International Baccalaureate · DP 1: Applications and Interpretation HL

Logarithms

⌚ ~194 min ▢ 22 blocks ✓ 90 exercises ➣ Prerequisites : Exponentials

I The Logarithm of a Number

I.1 Logarithms in Base 10

Recappreviously studied — click to reread

For a power of 10, the logarithm is simply the exponent read off the power. The table below shows this two-way reading:

$y$	$0.01$	$0.1$	$1$	$10$	$100$	$1000$
$y$ as a power of $10$	$10^{-2}$	$10^{-1}$	$10^{0}$	$10^{1}$	$10^{2}$	$10^{3}$
$\log y$	$-2$	$-1$	$0$	$1$	$2$	$3$

For example, $10^{-2} = 0.01$, so $\log 0.01 = -2$; and $10^{3} = 1000$, so $\log 1000 = 3$.
The two operations travel in opposite directions between a number $y$ and its exponent $x$: the logarithm extracts the exponent, and raising 10 to a power rebuilds the number.

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Definition — Logarithm

The (base 10) logarithm of a number $y$ (where $y > 0$) is the exponent to which 10 must be raised to obtain $y$. It is denoted $\log_{10} y$ and defined by the equivalence: $$\log_{10} y = x \quad \Leftrightarrow \quad 10^x = y$$ As on most calculators and in exams, $\log_{10} y$ is abbreviated $\log y$; so $\log y = \log_{10} y$. In other words, $\log y$ is the exponent $x$ such that $10^x = y$.

Why $\log 0$ and $\log(-5)$ are undefined

The logarithm $\log y$ is defined only for $y > 0$. Indeed, a power of 10 is always strictly positive: $10^x > 0$ for every real $x$. Since no power of 10 equals $0$ or a negative number, expressions such as $\log 0$ and $\log(-5)$ have no value.

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Proposition — Inverse Properties of Logarithms

For real numbers:

$10^{\log x} = x$ for any $x > 0$,
$\log (10^x) = x$ for any real $x$.

Example

Evaluate $\log 100$.

Answer

$\begin{aligned}[t]\log(100) &= \log\left(10^{2}\right) &&(100 = 10^2) \\&=2\\\end{aligned}$
Since $10^2 = 100$, the logarithm of $100$ (base 10) is $2$.

Skills to practice

Evaluating Logarithms without a Calculator

I.2 Logarithms in Base $a$

Recappreviously studied — click to reread

The logarithm in base $10$ of a number is the exponent to which $10$ must be raised in order to obtain that number.
We can use the same principle to define logarithms in other bases:

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Definition — Logarithm in Base a

For a positive number $a \neq 1$, the logarithm base $a$ of a number $y$ (where $y > 0$) is the exponent to which $a$ must be raised to obtain $y$. It is denoted as:$$\log_a y = x \quad \Leftrightarrow \quad a^x = y.$$

Proposition — Inverse Properties of Logarithms in Base a

Let $a>0$ with $a\neq 1$.

$a^{\log_a x} = x$ for any $x > 0$,
$\log_a (a^x) = x$ for any real $x$.

Example

Evaluate $\log_2 8$.

Answer

$\begin{aligned}[t]\log_2 8 &= \log_2 (2^3) &&(8 = 2^3)\\&=3\\\end{aligned}$

Method — Solving Exponential Equations with Base a

To solve $a^x = b$ (where $a > 0$, $a \neq 1$, $b > 0$):

Take the logarithm base $a$ of both sides: $\log_a (a^x) = \log_a b$
Simplify: $x = \log_a b$

Example

Solve $3^x = 81$.

Answer

$$\begin{aligned}3^x &= 81 \\ \log_3 (3^x) &= \log_3 (81) &&\text{(applying $\log_3$ to both sides)}\\ x &= \log_3 (3^4) &&(\log_3(3^x) = x,\ 81 = 3^4)\\ x &= 4\end{aligned}$$

Skills to practice

Evaluating Logarithms in Base $a$

I.3 The Natural Logarithm

The natural logarithm, often denoted $\ln x$, is the logarithm with base $e$ (where $e \approx 2.71828$ is the base of the natural exponential function). It is the inverse of the exponential function $e^x$ and plays a central role in calculus, growth models, and many scientific applications.

