IB Mathematics: Applications and Interpretation Quiz: Exponential And Logarithmic Models
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Exponential And Logarithmic ModelsQuestion 1 of 20

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, TT, of the coffee after tt minutes is modelled by T(t)=20+75ektT(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.

9.3 minutes
11.0 minutes
12.5 minutes
16.0 minutes
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Exponential And Logarithmic Models

Practice Exponential And Logarithmic Models in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Exponential And Logarithmic Models, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 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, TT, of the coffee after tt minutes is modelled by T(t)=20+75ektT(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.

  1. 9.3 minutes
  2. 11.0 minutes
  3. 12.5 minutes
  4. 16.0 minutes (correct answer)
Explanation: First, find kk using the given information: 60=20+75e5k60 = 20 + 75e^{-5k}. This gives 40=75e5k40 = 75e^{-5k}, so e5k=40/75=8/15e^{-5k} = 40/75 = 8/15. Taking natural logs, 5k=ln(8/15)-5k = \ln(8/15), so k=ln(8/15)50.1256k = -\frac{\ln(8/15)}{5} \approx 0.1256. Next, find tt when T(t)=30T(t)=30: 30=20+75e0.1256t30 = 20 + 75e^{-0.1256t}. This gives 10=75e0.1256t10 = 75e^{-0.1256t}, so e0.1256t=10/75=2/15e^{-0.1256t} = 10/75 = 2/15. Taking natural logs, 0.1256t=ln(2/15)-0.1256t = \ln(2/15), so t=ln(2/15)0.125616.04t = \frac{\ln(2/15)}{-0.1256} \approx 16.04 minutes. The closest answer is 16.0 minutes.

Question 2

The temperature TT in degrees Celsius inside a building tt hours after a power failure is modelled by T(t)=5+17e0.08tT(t) = 5 + 17e^{-0.08t}. How long does it take for the temperature to drop by 10°C from its initial temperature at t=0t=0?

  1. 8.7 hours
  2. 9.9 hours
  3. 12.5 hours
  4. 11.1 hours (correct answer)
Explanation: 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+17e0.08tT(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=0t = 0: T(0)=5+17e0=5+17(1)=22°CT(0) = 5 + 17e^{0} = 5 + 17(1) = 22°C. After dropping 10°C, the target temperature is 2210=12°C22 - 10 = 12°C. Now solve for when T(t)=12T(t) = 12: 12=5+17e0.08t12 = 5 + 17e^{-0.08t} 7=17e0.08t7 = 17e^{-0.08t} 717=e0.08t\frac{7}{17} = e^{-0.08t} Taking the natural logarithm of both sides: ln(717)=0.08t\ln\left(\frac{7}{17}\right) = -0.08t t=ln(7/17)0.08=0.9160.08=11.1 hourst = \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 ee.

Question 3

A scientist models the number of cells in a culture, NN, after tt days with the function N(t)=100e0.23tN(t) = 100e^{0.23t}. Another scientist proposes a model of the form N(t)=100(b)tN(t) = 100(b)^t. What is the value of bb that makes the two models equivalent?

  1. 0.23
  2. 1.23
  3. 1.26 (correct answer)
  4. e
Explanation: For the two models to be equivalent, we must have 100e0.23t=100(b)t100e^{0.23t} = 100(b)^t. Dividing both sides by 100 gives e0.23t=bte^{0.23t} = b^t. Using exponent rules, we can rewrite the left side as (e0.23)t=bt(e^{0.23})^t = b^t. For this equality to hold for all tt, the bases must be equal. Therefore, b=e0.23b = e^{0.23}. Using a calculator, e0.231.2586e^{0.23} \approx 1.2586. Rounded to three significant figures, the value of bb is 1.26.

Question 4

The population of a town, PP, is modelled by the function P(t)=5000×(1.02)2tP(t) = 5000 \times (1.02)^{2t}, where tt is the number of years since 2020. Which of the following statements correctly describes the annual percentage growth rate of the population?

  1. The annual percentage growth rate is 2%.
  2. The annual percentage growth rate is 4%.
  3. The annual percentage growth rate is 4.04%. (correct answer)
  4. The annual percentage growth rate is 102%.
Explanation: The model is P(t)=5000×(1.02)2tP(t) = 5000 \times (1.02)^{2t}. Using exponent rules, this can be rewritten as P(t)=5000×((1.02)2)tP(t) = 5000 \times ((1.02)^2)^t. Calculating (1.02)2(1.02)^2 gives 1.0404. So the model is P(t)=5000×(1.0404)tP(t) = 5000 \times (1.0404)^t. This is in the form P(t)=P0(1+r)tP(t) = P_0(1+r)^t, where 1+r1+r is the annual growth factor. Here, the annual growth factor is 1.0404. The annual growth rate rr is 1.04041=0.04041.0404 - 1 = 0.0404. As a percentage, this is 4.04%.

