) = 3000 \times k^t$. What is the value of $k$, correct to three significant figures?
   Answer: 1.32 The population triples every 4 hours. This means at $t=4$, the population $P(4)$ will be $3 \times 3000 = 9000$. We can substitute this into the model $P(t) = 3000 \times k^t$: $9000 = 3000 \times k^4$. Divide by 3000 to get $3 = k^4$. To find $k$, take the fourth root of both sides: $k = \sqrt[4]{3} = 3^{1/4}$. Calculating this value gives $k \approx 1.31607...$. Correct to three significant figures, $k = 1.32$.
9. A company finds that its profit $P$ in thousands of dollars is modelled by $P(t) = 50 - 45e^{-0.1t}$, where $t$ is the number of months since its launch. What is the company's profit in the long run, according to this model?
   Answer: $50,000 When you encounter an exponential model asking about "long-run" behavior, you need to find the limit as time approaches infinity. This tests your understanding of how exponential functions behave over extended periods. To find the company's long-run profit, evaluate $\lim_{t \to \infty} P(t) = \lim_{t \to \infty} (50 - 45e^{-0.1t})$. As $t$ increases without bound, the term $e^{-0.1t}$ approaches zero because the exponent becomes increasingly negative. Therefore, $45e^{-0.1t} \to 0$, and the profit approaches $50 - 45(0) = 50$ thousand dollars, which equals $50,000. Answer A ($5,000) likely comes from misreading the function or confusing the initial conditions. If you calculated $P(0) = 50 - 45e^0 = 50 - 45 = 5$, that gives the initial profit (a $5,000 loss), not the long-run profit. Answer B ($45,000) represents a common error where students focus on the coefficient 45 in the exponential term, perhaps thinking this represents the limiting value. However, this coefficient gets multiplied by zero in the limit. Answer C (profit increases indefinitely) misunderstands exponential decay. While profit does increase over time, it approaches an asymptote rather than growing without bound. The negative exponent ensures the function levels off. Remember: when analyzing exponential models with negative exponents, the exponential term disappears in the long run, leaving only the constant term. Always check whether the exponent is positive (explosive growth) or negative (approach to equilibrium).
10. A scientist models the number of cells in a culture, $N$, after $t$ days with the function $N(t) = 100e^{0.23t}$. Another scientist proposes a model of the form $N(t) = 100(b)^t$. What is the value of $b$ that makes the two models equivalent?
    Answer: 1.26 For the two models to be equivalent, we must have $100e^{0.23t} = 100(b)^t$. Dividing both sides by 100 gives $e^{0.23t} = b^t$. Using exponent rules, we can rewrite the left side as $(e^{0.23})^t = b^t$. For this equality to hold for all $t$, the bases must be equal. Therefore, $b = e^{0.23}$. Using a calculator, $e^{0.23} \approx 1.2586$. Rounded to three significant figures, the value of $b$ is 1.26.
11. The atmospheric pressure $P$, in kilopascals (kPa), at an altitude $h$, in kilometers, is modelled by $P(h) = 101e^{-h/8}$. At what altitude is the atmospheric pressure equal to 50 kPa?
    Answer: 5.63 km We need to solve the equation $50 = 101e^{-h/8}$ for $h$. First, divide by 101: $\frac{50}{101} = e^{-h/8}$. Take the natural logarithm of both sides: $\ln(\frac{50}{101}) = -\frac{h}{8}$. To find $h$, multiply by -8: $h = -8 \ln(\frac{50}{101})$. Using a calculator, $h \approx -8(-0.7033) \approx 5.63$ km.
12. The number of views, $V$, of a new online video is modelled by $V(t) = 150 \times (1.3)^t$, where $t$ is the number of hours after it was uploaded. To the nearest minute, how long does it take for the number of views to double?
