Exponential and logarithmic functions

Three situations where a quantity is multiplied again and again. The pictures come first; the maths drawn on them is generated from the same numbers that the solutions use.

$10$$20$$30$$40$$50$$20$$40$$60$$80$$100$$O$$t\text{ (min)}$$T\text{ (}^\circ\text{C)}$$T = 30 + 60\mathrm{e}^{-0.05t}$$(0,\ 90)$$T = 30$

Maths In Real Life 1

The cooling kopi

A hot drink does not cool at a steady rate: it loses heat fastest at the start and more slowly as it nears the temperature of the room. That slowing-down is the signature of an exponential model with a limiting value.

A cup of kopi is served in a food court where the room temperature stays at $30\text{ }^\circ\text{C}$. The temperature $T$ of the kopi, in $^\circ\text{C}$, $t$ minutes after it is served, is modelled by $T = 30 + 60\mathrm{e}^{-0.05t}$.

• (a)Find the temperature of the kopi when it is served.
• (b)Find the temperature after $10$ minutes, correct to $3$ significant figures.
• (c)How long does the kopi take to cool to $50\text{ }^\circ\text{C}$? Give your answer in minutes, correct to $3$ significant figures.
• (d)What value does the temperature approach as $t$ becomes very large? Explain why it never goes below this value.

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Colour keyfinal answer

(a)

$\displaystyle T(0)\;= 30 + 60\mathrm{e}^{0}$

Put $t = 0$: $\mathrm{e}^0 = 1$.

$\displaystyle \phantom{T(0)}\;= \textcolor{#010106}{90}$

(b)

$\displaystyle T(10)\;= 30 + 60\mathrm{e}^{-0.05 \times 10}$

Put $t = 10$ into the model.

$\displaystyle \phantom{T(10)}\;\approx \textcolor{#010106}{66.4}$

Calculator, then round to 3 significant figures.

(c)

$\displaystyle 30 + 60\mathrm{e}^{-0.05t} = 50 \iff \mathrm{e}^{-0.05t} = \dfrac{1}{3}$

Move the constant across, then divide by the number in front.

$\displaystyle \mathrm{e}^{-0.05t} = \dfrac{1}{3} \iff -0.05t = \ln \dfrac{1}{3}$

Take $\ln$ of both sides.

$\displaystyle -0.05t = \ln \dfrac{1}{3} \iff t = \dfrac{\ln \dfrac{1}{3}}{-0.05}$

Divide by the coefficient of $t$, which is negative.

$\displaystyle \approx \textcolor{#010106}{22.0}$

Calculator, then round to 3 significant figures.

(d)

As $t$ becomes very large, $\mathrm{e}^{-0.05t}$ gets closer and closer to $0$ (it is one over a huge power of $\mathrm{e}$), so $60\mathrm{e}^{-0.05t}$ is almost $0$.

Think of $\mathrm{e}^{-y}$ as $1 \div \mathrm{e}^y$.

So $T$ gets closer and closer to $30$ but never reaches it: the limiting value is $30$ and the graph has the horizontal asymptote $T = 30$.

Why it never goes below $30$: $\mathrm{e}^{-0.05t} > 0$ for every $t$, so $60\mathrm{e}^{-0.05t} > 0$ and $T > 30$ always.

A power of $\mathrm{e}$ is never zero or negative.

Answer

(a) $T(0) = 90$ °C

(b) $T(10) \approx 66.4$ °C

(c) $t \approx 22.0$ minutes

(d) $T$ approaches $30$ °C (the line $T = 30$ is an asymptote); since $\mathrm{e}^{-0.05t} > 0$, $T > 30$ always

$20$$40$$60$$80$$100$$5$$10$$15$$20$$O$$r$$L\text{ (dB)}$$L = 10\lg r$$(100,\ 20)$

Maths In Real Life 2

How loud is the hawker centre?

Sound is measured on a scale where adding the same number of decibels means multiplying the intensity. A logarithm is exactly the tool that turns multiplying into adding, which is why the decibel scale uses one.

The sound level $L$, in decibels (dB), of a sound is $L = 10\lg r$, where $r$ is the intensity of the sound as a multiple of the quietest sound a person can hear.

• (a)Show that a sound $100$ times as intense as the quietest sound has a level of $20$ dB.
• (b)When the intensity of a sound is multiplied by $2$, by how many decibels does its level rise? Give your answer correct to $3$ significant figures.
• (c)A food court measures $75$ dB. Find the intensity $r$, giving your answer in the form $a \times 10^n$ with $a$ correct to $3$ significant figures.

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Colour keyfinal answer

(a)

$\displaystyle L(100)\;= 10\lg 100 = 10 \times 2 = \textcolor{#010106}{20}$

Put $r = 100$: $\lg 100$ is the power of ten that gives 100.

(b)

Let $d$ be the rise in $L$ when $r$ becomes $2r$: $d = L(2r) - L(r)$.

$\displaystyle d\;= 10\lg (2r) - 10\lg r = 10(\lg 2 + \lg r) - 10\lg r = 10\lg 2$

Product law: $\lg (kr) = \lg k + \lg r$. The $\lg r$ terms cancel.

$\displaystyle \phantom{d}\;\approx \textcolor{#010106}{3.01}$

Calculator, then round to 3 significant figures.

