Laws of logarithms — Mathematics — Mathemagicians Library

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This leads to

Solving exponential equations
pH — the hydrogen ion scale · coming soon
Logarithmic graphs · coming soon

Multiplying by adding

Before calculators, multiplying two seven-digit numbers by hand took a long time, and one slip spoiled the answer. Astronomers and navigators did thousands of these calculations. In the early 1600s they were given a way to replace almost every multiplication with an addition. Adding is much faster, and much safer.

That sounds impossible. Multiplying and adding are different operations: $8 \times 32$ is 256, and $8 + 32$ is 40. So where could an addition hide inside a multiplication?

Look at the indices, not the numbers. $8 = 2^{3}$ and $32 = 2^{5}$ . Their product is $256 = 2^{8}$ . And $3 + 5 = 8$ . The figure shows the same thing on a line.

Powers of 2 above, their indices below. Each step to the right multiplies the number by 2 and adds 1 to the index. Going 3 steps and then 5 more lands on 256, because

8 \times 32 = 256

and

3 + 5 = 8

. Multiplying along the top is adding along the bottom.

The last lecture ended with a pattern like this. It found that $lo g_{2} 40 = lo g_{2} 20 + 1$ : doubling a number adds exactly 1 to its base-2 logarithm. It promised that this lecture would show why patterns like that always hold. Here is the reason, and two more laws that come with it.

Where the pattern comes from

Start from the one idea of the last lecture: a logarithm is an index. In base 2, $lo g_{2} 8 = 3$ and $lo g_{2} 32 = 5$ . So the pattern above says

lo g_{2} (8 \times 32) = lo g_{2} 8 + lo g_{2} 32.

Is that a coincidence of these numbers? To find out, we prove it for any two positive numbers $A$ and $B$ and any allowed base $a$ . The proof uses nothing new. It uses the definition of a logarithm and the first index law: multiplying powers of the same base adds the indices.

Step 1. Give the two logarithms names. Let $p = lo g_{a} A$ and $q = lo g_{a} B$ . By the definition, this means $a^{p} = A$ and $a^{q} = B$ .

Step 2. Multiply. $A B = a^{p} \times a^{q}$ . By the first index law, this is $a^{p + q}$ .

Step 3. Read it as a logarithm. $A B = a^{p + q}$ says that $a$ to the power $p + q$ gives $A B$ . By the definition, that is exactly $lo g_{a} (A B) = p + q$ . Now put the names back: $lo g_{a} (A B) = lo g_{a} A + lo g_{a} B$ .

That is it. That is the whole idea. The logarithm of a product is the sum of the logarithms, because logarithms are indices, and multiplying powers adds their indices. The law is the first index law, read through the logarithm.

Now the promise from the last lecture. Write 40 as $20 \times 2$ . Then $lo g_{2} 40 = lo g_{2} 20 + lo g_{2} 2 = lo g_{2} 20 + 1$ . Doubling adds $lo g_{2} 2$ , which is 1. In the same way, multiplying by 10 adds exactly 1 to a base-10 logarithm, because $lo g_{10} 10 = 1$ .

This is also how the old tables were used. To find $37.4 \times 4.82$ , a navigator looked up $lo g_{10} 37.4 \approx 1.5729$ and $lo g_{10} 4.82 \approx 0.6830$ . One addition gave about 2.2559. Then a second table, read backwards, gave the number with that logarithm: about 180.3. The exact product is 180.268, so the tables gave the first four figures correctly.

The explorer below turns this into lengths. On this ruler, each number sits at a distance equal to its base-10 logarithm. So laying two lengths end to end adds two logarithms.

Explore — a ruler that multiplies

Choose two numbers A and B. The teal bar has length log₁₀ A. The ink bar, placed after it, has length log₁₀ B. Read where the two bars end, then compare it with A × B.

A = 2.0

B = 4.0

A ruler on which each number sits at a distance equal to its base-10 logarithm. The teal length is $lo g_{10} 2$ and the ink length is $lo g_{10} 4$ . Laid end to end, they end at 8.

