Operations on the Set of Real Numbers

Lessons

Senior 4 · Mathematics

Operations on the Set of Real Numbers

5 lessons

Lesson 4 of 5

Decimal logarithms and their properties

25 min

Turning multiplication into addition

Multiply $1000 \times 100\,000$ . You do not multiply at all: you add the numbers of zeros, $3 + 5 = 8$ , and write $10^{8}$ . The exponents of 10 turn a product into a sum. The decimal logarithm extends this trick to every positive number: it tells you the power of 10 that the number is. With a table of these exponents, a hard product such as $2.37 \times 48.6$ becomes an easy sum.

That is exactly how scientists, surveyors and engineers calculated for centuries. In 1616 and 1617 Henry Briggs visited John Napier, the inventor of logarithms, in Edinburgh and proposed base 10, and tables of base-10 logarithms were used whenever more accuracy was needed than a slide rule could give. Today the "log" key on your calculator is the decimal logarithm.

Definition

Decimal (common) logarithm: for a real number $x \gt 0$ , the decimal logarithm of $x$ is the exponent to which 10 must be raised to give $x$ :

\log x = y \iff 10^y = x \qquad (x \gt 0).

Notation: $\log x$ , also written $\log_{10} x$ or $\lg x$ . Its domain is $]0, +\infty[$ and its range is all of $\mathbb{R}$ .

Examples: $\log 1000 = 3$ since $10^{3} = 1000$ ; $\log 0.01 = -2$ since $10^{-2} = 0.01$ ; $\log \sqrt{10} = \frac{1}{2}$ . The expressions $\log 0$ and $\log(-5)$ are not defined: no power of 10 is zero or negative.

The graph of $y = \log x$

$x$	0.001	0.01	0.1	1	10	100	1000
$\log x$	$- 3$	$- 2$	$- 1$	0	1	2	3

Read three facts from the graph. The function is increasing: if $0 \lt x_1 \lt x_2$ , then $\log x_1 \lt \log x_2$ . It is negative for $0 \lt x \lt 1$ , zero at $x = 1$ , and positive for $x \gt 1$ . And it grows very slowly: multiplying $x$ by 10 adds only 1 to $\log x$ . Because $\log$ undoes "10 to the power", its graph is the reflection of $y = 10^{x}$ in the line $y = x$ .

Important

Laws of logarithms. For $x \gt 0$ , $y \gt 0$ and any real $n$ :

Law	Example
$\log(xy) = \log x + \log y$	$\log 20 = \log 2 + \log 10 = \log 2 + 1$
$\log\frac{x}{y} = \log x - \log y$	$\log 0.3 = \log 3 - 1$
$\log(x^n) = n\log x$	$\log 8 = 3\log 2$
$\log 10^n = n$ and $10^{\log x} = x$	$10^{\log 7} = 7$
$\log 1 = 0$ , $\log 10 = 1$

Proof of the product law. Put $a = \log x$ and $b = \log y$ , so $x = 10^{a}$ and $y = 10^{b}$ . Then $xy = 10^a \cdot 10^b = 10^{a+b}$ , and by the definition $\log(xy) = a + b = \log x + \log y$ . The other laws follow in the same way from the laws of exponents.

Warning

Three things the laws do NOT say. $\log(x + y) \ne \log x + \log y$ ; $\log x \cdot \log y \ne \log(xy)$ ; and $\frac{\log x}{\log y} \ne \log\frac{x}{y}$ . For example, $\log(10 + 10) = \log 20 \approx 1.301$ , not $1 + 1 = 2$ .

Worked example 1: building logarithms from two values

Given $\log 2 \approx 0.3010$ and $\log 3 \approx 0.4771$ , find (a) $\log 5$ (b) $\log 18$ (c) $\log 0.18$ (d) $\log\sqrt{12}$ .

Solution Write each number using 2, 3 and 10 only.

$\log 5 = \log\frac{10}{2} = 1 - 0.3010 = 0.6990$ .
$\log 18 = \log(2 \times 3^2) = 0.3010 + 2(0.4771) = 1.2552$ .
$\log 0.18 = \log\frac{18}{100} = 1.2552 - 2 = -0.7448$ .
$\log\sqrt{12} = \frac{1}{2}\log(2^2 \times 3) = \frac{1}{2}(0.6020 + 0.4771) \approx 0.5396$ .