Question 5

The pH of vinegar is 2.2 and the pH of a tomato is 4.7. The pH is defined as pH=log10[H+]pH = -\log_{10}[H^+], where [H+][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.

  1. 2
  2. 3
  3. 316 (correct answer)
  4. 50119
Explanation: For vinegar: 2.2=log10[H+]vinegar2.2 = -\log_{10}[H^+]_{vinegar}, so [H+]vinegar=102.2[H^+]_{vinegar} = 10^{-2.2}. For the tomato: 4.7=log10[H+]tomato4.7 = -\log_{10}[H^+]_{tomato}, so [H+]tomato=104.7[H^+]_{tomato} = 10^{-4.7}. To find how many times greater the concentration is, we find the ratio: [H+]vinegar[H+]tomato=102.2104.7=102.2(4.7)=102.5\frac{[H^+]_{vinegar}}{[H^+]_{tomato}} = \frac{10^{-2.2}}{10^{-4.7}} = 10^{-2.2 - (-4.7)} = 10^{2.5}. Calculating this value gives 102.5316.227...10^{2.5} \approx 316.227.... Rounded to the nearest integer, the concentration is 316 times greater.

Question 6

The value of a car, VV in dollars, depreciates according to the model V(t)=25000(0.85)tV(t) = 25000(0.85)^t, where tt is the number of years since it was purchased. What is the value of log10(V)\log_{10}(V) after 5 years, to two decimal places?

  1. 3.87
  2. 4.05 (correct answer)
  3. 4.13
  4. 4.39
Explanation: First, calculate the value of the car after 5 years: V(5)=25000(0.85)511092.63V(5) = 25000(0.85)^5 \approx 11092.63. Next, take the base-10 logarithm of this value: log10(11092.63)4.0450...\log_{10}(11092.63) \approx 4.0450.... Rounded to two decimal places, the value is 4.05. Alternatively, use log properties: log10(V(5))=log10(25000(0.85)5)=log10(25000)+5log10(0.85)\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)4.39790.353=4.04494.3979 + 5(-0.0706) \approx 4.3979 - 0.353 = 4.0449, which rounds to 4.05.

Question 7

A company finds that its profit PP in thousands of dollars is modelled by P(t)=5045e0.1tP(t) = 50 - 45e^{-0.1t}, where tt is the number of months since its launch. What is the company's profit in the long run, according to this model?

  1. $5,000
  2. $45,000
  3. The profit will increase indefinitely.
  4. $50,000 (correct answer)
Explanation: 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 limtP(t)=limt(5045e0.1t)\lim_{t \to \infty} P(t) = \lim_{t \to \infty} (50 - 45e^{-0.1t}). As tt increases without bound, the term e0.1te^{-0.1t} approaches zero because the exponent becomes increasingly negative. Therefore, 45e0.1t045e^{-0.1t} \to 0, and the profit approaches 5045(0)=5050 - 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)=5045e0=5045=5P(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).

Question 8

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)=P0ektP(t) = P_0 e^{-kt}, where tt is the number of years since 2000, what was the estimated population in 2000?

  1. 1763
  2. 1920
  3. 2164 (correct answer)
  4. 3462
Explanation: Let t=10t=10 for 2010 and t=18t=18 for 2018. We have two equations: (1) 1200=P0e10k1200 = P_0 e^{-10k} and (2) 750=P0e18k750 = P_0 e^{-18k}. Divide (2) by (1): 7501200=P0e18kP0e10k\frac{750}{1200} = \frac{P_0 e^{-18k}}{P_0 e^{-10k}} which simplifies to 0.625=e8k0.625 = e^{-8k}. Taking the natural log, ln(0.625)=8k\ln(0.625) = -8k, so k=ln(0.625)80.05875k = -\frac{\ln(0.625)}{8} \approx 0.05875. Substitute this value of kk back into equation (1) to find P0P_0: 1200=P0e10(0.05875)1200 = P_0 e^{-10(0.05875)}. P0=1200e0.5875=1200e0.58752163.5P_0 = \frac{1200}{e^{-0.5875}} = 1200e^{0.5875} \approx 2163.5. The estimated population in 2000 (at t=0t=0) was about 2164.