    Answer: 159 minutes We need to find the time $t$ when the number of views is double the initial amount, which is $2 \times 150 = 300$. Set up the equation: $300 = 150 \times (1.3)^t$. Divide by 150: $2 = (1.3)^t$. To solve for $t$, take the natural logarithm of both sides: $\ln(2) = \ln(1.3^t)$, which simplifies to $\ln(2) = t \ln(1.3)$. Therefore, $t = \frac{\ln(2)}{\ln(1.3)} \approx 2.6419$ hours. To convert this to minutes, multiply by 60: $2.6419 \times 60 \approx 158.51$ minutes. To the nearest minute, the time is 159 minutes.
13. The pH of vinegar is 2.2 and the pH of a tomato is 4.7. The pH is defined as $pH = -\log_{10}[H^+]$, where $[H^+]$ is the hydrogen ion concentration. How many times greater is the hydrogen ion concentration in vinegar than in the tomato? Round your answer to the nearest integer.
    Answer: 316 For vinegar: $2.2 = -\log_{10}[H^+]_{vinegar}$, so $[H^+]_{vinegar} = 10^{-2.2}$. For the tomato: $4.7 = -\log_{10}[H^+]_{tomato}$, so $[H^+]_{tomato} = 10^{-4.7}$. To find how many times greater the concentration is, we find the ratio: $\frac{[H^+]_{vinegar}}{[H^+]_{tomato}} = \frac{10^{-2.2}}{10^{-4.7}} = 10^{-2.2 - (-4.7)} = 10^{2.5}$. Calculating this value gives $10^{2.5} \approx 316.227...$. Rounded to the nearest integer, the concentration is 316 times greater.
14. The equation $5^{2x-1} = 2^{x+2}$ is solved. Which of the following expressions gives the correct solution for $x$?
    Answer: $\frac{\ln(5) + 2\ln(2)}{2\ln(5) - \ln(2)}$ Taking the natural logarithm of both sides: $\ln(5^{2x-1}) = \ln(2^{x+2})$. Using logarithm properties: $(2x-1)\ln(5) = (x+2)\ln(2)$. Expanding: $2x\ln(5) - \ln(5) = x\ln(2) + 2\ln(2)$. Collecting terms with $x$: $2x\ln(5) - x\ln(2) = \ln(5) + 2\ln(2)$. Factoring: $x(2\ln(5) - \ln(2)) = \ln(5) + 2\ln(2)$. Therefore: $x = \frac{\ln(5) + 2\ln(2)}{2\ln(5) - \ln(2)}$.
15. The function $f(x) = a \ln(x-b)$ passes through the points (4, 0) and (5, 2). What is the value of $a$?
    Answer: $\frac{2}{\ln(2)}$ Substitute the two points into the function to get a system of two equations: (1) $0 = a \ln(4-b)$ and (2) $2 = a \ln(5-b)$. From equation (1), since $a$ cannot be 0 (otherwise equation 2 is false), we must have $\ln(4-b)=0$. This implies that $4-b = e^0 = 1$, which gives $b=3$. Now substitute $b=3$ into the second equation: $2 = a \ln(5-3) = a \ln(2)$. Solving for $a$ gives $a = \frac{2}{\ln(2)}$.
16. Anja invests $10,000 in Fund A which earns 4% annual interest compounded continuously. Boris invests $8,000 in Fund B which earns 6% annual interest compounded continuously. After how many years will the value of Boris's investment first equal the value of Anja's investment?
    Answer: 11.2 years The formula for continuous compounding is $A = Pe^{rt}$. Anja's investment is modelled by $A(t) = 10000e^{0.04t}$. Boris's investment is modelled by $B(t) = 8000e^{0.06t}$. We set them equal to find when the values are the same: $8000e^{0.06t} = 10000e^{0.04t}$. Divide both sides by 8000 and by $e^{0.04t}$: $\frac{e^{0.06t}}{e^{0.04t}} = \frac{10000}{8000}$. Using exponent rules, this becomes $e^{0.06t - 0.04t} = 1.25$, so $e^{0.02t} = 1.25$. Take the natural logarithm of both sides: $0.02t = \ln(1.25)$. Solve for $t$: $t = \frac{\ln(1.25)}{0.02} \approx 11.157$ years. The closest answer is 11.2 years.