A × B	log₁₀ A + log₁₀ B	Product
2 × 4	≈ 0.3010 + 0.6021 ≈ 0.9031	8
2.5 × 4	≈ 0.3979 + 0.6021 = 1	10
3 × 7	≈ 0.4771 + 0.8451 ≈ 1.3222	21
6 × 9	≈ 0.7782 + 0.9542 ≈ 1.7324	54

What should you have seen? Wherever you put A and B, the bars end exactly at $A \times B$ . Try $2.5 \times 4$ : the bars end at 10, because $lo g_{10} 2.5 + lo g_{10} 4 = lo g_{10} 10 = 1$ . Neither logarithm is a tidy number, but their sum is exactly 1. This ruler is the idea behind the slide rule, which engineers used to multiply for more than three hundred years.

Dividing, and powers

The first index law gave the first law of logarithms. The other two index laws give the other two, with the same three steps.

Dividing. Keep $a^{p} = A$ and $a^{q} = B$ . Then $\frac{A}{B} = \frac{a ^{p}}{a ^{q}} = a^{p - q}$ , by the second index law. Read it as a logarithm: $lo g_{a} \frac{A}{B} = p - q = lo g_{a} A - lo g_{a} B$ . Dividing the numbers subtracts their logarithms.

Powers. Now raise $A$ to a power $n$ . $A^{n} = (a^{p})^{n} = a^{p n}$ , by the third index law. Read it as a logarithm: $lo g_{a} (A^{n}) = p n = n lo g_{a} A$ . The power comes down to the front and multiplies the logarithm.

Check the power law with numbers. $lo g_{2} (8^{2}) = lo g_{2} 64 = 6$ , and $2 lo g_{2} 8 = 2 \times 3 = 6$ . ✓ The third law also works when $n$ is a fraction or negative, because the third index law does. For example, $lo g_{a} A = lo g_{a} A^{1/2} = \frac{1}{2} lo g_{a} A$ .

The laws of logarithms

For an allowed base $a$ (positive and not 1), positive numbers $A$ and $B$ , and any number $n$ :

$lo g_{a} (A B) = lo g_{a} A + lo g_{a} B$ (from $a^{p} \times a^{q} = a^{p + q}$ )

$lo g_{a} \frac{A}{B} = lo g_{a} A - lo g_{a} B$ (from $a^{p} \div a^{q} = a^{p - q}$ )

$lo g_{a} (A^{n}) = n lo g_{a} A$ (from $(a^{p})^{n} = a^{p n}$ )

Do not learn these as three new facts. Learn the one sentence behind them: a logarithm is an index. If you ever forget a law, let $p = lo g_{a} A$ , write the power, and use the index law. The law returns in three lines.

Two small facts that come for free

Two values appear in almost every question. Both come straight from the definition. For any allowed base $a$ :

$lo g_{a} a = 1$ , because $a^{1} = a$ . So $lo g_{10} 10 = 1$ and $lo g_{7} 7 = 1$ .
$lo g_{a} 1 = 0$ , because $a^{0} = 1$ . You met this in the last lecture.

They also agree with the laws, as they must. The first law gives $lo g_{a} (A \times 1) = lo g_{a} A + lo g_{a} 1 = lo g_{a} A + 0$ . And $A \times 1$ is $A$ , so nothing has changed. ✓ The second law gives one more useful fact: $lo g_{a} \frac{1}{B} = lo g_{a} 1 - lo g_{a} B = - lo g_{a} B$ . Turning a number upside down changes the sign of its logarithm. For example, $lo g_{2} \frac{1}{8} = - lo g_{2} 8 = - 3$ , as the last lecture found directly.

Stop and think

A student writes $lo g_{10} 2 + lo g_{10} 8 = lo g_{10} 10 = 1$ , because $2 + 8 = 10$ . Is this right?

Show the answer

No. The first law turns a sum of logarithms into the logarithm of a product: $lo g_{10} 2 + lo g_{10} 8 = lo g_{10} 16 \approx 1.204$ . The reason is in the proof. Adding the logarithms adds the indices, and adding indices means multiplying the powers, not adding them. There is no law for the logarithm of a sum. $lo g_{10} (2 + 8)$ is simply $lo g_{10} 10 = 1$ , and it cannot be split into two logarithms.