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Definition — Natural Logarithm

The natural logarithm, denoted $\ln x$, is the logarithm base $e$, defined for $x>0$:$$\ln x = \log_e(x).$$

Proposition — Inverse Properties of Natural Logarithms

$e^{\ln x} = x$ for $x > 0$,
$\ln (e^x) = x$ for any real $x$.

Notation. A logarithm always has a base. Three forms are in use:

$\log_a x$ — the logarithm in base $a$ (the base is written explicitly);
$\log x = \log_{10} x$ — the base-10 logarithm; as on calculators and in exams, the base 10 is omitted;
$\ln x = \log_e x$ — the natural logarithm (base $e$).

So a bare $\log$ means $\log_{10}$, and $\ln$ means $\log_e$.

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Skills to practice

Evaluating Natural Logarithms

II Laws of Logarithms

II.1 The Laws of Logarithms

Recappreviously studied — click to reread

Proposition — Laws of Logarithms

For $x>0$, $y>0$, $a>0$ with $a\neq 1$, and for any real number $k$:

Product Rule: $\log_a (xy) = \log _a x + \log _a y$
Quotient Rule: $\log _a \left(\dfrac{x}{y}\right) = \log _a x - \log_a y$
Power Rule: $\log_a (x^k) = k \log_a x$

Proof

$$\begin{aligned}\log_a(xy) &= \log_a\bigl(a^{\log_a x} \times a^{\log_a y}\bigr) &&(x = a^{\log_a x})\\ &= \log_a\bigl(a^{\log_a x + \log_a y}\bigr) &&\text{(exponent product law)}\\ &= \log_a x + \log_a y &&(\log_a(a^u) = u)\\ \\ \log_a\left(\frac{x}{y}\right) &= \log_a\left(\frac{a^{\log_a x}}{a^{\log_a y}}\right) &&(x = a^{\log_a x})\\ &= \log_a\bigl(a^{\log_a x - \log_a y}\bigr) &&\text{(exponent quotient law)}\\ &= \log_a x - \log_a y &&(\log_a(a^u) = u)\\ \\ \log_a(x^k) &= \log_a\left(\bigl(a^{\log_a x}\bigr)^k\right) &&(x = a^{\log_a x})\\ &= \log_a\bigl(a^{k \log_a x}\bigr) &&\text{(power of a power)}\\ &= k \log_a x\\ \end{aligned}$$

Example

Write as a single logarithm: $\log (5)+\log (3)$.

Answer

$\begin{aligned}[t]\log (5)+\log (3) &= \log (5\times 3) &&\text{(product rule)} \\&= \log 15\\\end{aligned}$

Example

For $a>0$ and $b>0$, write $\log\left(\dfrac{a^2 b}{10}\right)$ in terms of $\log a$ and $\log b$.

Answer

$$\begin{aligned}[t] \log\left(\dfrac{a^2 b}{10}\right) &= \log(a^2 b) - \log 10 &&\text{(quotient rule)}\\ &= \log(a^2) + \log b - \log 10 &&\text{(product rule)}\\ &= 2\log a + \log b - \log 10 &&\text{(power rule)}\\ &= 2\log a + \log b - 1 \end{aligned}$$ The laws also work in this direction: they break a single logarithm into a combination of simpler logarithms.

Skills to practice

The Laws of Logarithms

II.2 Using the Laws with $\ln$ and in Base $a$

In particular, for the natural logarithm, for $x>0$ and $y>0$:

$\ln (xy) = \ln x + \ln y$
$\ln \left(\dfrac{x}{y}\right) = \ln x - \ln y$
$\ln (x^k) = k \ln x$

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Example

Evaluate $\ln \left(e^4\right)$.

Answer

$\begin{aligned}[t]\ln\left(e^4\right) &= 4 \ln e &&\text{(power rule)}\\&=4 \times 1 &&(\ln e = 1)\\&=4\\\end{aligned}$

Skills to practice

Writing as a Single Natural Logarithm
Writing as a Single Logarithm in Base $a$

III Solving Exponential Equations

III.1 Solving Exponential Equations with Logarithms

Recappreviously studied — click to reread

Some exponential equations are solved by writing both sides with the same base (e.g., $2^x = 8 \Rightarrow 2^x = 2^3$). However, most real-world problems involve values that are not perfect powers. To solve these, we use the power rule of logarithms to "bring down" the exponent.