Characteristic, mantissa and the size of a number

Write a positive number in standard form, $N = a \times 10^k$ with $1 \le a \lt 10$ . Then

\log N = k + \log a, \qquad 0 \le \log a \lt 1 .

The integer $k$ is the characteristic and the decimal part $\log a$ is the mantissa. For example $\log 48\,300 = \log(4.83 \times 10^4) = 4 + 0.6839 = 4.6839$ . Old log tables listed only mantissas; the characteristic is read off from the position of the decimal point. A useful consequence: a whole number $N \ge 1$ with characteristic $k$ has $k + 1$ digits.

Worked example 2: a number too big for the calculator

How many digits has $2^{100}$ , and what are its first three digits? Use $\log 2 = 0.30103$ .

Solution $\log 2^{100} = 100 \log 2 = 30.103$ . The characteristic is 30, so $2^{100}$ has $31$ digits. The mantissa gives the leading digits: $10^{0.103} \approx 1.27$ , so $2^{100} \approx 1.27 \times 10^{30}$ , and it begins 126 or 127. (The exact value is $1\,267\,650\,600\,228\,229\,401\,496\,703\,205\,376$ , which begins 126; four-figure logarithms are not accurate enough to settle the third digit.)

Solving equations with decimal logarithms

Exponential equations. An equation $a^{x} = b$ ( $a \gt 0$ , $a \ne 1$ , $b \gt 0$ ) is solved by taking the logarithm of both sides, which brings the unknown down from the exponent:

a^x = b \iff x \log a = \log b \iff x = \frac{\log b}{\log a} .

The same idea gives the change of base rule, $\log_a b = \frac{\log b}{\log a}$ , so any logarithm can be found with the "log" key alone: $\log_2 50 = \frac{\log 50}{\log 2} \approx 5.644$ .

Logarithmic equations. First state the domain (every argument of $\log$ must be positive), combine the logs into one, remove the log with $\log A = \log B \iff A = B$ or $\log A = c \iff A = 10^c$ , and finally reject any root outside the domain.

Worked example 3: exponential equations

Solve, giving answers to 4 significant figures: (a) $3^{x} = 20$ (b) $5^{2x - 1} = 7^x$ .

Solution

$x = \frac{\log 20}{\log 3} = \frac{1.30103}{0.47712} \approx 2.727$ . Check: $3^{2.727} \approx 20.0$ . ✓
Take logs: $(2x - 1)\log 5 = x\log 7$ . Collect the $x$ terms: $x(2\log 5 - \log 7) = \log 5$ , so $x = \frac{\log 5}{2\log 5 - \log 7} = \frac{0.69897}{1.39794 - 0.84510} = \frac{0.69897}{0.55284} \approx 1.264 .$

Worked example 4: a logarithmic equation

Solve $\log(x + 3) + \log x = 1$ .

Solution Domain: $x + 3 \gt 0$ and $x \gt 0$ , so $x \gt 0$ .

Combine: $\log\big(x(x + 3)\big) = 1$ , so $x^{2} + 3 x = 10^{1}$ , that is $x^{2} + 3 x - 10 = 0$ , or $(x + 5) (x - 2) = 0$ .

$x = - 5$ is outside the domain ( $\log(-5)$ does not exist), so reject it. $x = 2$ : $\log 5 + \log 2 = \log 10 = 1$ . ✓ Solution set $S = {2}$ .

Worked example 5: pH in chemistry

The pH of a solution is defined by $\mathrm{pH} = -\log[\mathrm{H^+}]$ , where $[\mathrm{H^+}]$ is the hydrogen-ion concentration in $\mathrm{mol\,dm^{-3}}$ . (a) Find the pH when $[\mathrm{H^+}] = 3.2 \times 10^{-4}$ . (b) Find $[\mathrm{H^+}]$ for a solution of pH 8.3. (The values are illustrative.)