(c)

$\displaystyle 10\lg r = 75 \iff \lg r = 7.5$

Divide both sides by $10$.

$\displaystyle \lg r = 7.5 \iff r = 10^{7.5}$

Index form: base ten to the power of the log.

$\displaystyle \approx \textcolor{#010106}{3.16 \times 10^{7}}$

Calculator, then round to 3 significant figures.

Answer

(a) $L(100) = 20$

(b) the level rises by $10\lg 2 \approx 3.01$ dB

(c) $r = 10^{7.5} \approx 3.16 \times 10^{7}$

$2$$4$$6$$500$$1000$$1500$$O$$t$$V$$V = 1500 \times 0.8^{t}$$(0,\ 1500)$$(3,\ 768)$$V = 500$

Maths In Real Life 3

The e-scooter that loses value

A second-hand e-scooter loses a fixed fraction of its value each year, not a fixed amount. That makes its value an exponential decay: big drops early, small drops later, and a value that never quite reaches zero.

An e-scooter is bought for $1500. Its value, $V$ dollars, when it is $t$ years old is modelled by $V = 1500 \times 0.8^{t}$.

• (a)Find the value of the e-scooter when it is $3$ years old.
• (b)The owner plans to sell it as soon as its value is below $500. After how many whole years should it be sold?
• (c)State the value the model approaches in the long run, and say why.

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Colour keyroot or value foundfinal answer

(a)

$\displaystyle V(3)\;= 1500 \times 0.8^{3}$

Put $t = 3$ into the model.

$\displaystyle \phantom{V(3)}\;\approx \textcolor{#010106}{768}$

Calculator, then round to 3 significant figures.

(b)

$\displaystyle 1500 \times 0.8^{n} < 500 \iff 0.8^{n} < \dfrac{1}{3}$

Divide both sides by the number in front.

$\displaystyle 0.8^{n} < \dfrac{1}{3} \iff \lg (0.8^{n}) < \lg \dfrac{1}{3}$

Take logs of both sides. Log is an increasing function, so the inequality keeps its direction.

$\displaystyle \lg (0.8^{n}) < \lg \dfrac{1}{3} \iff n\lg 0.8 < \lg \dfrac{1}{3}$

Power law: bring $n$ down.

$\displaystyle n\lg 0.8 < \lg \dfrac{1}{3} \iff n > \dfrac{\lg \dfrac{1}{3}}{\lg 0.8}$

$\lg 0.8$ is negative, so dividing by it reverses the inequality.

$\displaystyle \dfrac{\lg \dfrac{1}{3}}{\lg 0.8} \approx \textcolor{#010105}{4.923}$

Calculator. Keep one extra significant figure so the whole number is safe.

$\displaystyle n > 4.923 \Rightarrow n \ge \textcolor{#010106}{5}$

$n$ is a whole number, so the least value is the next whole number up.

So the least whole number is $n = 5$.

$\displaystyle 1500 \times 0.8^{5} \approx 492 < 500 \text{ and } 1500 \times 0.8^{4} \approx 614 > 500$

Check the whole number and its neighbour, each value to 3 s.f.

(c)

Because $0.8 < 1$, each step multiplies by a number smaller than one, so $0.8^{t}$ gets closer and closer to $0$ as $t$ grows.

A decay factor below one shrinks every step.

So $V$ approaches $0$ but never reaches it: the limiting value is $0$.

Answer

(a) $V(3) \approx 768$

(b) $n = 5$ (the least whole number)

(c) $V$ approaches $0$

An exponential function has the unknown in the index: $y = a^x$. Compare $x^2$, where the unknown is the base, with $2^x$, where it is the power. The base $a$ is a fixed number with $\textcolor{#010101}{a > 0}$ and $a \ne 1$. If $a$ were $1$ the function would be the constant $1$, and a negative $a$ would give no real value for a power such as $(-2)^{1/2}$. With $\textcolor{#010101}{a > 0}$, $a^x$ is positive for every real $x$, so the graph lives entirely above the $x$-axis.

One special base appears everywhere in science: the number $\mathrm{e} \approx 2.718$. It is the base for which the graph of $y = \mathrm{e}^x$ has gradient exactly $1$ at the point $(0, 1)$. On a calculator it is the second function of the $\ln$ key. Write it as $\mathrm{e}$, upright, to tell it from a variable.

Ask yourself: to what power must $2$ be raised to give $32$? The answer is $5$, and we write $\log_2 32 = 5$. A logarithm is that power. In symbols, $\log_a x = y$ means $a^y = x$. The base stays the base, the log is the index, and the number sits alone on the other side.

For $\log_a x$ to make sense we need $\textcolor{#010101}{a > 0}$ and $a \ne 1$ (the same rule as for $a^x$) and $x > 0$: no power of a positive base is ever zero or negative. Two bases have their own names: $\lg x$ means $\log_{10} x$ and $\ln x$ means $\log_{\mathrm{e}} x$.

A log is a power, so the laws of logs are the index laws in disguise. Multiplying two numbers adds their powers of a base, so a log of a product is a sum of logs. Dividing subtracts the powers. Raising to a power multiplies the power. These laws let you replace hard arithmetic with easy arithmetic.