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Method — Solving $a^x \equal b$

To solve an equation $a^x = b$ (where $a > 0$, $a \neq 1$, $b > 0$) in which the unknown is in the exponent:

Isolate the exponential term ($a^x$) if necessary.
Take the logarithm ($\log$ or $\ln$) of both sides.
Use the Power Rule: $\log(a^x) = x \log a$.
Solve for $x$ by dividing.

Example

Solve $2^x = 7$.

Answer

$$\begin{aligned}2^x &= 7 \\ \log(2^x) &= \log 7 &&\text{(taking the logarithm of both sides)}\\ x \log 2 &= \log 7 &&\text{(power rule)}\\ x &= \frac{\log 7}{\log 2} &&\text{(dividing)}\\ \end{aligned}$$

Skills to practice

Solving $a^x \equal b$ with a Logarithm
Solving Equations with $e^x$

III.2 Solving Exponential Models with Logarithms

Example

A radioactive element decays according to the formula $R(t) = 100 e^{-0.05t}$, where $R$ is the radiation in counts per minute and $t$ is time in years.

State the initial radiation.
Find the half-life (time taken for radiation to halve).

Answer

Initial radiation: At $t=0$, $$ R(0) = 100 e^0 = 100 \text{ cpm}. $$
Half-life: We find $t$ such that $R(t) = 50$. $$ 50 = 100 e^{-0.05t} \implies 0.5 = e^{-0.05t}. $$ Taking $\ln$ of both sides: $$ \ln(0.5) = -0.05t \implies t = \frac{\ln(0.5)}{-0.05} \approx 13.86 \text{ years}. $$

Example

The number of fish in a lake, $F$, can be modelled by the function$$ F(t) = \frac{800}{1 + Ce^{-0.6t}} $$ where $t$ is the number of months after fish were introduced to the lake.

Initially, 50 fish were introduced to the lake. Find the value of $C$.
Write down the limiting capacity for the number of fish in the lake.
Calculate the number of months it takes until there are 500 fish in the lake.

Answer

Find $C$: At $t=0$, $F=50$. $$ 50 = \frac{800}{1 + Ce^0} = \frac{800}{1+C}. $$ Hence $$ 1 + C = \frac{800}{50} = 16 \implies C = 15. $$ So, $F(t) = \frac{800}{1 + 15e^{-0.6t}}$.
Limiting capacity: $$ L = 800 \text{ fish}. $$
Find $t$ when $F=500$: $$ 500 = \frac{800}{1 + 15e^{-0.6t}} \implies 1 + 15e^{-0.6t} = \frac{800}{500} = 1.6. $$ $$ 15e^{-0.6t} = 0.6 \implies e^{-0.6t} = 0.04. $$ $$ -0.6t = \ln(0.04) \implies t = \frac{\ln(0.04)}{-0.6} \approx 5.36 \text{ months}. $$

Skills to practice

Finding When an Exponential Model Reaches a Value
Solving Logistic Models with Logarithms

III.3 Logarithmic Scales in Science

Recappreviously studied — click to reread

Logarithms are used to describe quantities that vary over many orders of magnitude. In the scientific formulas below, the base-10 logarithm is written in full as $\log_{10}$. Some important examples in science include:

pH scale in chemistry: $\text{pH} = -\log_{10} [H^+]$
(where $[H^+]$ is the numerical value of the hydrogen ion concentration in moles per litre)
Richter scale for earthquakes (simplified model): $\text{Magnitude} = \log_{10}\left(\dfrac{I}{I_0}\right)$
(where $I$ is the intensity of the earthquake, $I_0$ is a reference intensity)
Decibel (dB) scale for sound: $L = 10 \log_{10}\left(\dfrac{P}{P_0}\right)$
(where $P$ is the power/intensity measured, $P_0$ is a reference level)

On calculators, the $\log$ key gives $\log_{10}$.

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Example

The pH of a solution is $3.2$. Find the hydrogen ion concentration $[H^+]$.