Solution

$\mathrm{pH} = -\left(\log 3.2 + \log 10^{-4}\right) = -(0.5051 - 4) = 3.49$ (2 d.p.).
$\log[\mathrm{H^+}] = -8.3$ , so $[\mathrm{H^+}] = 10^{-8.3} = 10^{0.7} \times 10^{-9} \approx 5.0 \times 10^{-9}\,\mathrm{mol\,dm^{-3}}$ .

A change of 1 in pH means a tenfold change in $[\mathrm{H^+}]$ : that is the decimal logarithm at work.

Tip

To remember the definition, say it as a sentence: " $\log x$ is the power of ten that gives $x$ ." Then $\log 0.001$ is easy: $0.001 = 10^{-3}$ , so the answer is $- 3$ . And always write the domain before solving a logarithmic equation; it is where most marks are lost.

KoonPhysics: Logarithmic scale: the slide rule. Watch for how sliding one scale along another adds two lengths, and why adding logarithms multiplies the numbers.

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Practice

Use $\log 2 \approx 0.3010$ and $\log 3 \approx 0.4771$ where needed; otherwise a calculator is allowed from question 4.

Without a calculator: (a) $\log 0.0001$ (b) $\log\sqrt[3]{100}$ (c) $10^{2\log 3}$ .
Find (a) $\log 12$ (b) $\log 0.06$ (c) $\log 45$ .
Write as a single logarithm and evaluate: $2\log 5 + \log 8 - \log 2$ .
How many digits has $7^{40}$ ?
Solve $2^{x} = 45$ to 4 significant figures.
Solve $3^{x + 1} = 5^{x - 1}$ to 4 significant figures.
Solve $\log x + \log(x - 9) = 1$ .
A solution has pH 4.7. Find $[\mathrm{H^+}]$ to 2 significant figures.

Show answersHide answers

(a) $- 4$ (b) $\log 10^{\frac{2}{3}} = \frac{2}{3}$ (c) $10^{\log 9} = 9$ .
(a) $2 (0.3010) + 0.4771 = 1.0791$ (b) $\log 6 - 2 = 0.7781 - 2 = -1.2219$ (c) $\log\frac{9 \times 10}{2} = 0.9542 + 1 - 0.3010 = 1.6532$ .
$\log\frac{25 \times 8}{2} = \log 100 = 2$ .
$40\log 7 \approx 33.80$ : characteristic 33, so 34 digits.
$x = \frac{\log 45}{\log 2} \approx 5.492$ .
$(x + 1)\log 3 = (x - 1)\log 5$ , so $x(\log 5 - \log 3) = \log 5 + \log 3$ and $x = \frac{\log 15}{\log\frac{5}{3}} \approx 5.301$ .
Domain $x \gt 9$ . $x (x - 9) = 10$ , so $x^{2} - 9 x - 10 = 0$ , $(x - 10) (x + 1) = 0$ . Only $x = 10$ is in the domain.
$[\mathrm{H^+}] = 10^{-4.7} \approx 2.0 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ .

Quick check

Key points

$\log x = y \iff 10^y = x$ , defined only for $x \gt 0$ .
$\log(xy) = \log x + \log y$ , $\log\frac{x}{y} = \log x - \log y$ , $\log x^n = n\log x$ .
A whole number whose logarithm has characteristic $k$ has $k + 1$ digits.
$a^x = b \Rightarrow x = \frac{\log b}{\log a}$ ; for log equations, state the domain first.

Sources and credits

Written by Ganzaa: the explanations, the graph, the worked examples, the practice items and the questions.
The history of Briggs's base-10 logarithms and the terms characteristic and mantissa follow Common logarithm, Wikipedia contributors, Wikipedia, CC BY-SA 4.0, adapted by Ganzaa.
Photo: Rechenschieber (slide rule), Berthold Werner, Wikimedia Commons, CC BY-SA 4.0.
Video: Logarithmic scale: the slide rule, KoonPhysics, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Turning multiplication into addition

2The graph of y=log⁡x y = \log x y=logx

3Characteristic, mantissa and the size of a number

4Solving equations with decimal logarithms

Turning multiplication into addition

The graph of $y = \log x$

Characteristic, mantissa and the size of a number

Solving equations with decimal logarithms