Exponential models, half-life and logarithmic models Problem set A Worked solutions

The mass of a sample is \(M=M_0e^{-0.035t}\), with t in hours. Find the half-life.

• Half-life = −ln 2 ÷ k
• = −ln 2 ÷ (−0.035) = 19.8 hours

Starting from \(y=y_0e^{kt}\), where k < 0, show that the half-life is \(-\dfrac{\ln2}{k}\).

• Set y = \(\tfrac12y_0\): \(\tfrac12y_0=y_0e^{kt}\)
• Divide by \(y_0\): \(\tfrac12=e^{kt}\)
• Take ln: \(kt=\ln\tfrac12=-\ln2\), so \(t=-\ln2/k\)

A liquid is at 90°C in a room at 30°C. The difference from the room temperature halves every 8 minutes. Write the temperature as \(T=d+Ae^{kt}\), with t in minutes.

• d = 30 and A = 90 − 30 = 60
• k = −ln 2 ÷ 8 = -0.0866
• \(T=30+60e^{-0.0866t}\)

A model \(y=a+b\ln x\) passes through (2, 3) and (20, 9). Find a and b.

• \(3=a+b\ln2\) and \(9=a+b\ln20\)
• Solve the pair on the calculator: a = 1.19, b = 2.61
• Check: \(1.19+2.61\ln 20=9\)

A model is \(y=2.4+3.1\ln x\).

1. (a) Find y when x = 12.

   • \(y=2.4+3.1\ln 12=10.1\)

2. (b) Find x when y = 15.

   • \(2.4+3.1\ln x=15\)
   • Solve for x: x = 58.2

A radioactive sample is placed in a detector. The count rate is 200 counts per minute at t = 0, where t is in minutes. The room has a background count of 40 counts per minute. The count above the background halves every 30 minutes.

1. (a) Find a model \(R=d+Ae^{kt}\) for the count rate.

   • d = 40, A = 200 − 40 = 160, k = −ln 2 ÷ 30 = -0.0231
   • \(R=40+160e^{-0.0231t}\)

2. (b) Find when the count rate falls to 55 counts per minute.

   • \(40+160e^{-0.0231t}=55\)
   • Solve for t: t = 102 minutes

A drug has concentration 12 mg/l in the blood at t = 0, where t is in hours. The concentration halves every 6 hours.

1. (a) Write a model \(C=Ae^{kt}\) for the concentration.

   • Background 0, so d = 0 and A = 12
   • k = −ln 2 ÷ 6 = -0.116
   • \(C=12e^{-0.116t}\)

2. (b) Find the concentration after 15 hours.

   • t = 15: \(C=12e^{-0.116\times15}=2.12\)

The height h metres of a young tree at age t years is modelled by \(h=a+b\ln t\). The tree is 1.1 m tall at 2 years and 3.4 m tall at 10 years.

1. (a) Find a and b.

   • \(1.1=a+b\ln 2\) and \(3.4=a+b\ln 10\)
   • Solve the pair: a = 0.109, b = 1.43

2. (b) Find the height at 6 years.

   • \(0.109+1.43\ln 6=2.67\)

3. (c) Find the age at which the tree is 5 m tall.

   • \(0.109+1.43\ln x=5\)
   • Solve for x: x = 30.6

The number of cases of an infection grows by 8% each day. It is modelled by \(N=N_0e^{kt}\), where t is in days.

1. (a) Find k.

   • \(e^k=1.08,\ k=\ln1.08=0.0770\)

2. (b) Find the time for the number of cases to triple.

   • \(e^{0.0770t}=3\)
   • Solve for t: t = 14.3

A wooden beam from an old barn has 45% of the carbon-14 of living wood. The half-life of carbon-14 is 5730 years. The amount is \(M=M_0e^{kt}\), where t is in years.

1. (a) Find k.

   • \(k=-\ln2\div5730=-0.000121\)

2. (b) Find the age of the beam.

   • \(e^{-0.000121t}=0.45\)
   • Solve for t: t = 6600

The graph shows the temperature T°C of a metal bar t minutes after it leaves a furnace. It is modelled by \(T=d+Ae^{kt}\).

1. (a) Write down d and A.

   • The curve levels off at 20, so d = 20. At x = 0 the value is 80, so A = 80 − 20 = 60

2. (b) Find k.

   • \(47=20+60e^{5k}\)
   • \(e^{5k}=27\div60\), so k = ln(0.45) ÷ 5 = -0.160

3. (c) Find T when t = 12.

   • \(20+60e^{-0.160\times12}=28.8\)

A model is \(y=a+b\ln x\).

1. (a) Show that doubling x adds \(b\ln2\) to y, whatever x is.

   • New y: \(a+b\ln(2x)=a+b\ln 2+b\ln x\)
   • Subtract the old y: \(b\ln 2\), which does not depend on x

2. (b) Find the increase in y when x doubles and b = 3.1.

   • \(3.1\ln 2=2.15\)