Answer

We know: $$\begin{aligned} \text{pH} &= -\log_{10}[H^+] \\ 3.2 &= -\log_{10}[H^+] &&\text{(substituting the value)}\\ -3.2 &= \log_{10}[H^+] &&\text{(multiplying both sides by $-1$)}\\ 10^{-3.2} &= 10^{\log_{10}[H^+]} &&\text{(exponentiating both sides, base 10)}\\ [H^+] &= 10^{-3.2} \\ [H^+] &\approx 6.31 \times 10^{-4}\ \text{mol/L} &&\text{(using a calculator)} \end{aligned}$$

Skills to practice

Logarithmic Scales in Science

IV Logarithmic Functions

IV.1 The Logarithmic Function in Base $a$

Recappreviously studied — click to reread

Definition — Logarithmic Function in Base a

The logarithmic function in base a is $f(x) = \log_a(x)$, where $a$ is a positive constant and $a \neq 1$.

Domain: $(0, +\infty)$
Range: $(-\infty, +\infty)$

Proposition — Graph of the Logarithmic Function in Base a

The graph of $\textcolor{colordef}{y = \log_a(x)}$ is the reflection of $\textcolor{colorprop}{y = a^x}$ over the line $\textcolor{olive}{y = x}$. It has a vertical asymptote at $x=0$ and passes through the point $(1,0)$.

If $a > 1$, the function is increasing.
If $0 < a < 1$, the function is decreasing.

$\quad$

Example

Study the function $f(x) = \log(x - 1)$: domain, asymptote, intercept, and sketch.

Domain: the logarithm needs a positive input: $x - 1 > 0$, so the domain is $(1, +\infty)$.
Vertical asymptote: at the edge of the domain, $x = 1$ (the whole curve of $\log$ is shifted $1$ unit right, and its asymptote moves with it).
$x$-intercept: $f(x) = 0$ when $\log(x-1) = 0$, i.e. $x - 1 = 1$ (since $\log 1 = 0$): the point $(2, 0)$.
One exact point: $f(11) = \log(10) = 1$ (ten to the power $1$): the point $(11, 1)$.

Skills to practice

Finding Domains in Base $a$

IV.2 The Natural Logarithm Function

Definition — Natural Logarithm Function

The natural logarithm function is $x\mapsto \ln x$.

Domain: $(0, +\infty)$
Range: $(-\infty, +\infty)$

Proposition — Graph of the Natural Logarithm Function

The graph of $\textcolor{colordef}{y = \ln x}$ is the reflection of $\textcolor{colorprop}{y = e^x}$ over the line $\textcolor{olive}{y = x}$. It has a vertical asymptote at $x=0$, passes through (1,0), and is increasing.

Skills to practice

Finding Domains of Logarithmic Functions
Finding Inverses and Asymptotes
Graphing and Studying Logarithmic Functions

V Logarithmic Models and Scales

V.1 Logarithmic Models

Definition — Logarithmic Model

A logarithmic model is a function of the form$$ f(x) = a + b\ln x, \qquad x>0,\; b\neq 0. $$

$\ln x$ denotes the natural logarithm (base $e$).
$a$ is the value of the function at $x=1$, since $f(1)=a$.
$b$ controls the rate of increase/decrease.

Proposition — Properties

The graph has a vertical asymptote at the $y$-axis ($x=0$).
The function has one root (an $x$-intercept) given by $$ f(x)=0 \iff x=e^{-\frac{a}{b}}. $$
If $b > 0$, the function is increasing on $(0,+\infty)$. If $b < 0$, the function is decreasing on $(0,+\infty)$.

Example

The sound intensity level $L$, in decibels (dB), of a sound of intensity $I$, in watts per square metre (Wm$^{-2}$), is $L = 10 \log_{10}\left(\dfrac{I}{10^{-12}}\right)$, where $I$ stands for its numerical value in Wm$^{-2}$. The level can also be written as a logarithmic model $$ L(I) = a + b \ln I. $$

A sound intensity of $1 \text{ Wm}^{-2}$ produces a sound intensity level of $120 \text{ dB}$. Write down the value of $a$.
A sound intensity of $10 \text{ Wm}^{-2}$ produces a sound intensity level of $130 \text{ dB}$. Find the value of $b$.
Find the sound intensity, in Wm$^{-2}$, of a car alarm that has a sound intensity level of $105 \text{ dB}$.

