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3.13b · Exponential Functions · Sub-skill

e and the exponential function

Meet the number e, see why $e^x$ is special, and use it in models of continuous growth and decay.

See it first

Watch it happen

Scroll to the graphic and it plays by itself — press Replay to watch it again.

Build it up, step by step

e and the exponential function — the key ideas

Click each step below to reveal it — work through them in order the first time round.

Step 1 · The number e

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$e \approx 2.71828$ is an irrational constant, like $\pi$. It appears when growth happens continuously rather than in steps. You can find powers of $e$ with the $e^x$ button on your calculator: e.g. $e^2 \approx 7.389$.

Step 2 · The function $y = e^x$

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The graph of $y = e^x$ passes through $(0, 1)$, is always above the $x$-axis and rises ever more steeply. Its special property: the gradient at any point equals the $y$-value there. So at $x = 1$ the gradient is $e \approx 2.72$, and at $x = 2$ it is $e^2 \approx 7.39$.

Step 3 · Models using e

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Continuous growth or decay is modelled by $N = Ae^{kt}$: $A$ is the starting amount and $k$ is the growth rate (positive for growth, negative for decay). For example $N = 200e^{0.05t}$ starts at 200 and grows continuously at about 5% per unit time.

Step 4 · Using your calculator

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Use the $e^x$ key (not 2.718 typed in by hand). Take care with brackets: for $e^{0.05 \times 10}$ enter the whole exponent in brackets. Give the answer to a sensible number of significant figures for the context.

Worked example

Find $e^2$ and the gradient of $y = e^x$ at $x = 1$.

$e^2 \approx$ 7.389.

The gradient of $y = e^x$ equals the height of the curve, so at $x = 1$ it is $e^1 \approx$ 2.718.

Test yourself

Past-paper style question

A population is modelled by $N = 200e^{0.05t}$, where $t$ is the time in years.

(a) Write down the initial population.
(b) Find the population after 10 years, to the nearest whole number.
(c) What does the 0.05 tell you? [5 marks]

Show the answer

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(a) When $t = 0$, $N = 200e^0 = $ 200. (1 mark)

(b) $N = 200e^{0.5} = 200 \times 1.6487 = 329.7 \approx$ 330. (2 marks)

(c) The population is growing continuously at a rate of 5% per year (approximately). (2 marks)

Practice

e and the exponential function worksheet

Five short questions. Work through them, then reveal the mark scheme to check.

  1. Use your calculator to find $e^3$ to 4 significant figures.
  2. Find $e^{-1}$ to 4 significant figures.
  3. What is the gradient of $y = e^x$ at $x = 2$?
  4. A model is $N = 50e^{0.2t}$. Find $N$ when $t = 5$ to 1 decimal place.
  5. What is the value of $e^0$?

Mark scheme

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  1. 20.09.
  2. 0.3679.
  3. The gradient equals the height: $e^2 \approx$ 7.389.
  4. $50e^{1} = 50 \times 2.71828 = $ 135.9.
  5. 1.
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