17. The population of a species of frog is decreasing exponentially. In 2010, the population was 1200. In 2018, the population was 750. Assuming the population follows a model of the form $P(t) = P_0 e^{-kt}$, where $t$ is the number of years since 2000, what was the estimated population in 2000?
    Answer: 2164 Let $t=10$ for 2010 and $t=18$ for 2018. We have two equations: (1) $1200 = P_0 e^{-10k}$ and (2) $750 = P_0 e^{-18k}$. Divide (2) by (1): $\frac{750}{1200} = \frac{P_0 e^{-18k}}{P_0 e^{-10k}}$ which simplifies to $0.625 = e^{-8k}$. Taking the natural log, $\ln(0.625) = -8k$, so $k = -\frac{\ln(0.625)}{8} \approx 0.05875$. Substitute this value of $k$ back into equation (1) to find $P_0$: $1200 = P_0 e^{-10(0.05875)}$. $P_0 = \frac{1200}{e^{-0.5875}} = 1200e^{0.5875} \approx 2163.5$. The estimated population in 2000 (at $t=0$) was about 2164.
18. The value of a car, $V$ in dollars, depreciates according to the model $V(t) = 25000(0.85)^t$, where $t$ is the number of years since it was purchased. What is the value of $\log_{10}(V)$ after 5 years, to two decimal places?
    Answer: 4.05 First, calculate the value of the car after 5 years: $V(5) = 25000(0.85)^5 \approx 11092.63$. Next, take the base-10 logarithm of this value: $\log_{10}(11092.63) \approx 4.0450...$. Rounded to two decimal places, the value is 4.05. Alternatively, use log properties: $\log_{10}(V(5)) = \log_{10}(25000(0.85)^5) = \log_{10}(25000) + 5\log_{10}(0.85)$. This gives $4.3979 + 5(-0.0706) \approx 4.3979 - 0.353 = 4.0449$, which rounds to 4.05.
19. An ecologist is studying the relationship between the body mass ($M$, in kg) and the brain mass ($B$, in g) of different mammal species. When the ecologist plots $\log_{10}(B)$ against $\log_{10}(M)$, the data points lie close to a straight line with the equation $y = 0.75x + 0.9$. Which of the following functions correctly models the relationship between $B$ and $M$?
    Answer: $B = 7.94 M^{0.75}$ The linear relationship is between the logarithms of the variables: $\log_{10}(B) = 0.75 \log_{10}(M) + 0.9$. Using logarithm properties, we can write $\log_{10}(B) = \log_{10}(M^{0.75}) + 0.9$. To solve for $B$, we exponentiate both sides with base 10: $10^{\log_{10}(B)} = 10^{\log_{10}(M^{0.75}) + 0.9}$. This simplifies to $B = 10^{\log_{10}(M^{0.75})} \times 10^{0.9}$, which is $B = M^{0.75} \times 10^{0.9}$. Since $10^{0.9} \approx 7.94$, the model is $B = 7.94 M^{0.75}$.
20. A sample of a radioactive isotope initially has a mass of 200 mg. After 10 days, its mass is 150 mg. The decay is modelled by the function $M(t) = 200e^{-kt}$, where $t$ is the time in days. What is the half-life of this isotope, to the nearest tenth of a day?
    Answer: 24.1 days First, find the decay constant $k$. We are given that $M(10) = 150$. So, $150 = 200e^{-10k}$. $0.75 = e^{-10k}$. Taking the natural logarithm of both sides gives $\ln(0.75) = -10k$, so $k = -\frac{\ln(0.75)}{10} \approx 0.028768$. The half-life $T$ is the time it takes for the mass to reduce to half its initial value, which is 100 mg. We solve $100 = 200e^{-kT}$, which simplifies to $0.5 = e^{-kT}$. Taking the natural logarithm gives $\ln(0.5) = -kT$. Therefore, $T = \frac{\ln(0.5)}{-k} = \frac{\ln(0.5)}{-0.028768} \approx 24.09$ days. To the nearest tenth, this is 24.1 days.

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