Answer

Substitute $I = 1$ and $L = 120$: since $\ln 1 = 0$, $120 = a + b \times 0$, so $a = 120$.
Substitute $I = 10$ and $L = 130$: $$\begin{aligned}130 &= 120 + b \ln 10 &&\text{(substitution)}\\ b &= \frac{10}{\ln 10} \approx 4.34 &&\text{(dividing by } \ln 10 \gt 0\text{)} \end{aligned}$$ So $L(I) = 120 + \dfrac{10}{\ln 10} \ln I$.
With the exact value of $b$: $$\begin{aligned}105 &= 120 + \frac{10}{\ln 10} \ln I &&\text{(substitution)}\\ \ln I &= -\frac{15 \ln 10}{10} = -1.5 \ln 10 &&\text{(isolating } \ln I\text{)}\\ I &= \left(e^{\ln 10}\right)^{-1.5} = 10^{-1.5} \approx 0.0316 \text{ Wm}^{-2} &&\text{(inverse property of } \ln\text{)} \end{aligned}$$ Check with the decibel formula: $10 \log_{10}\left(\dfrac{10^{-1.5}}{10^{-12}}\right) = 10 \times 10.5 = 105$ dB.

Skills to practice

Applying Logarithmic Models

V.2 Logarithmic Scales

Discover

A culture of bacteria is counted every hour: $100$ bacteria at the start, and the number is multiplied by $3$ every hour, up to $656\,100$ after $8$ hours. On an ordinary graph, drawn with the vertical scale needed for the last value, the first points are all squeezed against the horizontal axis.

How could the vertical axis be graduated so that every point is clearly visible? What shape would the points then form?

Answer

Each hour multiplies the number by $3$. On a vertical axis where each number $N$ is placed at the height $\log N$ (the axis below starts at the mark $100$, at the height $\log 100 = 2$), a multiplication becomes an addition: $\log(3N) = \log N + \log 3$. Each hour then raises the point by the same height $\log 3 \approx 0.48$, and the values $100$, $1000$, $10\,000$, …\ are equally spaced. The points are aligned:

Definition — Logarithmic Scale and Logarithmic Frames

On a logarithmic scale, each number $y \gt 0$ is placed at the signed position $\log y$ from the mark $1$ (above it if $\log y \gt 0$, below it if $\log y \lt 0$), the unit of length being called a decade: the numbers $0.1$, $1$, $10$, $100$, $1000$ are equally spaced, and between $1$ and $10$ the marks $2, 3, \dots, 9$ stand at $\log 2 \approx 0.30$, $\log 3 \approx 0.48$, …, $\log 9 \approx 0.95$ decade; the same pattern repeats in every decade. Zero and negative numbers have no place on such a scale.

A semi-logarithmic frame has an ordinary horizontal axis and a logarithmic vertical axis.
A logarithmic frame (or log-log frame) has two logarithmic axes.

Method — Placing and Reading a Value on a Logarithmic Scale

To place $y \gt 0$: compute $\log y$. It is the signed position, in decades, from the mark $1$; its whole part tells between which powers of ten the point lies, and its decimal part how far it stands inside that decade.
To read the point at the distance $h$ decades from the mark $1$ (above it if $h \gt 0$, below it if $h \lt 0$): it represents $y = 10^h$.

For example, $\log 350 \approx 2.54$: the number $350$ lies between $100$ and $1000$, $0.54$ decade above $100$ (see the figure of the definition). For $0 \lt y \lt 1$ the logarithm is negative: $\log 0.035 \approx -1.46$ lies between $-2$ and $-1$, so $0.035$ lies between $10^{-2} = 0.01$ and $10^{-1} = 0.1$; the power of ten to take is the one below, $10^{-2}$, and the point stands $2 - 1.46 = 0.54$ decade above it. Conversely, the point $1.7$ decades above $1$ represents $10^{1.7} \approx 50$.

The figure below places the intensity ratio $\frac{I}{I_0}$ of an earthquake on a logarithmic scale: the ticks $10^0, 10^1, \dots, 10^6$ are drawn equally spaced.

On this axis, the ticks are equally spaced, yet each step from one tick to the next multiplies the intensity ratio by $10$. Equal steps on the axis correspond to equal factors of $10$: the magnitude $\log_{10}\left(\frac{I}{I_0}\right)$ increases by exactly $1$ at each tick. Logarithms turn multiplicative scales into additive ones — this is why the pH, Richter and decibel scales can display enormous ranges of values on one readable axis.

