Exponential Functions and the Number e

In short: An exponential function has the variable in the power, like $y = 2^{x}$ , so it grows or decays by the same factor each step. Its graph passes through $(0, 1)$ with the asymptote $y = 0$ , and the special base $e \approx 2.718$ has its own gradient: $\frac{d}{d x} e^{x} = e^{x}$ .

Starting point: Years 9 and 10

Index laws reminder

$a^{m} a^{n} = a^{m + n} \frac{a ^{m}}{a ^{n}} = a^{m - n} (a^{m})^{n} = a^{mn} a^{0} = 1 a^{- n} = \frac{1}{a ^{n}}$

Exponential functions

In an exponential function, the variable is in the power:

$y = a^{x} (a > 0, a \neq = 1)$

$x$	$- 2$	$- 1$	0	1	2	3
$2^{x}$	$\frac{1}{4}$	$\frac{1}{2}$	1	2	4	8

$y = (\frac{1}{2})^{x} = 2^{- x}$ is the mirror image in the y-axis. It shows decay.

Growth and decay

Something that grows by a fixed percentage follows $A = P (1 + r)^{n}$ . Something that shrinks by a fixed percentage follows $A = P (1 - r)^{n}$ .

Solving simple exponential equations

Write both sides with the same base, then match the powers.

$3^{x + 1} = 81 = 3^{4} \Rightarrow x + 1 = 4 \Rightarrow x = 3$

Exponentials

Index laws

Add, subtract, multiply powers

y = aˣ

Through (0, 1)
Asymptote y = 0
a > 1 grows
0 < a < 1 decays

Percent change

P(1 + r)ⁿ
P(1 - r)ⁿ

Number e

About 2.718
Gradient equals height
d/dx eˣ = eˣ

Year 12Exponential models and more derivatives

Continuous growth or decay is modelled by

$N = N_{0} e^{k t}$

$k > 0$ for growth and $k < 0$ for decay. $N_{0}$ is the amount at $t = 0$ .

Its derivative is $\frac{d N}{d t} = k N_{0} e^{k t} = k N$ : the rate of change is proportional to the amount. More bacteria make new bacteria faster.

The chain rule gives $\frac{d}{d x} e^{f (x)} = f^{'} (x) e^{f (x)}$ . For example, $\frac{d}{d x} e^{3 x^{2}} = 6 x e^{3 x^{2}}$ .

Reading the percentage. In $P = A (1.2)^{t}$ the population is multiplied by 1.2 each step, a 20% increase. In $P = A (1.25)^{- t} = A (0.8)^{t}$ it is multiplied by 0.8, a 20% decrease.

Finding $k$ from a doubling time or half-life. If an amount doubles every 5 years, $2 = e^{5 k}$ , so $k = \frac{ln 2}{5} \approx 0.139$ . If it halves every 8 hours, $\frac{1}{2} = e^{8 k}$ , so $k = \frac{ln 0.5}{8} = - \frac{ln 2}{8}$ . Taking logs to solve is explained in the Logarithms chapter. An amount that halves every 8 hours falls to one eighth after three halvings, 24 hours.

Models with a constant. Cooling follows $T = A e^{- k t} + C$ , where $C$ is the room temperature it settles to. For $T = 300 e^{- 0.1 t} + 20$ : it starts at 320 and approaches 20. To find when $T = 80$ , solve $300 e^{- 0.1 t} = 60$ . The rate of cooling is $\frac{d T}{d t} = - 30 e^{- 0.1 t}$ .

Quick summary

In an exponential function the variable is the power: $y = a^{x}$ .
$y = a^{x}$ passes through $(0, 1)$ with asymptote $y = 0$ . $a > 1$ grows, $0 < a < 1$ decays.
Percentage growth: $A = P (1 + r)^{n}$ . Percentage decay: $A = P (1 - r)^{n}$ .
Solve $a^{f (x)} = a^{k}$ by matching powers.
$e \approx 2.718$ and $\frac{d}{d x} e^{x} = e^{x}$ .

Quiz

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What is $2^{5}$ ?
1. A 10
2. B 32
3. C 25
4. D 64
Show answer
B: 32
$2 \times 2 \times 2 \times 2 \times 2 = 32$ .
Simplify $\frac{5 ^{7} \times 5 ^{2}}{5 ^{4}}$ .
1. A $5^{14}$
2. B $5^{5}$
3. C $5^{13}$
4. D $5^{3}$
Show answer
B: $5^{5}$
$7 + 2 - 4 = 5$ .
Every graph $y = a^{x}$ with $a > 0$ passes through:
1. A $(1, 0)$
2. B $(0, 1)$
3. C $(0, 0)$
4. D $(1, 1)$
Show answer
B: $(0, 1)$
$a^{0} = 1$ .