What the laws do not say

The laws are about three things inside one logarithm: multiplying, dividing and powers. They say nothing about adding or subtracting inside a logarithm. They also say nothing about multiplying or dividing two logarithms. This is where most mistakes come from, because some of these false rules look very like the real ones. Sort the statements below. Each one uses positive numbers $x$ and $y$ .

Sorter — a law, or a look-alike?

Tap a statement, then tap a box: always true, or not true in general. Read the reason after each one.

Ten statements about logarithms of positive numbers $x$ and $y$ , sorted by whether they are always true.

Always true. $lo g_{10} (x y) = lo g_{10} x + lo g_{10} y$ (first law). $lo g_{2} \frac{x}{8} = lo g_{2} x - 3$ (second law, with $lo g_{2} 8 = 3$ ). $lo g_{3} (x^{4}) = 4 lo g_{3} x$ (third law). $lo g_{5} \frac{1}{x} = - lo g_{5} x$ (second law, with $lo g_{5} 1 = 0$ ). $lo g_{2} x = \frac{1}{2} lo g_{2} x$ (third law, with $x = x^{1/2}$ ).
Not true in general. $lo g_{10} (x + y) = lo g_{10} x + lo g_{10} y$ : with $x = y = 1$ , the left is $lo g_{10} 2 \approx 0.301$ and the right is 0. $lo g_{10} x \times lo g_{10} y = lo g_{10} (x y)$ : with $x = y = 10$ , the left is 1 and the right is 2. $(lo g_{10} x)^{2} = 2 lo g_{10} x$ : with $x = 1000$ , the left is 9 and the right is 6. $lo g_{10} x \div lo g_{10} y = lo g_{10} \frac{x}{y}$ : with $x = 100$ and $y = 10$ , the left is 2 and the right is 1. $lo g_{10} (x - y) = lo g_{10} x - lo g_{10} y$ : with $x = 20$ and $y = 10$ , the left is 1 and the right is about 0.301.

What should you have learned? Every false statement can be broken by one pair of numbers. That is the quickest way to test a rule you are not sure of. If one test fails, the rule is false. A test that passes proves nothing, as the spot-the-error example in the last lecture showed.

Using the laws

The laws are used in two directions. Combining turns several logarithms into one. Expanding splits one logarithm into simpler pieces. In every example, the base stays the same throughout.

Worked Example 1 — Writing as a single logarithm

Write as a single logarithm, and evaluate where possible: (a) $lo g_{10} 3 + lo g_{10} 5 - lo g_{10} 2$ (b) $2 lo g_{10} 5 + lo g_{10} 4$ .

(a) The first law combines the first two: $lo g_{10} 3 + lo g_{10} 5 = lo g_{10} 15$ . The second law then gives $lo g_{10} 15 - lo g_{10} 2 = lo g_{10} 7.5$ .

(b) A number in front of a logarithm must go inside first, by the third law: $2 lo g_{10} 5 = lo g_{10} (5^{2}) = lo g_{10} 25$ . Now the first law: $lo g_{10} 25 + lo g_{10} 4 = lo g_{10} 100 = 2$ .

Check (b) with a calculator: $2 \times 0.69897 + 0.60206 = 2.00000$ . ✓ Neither logarithm was a whole number, but the answer is exactly 2.

Worked Example 2 — Exact values without a calculator

Find (a) $lo g_{2} 48 - lo g_{2} 3$ (b) $lo g_{6} 4 + lo g_{6} 9$ .

(a) Neither $lo g_{2} 48$ nor $lo g_{2} 3$ is a whole number. Combine them first: $lo g_{2} \frac{48}{3} = lo g_{2} 16$ . Then ask the question: 2 to the power of what gives 16? So the answer is 4.

(b) $lo g_{6} (4 \times 9) = lo g_{6} 36 = 2$ , because $6^{2} = 36$ .

The thinking: when the separate logarithms are hard, combine them and hope the number inside becomes a power of the base. Questions without a calculator are usually written so that it does.

Worked Example 3 — Building logarithms from one value

Given $lo g_{10} 2 \approx 0.30103$ , find $lo g_{10} 8$ , $lo g_{10} 5$ , $lo g_{10} 50$ and $lo g_{10} 0.4$ without a calculator.