Question 9

The concentration of a drug in a patient's bloodstream, CC, in mg/L, tt hours after administration is modelled by C(t)=15e0.2tC(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.

  1. 8 hours
  2. 9 hours
  3. 10 hours (correct answer)
  4. 11 hours
Explanation: We need to find the time tt when the concentration C(t)C(t) drops to 2 mg/L. We solve the equation 2=15e0.2t2 = 15e^{-0.2t}. Divide by 15: 215=e0.2t\frac{2}{15} = e^{-0.2t}. Take the natural logarithm of both sides: ln(215)=0.2t\ln(\frac{2}{15}) = -0.2t. Solve for tt: t=ln(2/15)0.22.01490.210.07t = \frac{\ln(2/15)}{-0.2} \approx \frac{-2.0149}{-0.2} \approx 10.07 hours. The drug is effective from t=0t=0 until this time. The duration is 10.07 hours. To the nearest hour, this is 10 hours.

Question 10

The population of a town, PP, is modelled by the function P(t)=5000×(1.02)2tP(t) = 5000 \times (1.02)^{2t}, where tt is the number of years since 2020. Which of the following statements correctly describes the annual percentage growth rate of the population?

  1. The annual percentage growth rate is 2%.
  2. The annual percentage growth rate is 4%.
  3. The annual percentage growth rate is 4.04%. (correct answer)
  4. The annual percentage growth rate is 102%.
Explanation: The model is P(t)=5000×(1.02)2tP(t) = 5000 \times (1.02)^{2t}. Using exponent rules, this can be rewritten as P(t)=5000×((1.02)2)tP(t) = 5000 \times ((1.02)^2)^t. Calculating (1.02)2(1.02)^2 gives 1.0404. So the model is P(t)=5000×(1.0404)tP(t) = 5000 \times (1.0404)^t. This is in the form P(t)=P0(1+r)tP(t) = P_0(1+r)^t, where 1+r1+r is the annual growth factor. Here, the annual growth factor is 1.0404. The annual growth rate rr is 1.04041=0.04041.0404 - 1 = 0.0404. As a percentage, this is 4.04%.

Question 11

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?

  1. 2.2 years
  2. 11.2 years (correct answer)
  3. 11.8 years
  4. 25.0 years
Explanation: The formula for continuous compounding is A=PertA = Pe^{rt}. Anja's investment is modelled by A(t)=10000e0.04tA(t) = 10000e^{0.04t}. Boris's investment is modelled by B(t)=8000e0.06tB(t) = 8000e^{0.06t}. We set them equal to find when the values are the same: 8000e0.06t=10000e0.04t8000e^{0.06t} = 10000e^{0.04t}. Divide both sides by 8000 and by e0.04te^{0.04t}: e0.06te0.04t=100008000\frac{e^{0.06t}}{e^{0.04t}} = \frac{10000}{8000}. Using exponent rules, this becomes e0.06t0.04t=1.25e^{0.06t - 0.04t} = 1.25, so e0.02t=1.25e^{0.02t} = 1.25. Take the natural logarithm of both sides: 0.02t=ln(1.25)0.02t = \ln(1.25). Solve for tt: t=ln(1.25)0.0211.157t = \frac{\ln(1.25)}{0.02} \approx 11.157 years. The closest answer is 11.2 years.

Question 12

A biological culture contains 3000 bacteria. The population is observed to triple every 4 hours. The population PP after tt hours is modelled by P(t)=3000×ktP(t) = 3000 \times k^t. What is the value of kk, correct to three significant figures?

  1. 1.32 (correct answer)
  2. 1.73
  3. 3.00
  4. 9.00
Explanation: The population triples every 4 hours. This means at t=4t=4, the population P(4)P(4) will be 3×3000=90003 \times 3000 = 9000. We can substitute this into the model P(t)=3000×ktP(t) = 3000 \times k^t: 9000=3000×k49000 = 3000 \times k^4. Divide by 3000 to get 3=k43 = k^4. To find kk, take the fourth root of both sides: k=34=31/4k = \sqrt[4]{3} = 3^{1/4}. Calculating this value gives k1.31607...k \approx 1.31607.... Correct to three significant figures, k=1.32k = 1.32.

Question 13

The magnitude of an earthquake, MM, on the Richter scale is given by M=23log10(EE0)M = \frac{2}{3}\log_{10}(\frac{E}{E_0}), where EE is the energy released and E0E_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?