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Proposition — Exponential Functions in a Semi-Logarithmic Frame

Let $k \gt 0$ and $a \gt 0$. In a semi-logarithmic frame, the points of the curve of $x \mapsto k \times a^x$ are aligned: their height $Y = \log y$ satisfies $$\textcolor{colorprop}{Y = \log k + x \log a},$$ a straight line that rises by $\log a$ decade when $x$ increases by $1$ (it falls when $0 \lt a \lt 1$), and that meets the vertical axis at the mark $k$. Conversely, points aligned in a semi-logarithmic frame follow a model $y = k \times a^x$.

Proof

$$\begin{aligned}Y = \log(k \times a^x) &= \log k + \log(a^x) &&\text{(logarithm of a product)}\\ &= \log k + x \log a &&\text{(power rule)}\end{aligned}$$ This is an affine expression of $x$: the points are aligned, and for $x = 0$ the height is $\log k$, the height of the mark $k$. Conversely, if $Y = p\,x + q$, then $y = 10^{p x + q} = 10^q \times (10^p)^x$: the model $k \times a^x$ with $k = 10^q$ and $a = 10^p$.

Method — Finding an Exponential Model from a Straight Line

When points lie on a straight line in a semi-logarithmic frame:

Read two points of the line, $(x_1, y_1)$ and $(x_2, y_2)$, the values $y_1$ and $y_2$ being read on the logarithmic scale.
Compute the rise per unit, $\log a = \dfrac{\log y_2 - \log y_1}{x_2 - x_1}$, then $a = 10^{\log a}$.
Compute $k$, the value at $x = 0$: $k = \dfrac{y_1}{a^{x_1}}$ (read directly if $x_1 = 0$).

Example

A capacitor discharges through a resistor. The voltage $U$ (in volts) across it, measured every second from $t = 0$ to $t = 5$ and plotted in a semi-logarithmic frame, gives six points on the straight line through $(0, 12)$ and $(5, 0.75)$.

Determine a model $U = k \times a^t$, and the time after which the voltage is halved.

Answer

$$\begin{aligned}\log a &= \frac{\log 0.75 - \log 12}{5 - 0} &&\text{(rise per second)}\\ &= \frac{\log(0.75 \div 12)}{5} = \frac{\log 0.0625}{5} &&\text{(logarithm of a quotient)}\\ a &= 10^{\frac{\log 0.0625}{5}} \approx 0.574 &&\text{(definition of the logarithm)} \end{aligned}$$ The line meets the vertical axis at $12$, so $k = 12$: $U \approx 12 \times 0.574^t$. The voltage is halved when $0.574^t = 0.5$, that is $t = \dfrac{\log 0.5}{\log 0.574} \approx 1.25$ s. Check: in $5$ s, four halvings divide the voltage by $2^4 = 16$, and indeed $12 \div 16 = 0.75$.

Logarithmic Frames and Power Functions

In a logarithmic frame, both coordinates are replaced by their logarithms. For $x \gt 0$, the curve of $x \mapsto k \times x^a$ ($k \gt 0$) becomes a straight line of slope $a$ counted in decades ($a$ decades up for one decade to the right), whatever the length of a decade on each axis, since $\log y = \log k + a \log x$. Below, $y = 3x^2$ for $1 \le x \le 100$: when $x$ is multiplied by $10$ (one decade to the right), $y$ is multiplied by $100$ (two decades up).

Such frames are used when both quantities span several orders of magnitude: the frequency response of a loudspeaker or of an amplifier (frequencies from $20$ Hz to $20\,000$ Hz), the size and the energy needs of animals, the brightness of stars and their distance.

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Skills to practice

Reading Logarithmic Scales and Frames
Placing Values and Finding a Model
Solving Problems with Logarithmic Scales

VI Going further

Skills to practice

Reasoning on Logarithmic Graphs
Solving Problems
Reasoning on Models and Scales
Quiz

\(y\)	\(0.01\)	\(0.1\)	\(1\)	\(10\)	\(100\)	\(1000\)
\(y\) as a power of \(10\)	\(10^{-2}\)	\(10^{-1}\)	\(10^{0}\)	\(10^{1}\)	\(10^{2}\)	\(10^{3}\)
\(\log y\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)