$8 = 2^{3}$ , so $lo g_{10} 8 = 3 lo g_{10} 2 \approx 0.90309$ .

5 is not a power of 2. But $5 = \frac{10}{2}$ , and $lo g_{10} 10 = 1$ . So $lo g_{10} 5 = 1 - lo g_{10} 2 \approx 0.69897$ . This works because 2 and 5 multiply to 10, so their logarithms must add to 1.

$50 = 5 \times 10$ , so $lo g_{10} 50 = lo g_{10} 5 + 1 \approx 1.69897$ . Another route gives the same: $50 = \frac{100}{2}$ , so $lo g_{10} 50 = 2 - 0.30103 = 1.69897$ .

$0.4 = \frac{4}{10}$ , so $lo g_{10} 0.4 = 2 lo g_{10} 2 - 1 \approx - 0.39794$ . It is negative because 0.4 is less than 1.

Check with a calculator: $lo g_{10} 8 \approx 0.90309$ , $lo g_{10} 5 \approx 0.69897$ , $lo g_{10} 0.4 \approx - 0.39794$ . ✓ One remembered value gave four more.

Worked Example 4 — Expanding, and combining with letters

(a) Expand $lo g_{10} \frac{x ^{2} y}{z ^{3}}$ , where $x$ , $y$ and $z$ are positive. (b) Write $2 lo g_{3} x - \frac{1}{2} lo g_{3} y + 1$ as a single logarithm.

(a) The top is a product and the whole is a quotient. So the first two laws give $lo g_{10} (x^{2}) + lo g_{10} y - lo g_{10} (z^{3})$ . The third law brings each power down: $2 lo g_{10} x + lo g_{10} y - 3 lo g_{10} z$ .

(b) The "+ 1" is not a logarithm yet. Write it as one in base 3: $1 = lo g_{3} 3$ . Then the third law: $2 lo g_{3} x = lo g_{3} (x^{2})$ and $\frac{1}{2} lo g_{3} y = lo g_{3} y$ . Combine: $lo g_{3} \frac{3 x ^{2}}{y}$ .

Check (a) with $x = 2$ , $y = 5$ , $z = 1$ : the left is $lo g_{10} 20 \approx 1.30103$ , and the right is $2 \times 0.30103 + 0.69897 - 0 = 1.30103$ . ✓

Worked Example 5 — How many digits?

How many digits does $2^{64}$ have? Use $lo g_{10} 2 \approx 0.30103$ .

The number of digits depends on the power of 10 a number lies between. A number between $1 0^{19}$ and $1 0^{20}$ has 20 digits, just as a number between $1 0^{1}$ and $1 0^{2}$ has two. So find $lo g_{10} (2^{64})$ .

By the third law, $lo g_{10} (2^{64}) = 64 lo g_{10} 2 \approx 64 \times 0.30103 \approx 19.27$ . So $2^{64} \approx 1 0^{19.27}$ , which is between $1 0^{19}$ and $1 0^{20}$ . It has 20 digits.

Check: $2^{64} = 18446744073709551616$ . Count them: 20 digits. ✓ The law answered a question about a number too long to work out by hand.

Three classic mistakes

Trap 1 — Splitting the logarithm of a sum

$lo g_{10} (x + 3)$ cannot be written as $lo g_{10} x + lo g_{10} 3$ . The first law is about $lo g_{10} (3 x)$ , a product. The logarithm of a sum has no law, so leave it as it is. If an expression inside a logarithm can be factorised, factorise it first: $lo g_{10} (x^{2} + 3 x) = lo g_{10} (x (x + 3)) = lo g_{10} x + lo g_{10} (x + 3)$ .

Trap 2 — Treating a quotient of logarithms as the second law

$\frac{lo g _{10} 8}{lo g _{10} 2}$ is not $lo g_{10} 4$ . It is $\frac{3 lo g _{10} 2}{lo g _{10} 2} = 3$ , by the third law. The second law is about $lo g_{10} \frac{8}{2}$ , one logarithm of a quotient. A quotient of two logarithms is something different. The next lecture shows what it means, and why it is useful.

