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Exponentials & Logarithms · Topic 6 of 7

Exponential Growth & Decay

Video lesson · from 47:212 worked examples

One lesson video covers all of Exponentials & Logarithms, so it opens at 47:21 for this topic — not from the beginning.

Theory

We previously learned that exponential functions are sometimes known as growth or decay functions. These often occur in models of real-life situations.

For example, radioactive decay can be modelled using an exponential function. An important measurement is the half-life of radioactive substance, which is the time taken for the mass of the radioactive substance to halve.

⚠️ Common Examiner Traps

  • Decay needs a negative exponent: or a base below 1. Getting the sign wrong turns decay into growth and the answer becomes nonsense.
  • Take logs to find the time: when the unknown is in the exponent, logarithms are the only route.
  • Half-life means half the original: set the expression equal to 0.5A00.5A_0 — you do not need to know A0A_0 itself, since it cancels.
  • Round sensibly and in context: a time usually needs rounding up to be sure the condition is met. Say what your answer means.

Worked examples

Example 1

The mass GG grams of a radioactive sample after time tt years is given by the formula G=100e−3tG = 100e^{-3t}.

  • What is the initial mass of radioactive substance in the sample?
  • Find the half-life of the radioactive substance.

Initial Mass (t=0t=0):

G=100e−3(0)=100e0=100×1=100 grams\begin{aligned} G &= 100e^{-3(0)} \\ &= 100e^0 \\ &= 100 \times 1 = 100 \text{ grams} \end{aligned}

Half-life: We want to find tt when G=50G = 50.

50=100e−3t50100=e−3t0.5=e−3tln⁡(0.5)=−3tt=ln⁡(0.5)−3≈0.231 years\begin{aligned} 50 &= 100e^{-3t} \\ \frac{50}{100} &= e^{-3t} \\ 0.5 &= e^{-3t} \\ \ln(0.5) &= -3t \\ t &= \frac{\ln(0.5)}{-3} \approx 0.231 \text{ years} \end{aligned}

Example 2

The world population, in billions, tt years after 1950 is given by P=2.54e0.0178tP = 2.54e^{0.0178t}.

  • What was the world population in 1950?
  • Find, to the nearest year, the time taken for the world population to double.

1950 (t=0t=0):

P=2.54e0=2.54 billionP = 2.54e^0 = 2.54 \text{ billion}

Time to double: We want to find tt when P=2.54×2=5.08P = 2.54 \times 2 = 5.08.

5.08=2.54e0.0178t2=e0.0178tln⁡2=0.0178tt=ln⁡20.0178≈38.94\begin{aligned} 5.08 &= 2.54e^{0.0178t} \\ 2 &= e^{0.0178t} \\ \ln 2 &= 0.0178t \\ t &= \frac{\ln 2}{0.0178} \approx 38.94 \end{aligned}

To the nearest year, this is 39 years.