  1. 32
  2. 100
  3. 1000 (correct answer)
  4. 31623
Explanation: Let E1E_1 be the energy of the first earthquake and E2E_2 be the energy of the aftershock. We have two equations: 6=23log10(E1E0)6 = \frac{2}{3}\log_{10}(\frac{E_1}{E_0}) and 4=23log10(E2E0)4 = \frac{2}{3}\log_{10}(\frac{E_2}{E_0}). From the first equation, 9=log10(E1E0)9 = \log_{10}(\frac{E_1}{E_0}), so E1=E0×109E_1 = E_0 \times 10^9. From the second equation, 6=log10(E2E0)6 = \log_{10}(\frac{E_2}{E_0}), so E2=E0×106E_2 = E_0 \times 10^6. The ratio of the energies is E1E2=E0×109E0×106=1096=103=1000\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.

Question 14

A scientist models the number of cells in a culture, NN, after tt days with the function N(t)=100e0.23tN(t) = 100e^{0.23t}. Another scientist proposes a model of the form N(t)=100(b)tN(t) = 100(b)^t. What is the value of bb that makes the two models equivalent?

  1. 0.23
  2. 1.23
  3. 1.26 (correct answer)
  4. e
Explanation: For the two models to be equivalent, we must have 100e0.23t=100(b)t100e^{0.23t} = 100(b)^t. Dividing both sides by 100 gives e0.23t=bte^{0.23t} = b^t. Using exponent rules, we can rewrite the left side as (e0.23)t=bt(e^{0.23})^t = b^t. For this equality to hold for all tt, the bases must be equal. Therefore, b=e0.23b = e^{0.23}. Using a calculator, e0.231.2586e^{0.23} \approx 1.2586. Rounded to three significant figures, the value of bb is 1.26.

Question 15

The temperature TT in degrees Celsius inside a building tt hours after a power failure is modelled by T(t)=5+17e0.08tT(t) = 5 + 17e^{-0.08t}. How long does it take for the temperature to drop by 10°C from its initial temperature at t=0t=0?

  1. 8.7 hours
  2. 9.9 hours
  3. 12.5 hours
  4. 11.1 hours (correct answer)
Explanation: 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+17e0.08tT(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=0t = 0: T(0)=5+17e0=5+17(1)=22°CT(0) = 5 + 17e^{0} = 5 + 17(1) = 22°C. After dropping 10°C, the target temperature is 2210=12°C22 - 10 = 12°C. Now solve for when T(t)=12T(t) = 12: 12=5+17e0.08t12 = 5 + 17e^{-0.08t} 7=17e0.08t7 = 17e^{-0.08t} 717=e0.08t\frac{7}{17} = e^{-0.08t} Taking the natural logarithm of both sides: ln(717)=0.08t\ln\left(\frac{7}{17}\right) = -0.08t t=ln(7/17)0.08=0.9160.08=11.1 hourst = \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 ee.

Question 16

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)=200ektM(t) = 200e^{-kt}, where tt is the time in days. What is the half-life of this isotope, to the nearest tenth of a day?

  1. 19.4 days
  2. 20.0 days
  3. 24.1 days (correct answer)
  4. 31.1 days
Explanation: First, find the decay constant kk. We are given that M(10)=150M(10) = 150. So, 150=200e10k150 = 200e^{-10k}. 0.75=e10k0.75 = e^{-10k}. Taking the natural logarithm of both sides gives ln(0.75)=10k\ln(0.75) = -10k, so k=ln(0.75)100.028768k = -\frac{\ln(0.75)}{10} \approx 0.028768. The half-life TT is the time it takes for the mass to reduce to half its initial value, which is 100 mg. We solve 100=200ekT100 = 200e^{-kT}, which simplifies to 0.5=ekT0.5 = e^{-kT}. Taking the natural logarithm gives ln(0.5)=kT\ln(0.5) = -kT. Therefore, T=ln(0.5)k=ln(0.5)0.02876824.09T = \frac{\ln(0.5)}{-k} = \frac{\ln(0.5)}{-0.028768} \approx 24.09 days. To the nearest tenth, this is 24.1 days.

Question 17

An ecologist is studying the relationship between the body mass (MM, in kg) and the brain mass (BB, in g) of different mammal species. When the ecologist plots log10(B)\log_{10}(B) against log10(M)\log_{10}(M), the data points lie close to a straight line with the equation y=0.75x+0.9y = 0.75x + 0.9. Which of the following functions correctly models the relationship between BB and MM?

  1. B=0.75M+7.94B = 0.75M + 7.94
  2. B=0.9M0.75B = 0.9 M^{0.75}
  3. B=7.94×(5.62)MB = 7.94 \times (5.62)^M
  4. B=7.94M0.75B = 7.94 M^{0.75} (correct answer)
Explanation: The linear relationship is between the logarithms of the variables: log10(B)=0.75log10(M)+0.9\log_{10}(B) = 0.75 \log_{10}(M) + 0.9. Using logarithm properties, we can write log10(B)=log10(M0.75)+0.9\log_{10}(B) = \log_{10}(M^{0.75}) + 0.9. To solve for BB, we exponentiate both sides with base 10: 10log10(B)=10log10(M0.75)+0.910^{\log_{10}(B)} = 10^{\log_{10}(M^{0.75}) + 0.9}. This simplifies to B=10log10(M0.75)×100.9B = 10^{\log_{10}(M^{0.75})} \times 10^{0.9}, which is B=M0.75×100.9B = M^{0.75} \times 10^{0.9}. Since 100.97.9410^{0.9} \approx 7.94, the model is B=7.94M0.75B = 7.94 M^{0.75}.

Question 18

The number of views, VV, of a new online video is modelled by V(t)=150×(1.3)tV(t) = 150 \times (1.3)^t, where tt 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?

  1. 23 minutes
  2. 159 minutes (correct answer)
  3. 180 minutes
  4. 200 minutes
Explanation: We need to find the time tt when the number of views is double the initial amount, which is 2×150=3002 \times 150 = 300. Set up the equation: 300=150×(1.3)t300 = 150 \times (1.3)^t. Divide by 150: 2=(1.3)t2 = (1.3)^t. To solve for tt, take the natural logarithm of both sides: ln(2)=ln(1.3t)\ln(2) = \ln(1.3^t), which simplifies to ln(2)=tln(1.3)\ln(2) = t \ln(1.3). Therefore, t=ln(2)ln(1.3)2.6419t = \frac{\ln(2)}{\ln(1.3)} \approx 2.6419 hours. To convert this to minutes, multiply by 60: 2.6419×60158.512.6419 \times 60 \approx 158.51 minutes. To the nearest minute, the time is 159 minutes.

Question 19

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?

  1. 2.2 years
  2. 11.2 years (correct answer)
  3. 11.8 years
  4. 25.0 years
Explanation: The formula for continuous compounding is A=PertA = Pe^{rt}. Anja's investment is modelled by A(t)=10000e0.04tA(t) = 10000e^{0.04t}. Boris's investment is modelled by B(t)=8000e0.06tB(t) = 8000e^{0.06t}. We set them equal to find when the values are the same: 8000e0.06t=10000e0.04t8000e^{0.06t} = 10000e^{0.04t}. Divide both sides by 8000 and by e0.04te^{0.04t}: e0.06te0.04t=100008000\frac{e^{0.06t}}{e^{0.04t}} = \frac{10000}{8000}. Using exponent rules, this becomes e0.06t0.04t=1.25e^{0.06t - 0.04t} = 1.25, so e0.02t=1.25e^{0.02t} = 1.25. Take the natural logarithm of both sides: 0.02t=ln(1.25)0.02t = \ln(1.25). Solve for tt: t=ln(1.25)0.0211.157t = \frac{\ln(1.25)}{0.02} \approx 11.157 years. The closest answer is 11.2 years.

Question 20

The loudness of a sound, LL, in decibels (dB), is given by L=10log10(II0)L = 10 \log_{10}(\frac{I}{I_0}), where II is the intensity of the sound and I0=1012 W/m2I_0 = 10^{-12} \text{ W/m}^2. A rock concert has a loudness of 115 dB. Find the intensity, II, of the sound at the concert.

  1. 1.15×1011 W/m21.15 \times 10^{-11} \text{ W/m}^2
  2. 9.88×108 W/m29.88 \times 10^{-8} \text{ W/m}^2
  3. 0.316 W/m20.316 \text{ W/m}^2 (correct answer)
  4. 3.16 W/m23.16 \text{ W/m}^2
Explanation: Substitute L=115L=115 into the formula: 115=10log10(I1012)115 = 10 \log_{10}(\frac{I}{10^{-12}}). First, divide by 10: 11.5=log10(I1012)11.5 = \log_{10}(\frac{I}{10^{-12}}). By the definition of logarithms, this means 1011.5=I101210^{11.5} = \frac{I}{10^{-12}}. To solve for II, multiply both sides by 101210^{-12}: I=1011.5×1012=1011.512=100.5I = 10^{11.5} \times 10^{-12} = 10^{11.5 - 12} = 10^{-0.5}. Using a calculator, I=100.50.316I = 10^{-0.5} \approx 0.316 W/m2^2.