Trap 3 — Bringing down a power that is not on the whole inside

$lo g_{10} (3 x^{2})$ is not $2 lo g_{10} (3 x)$ . The power 2 is only on $x$ . Split the product first: $lo g_{10} 3 + 2 lo g_{10} x$ . In the same way, $(lo g_{10} x)^{2}$ is not $2 lo g_{10} x$ , because the square is outside the logarithm. And never combine logarithms with different bases: $lo g_{2} x + lo g_{3} x$ is not one logarithm.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Write as a single logarithm: (a) $lo g_{10} 4 + lo g_{10} 7$ (b) $lo g_{2} 40 - lo g_{2} 5$ , and evaluate it (c) $3 lo g_{5} 2 + lo g_{5} 3 - lo g_{5} 12$ .

Hint 1
A plus sign means multiply inside, and a minus sign means divide inside.

Hint 2
In (c), move the 3 inside first: $3 lo g_{5} 2 = lo g_{5} 8$ .

Full answer
(a) $lo g_{10} 28$ . (b) $lo g_{2} \frac{40}{5} = lo g_{2} 8 = 3$ . (c) $lo g_{5} 8 + lo g_{5} 3 - lo g_{5} 12 = lo g_{5} \frac{8 \times 3}{12} = lo g_{5} 2$ . Check (b): $lo g_{2} 40 \approx 5.3219$ and $lo g_{2} 5 \approx 2.3219$ ; the difference is 3. ✓
Without a calculator, find (a) $lo g_{3} 54 - lo g_{3} 2$ (b) $lo g_{4} 2 + lo g_{4} 32$ (c) $lo g_{10} 25 + lo g_{10} 40$ (d) $lo g_{2} 12 - lo g_{2} 3 + lo g_{2} 0.5$ .

Hint 1
Combine first. Then ask: the base to the power of what gives the number inside?

Hint 2
(b) gives $lo g_{4} 64$ . (d) gives $lo g_{2} \frac{12 \times 0.5}{3}$ .

Full answer
(a) $lo g_{3} 27 = 3$ . (b) $lo g_{4} 64 = 3$ , because $4^{3} = 64$ . (c) $lo g_{10} 1000 = 3$ . (d) $lo g_{2} 2 = 1$ . Check (b) separately: $lo g_{4} 2 = \frac{1}{2}$ and $lo g_{4} 32 = \frac{5}{2}$ , because $32 = 2^{5} = 4^{5/2}$ . The sum is 3. ✓
Given $lo g_{10} 2 \approx 0.30103$ and $lo g_{10} 3 \approx 0.47712$ , find (a) $lo g_{10} 6$ (b) $lo g_{10} 1.5$ (c) $lo g_{10} 72$ (d) $lo g_{10} 0.09$ .

Hint 1
Write each number using only 2, 3 and 10, by multiplying, dividing and powers.

Hint 2
$1.5 = \frac{3}{2}$ , $72 = 2^{3} \times 3^{2}$ , and $0.09 = \frac{3 ^{2}}{100}$ .

Full answer
(a) $0.30103 + 0.47712 = 0.77815$ . (b) $0.47712 - 0.30103 = 0.17609$ . (c) $3 \times 0.30103 + 2 \times 0.47712 = 1.85733$ . (d) $2 \times 0.47712 - 2 = - 1.04576$ . So each value is approximately these. Check with a calculator: $lo g_{10} 72 \approx 1.85733$ and $lo g_{10} 0.09 \approx - 1.04576$ . ✓ The answer to (d) is negative because 0.09 is less than 1.
The loudness of a sound is measured in decibels (dB). A sound with intensity $I$ has a level $L = 10 lo g_{10} \frac{I}{I _{0}}$ , where $I_{0}$ is a fixed, very quiet reference intensity. (a) Show that doubling the intensity always adds about 3 dB, whatever the starting level. (b) One machine makes 70 dB. Two identical machines run together, so the intensity doubles. What is the level? (c) How many of these machines are needed to reach 90 dB?

Hint 1
(a) Write the new level $10 lo g_{10} \frac{2 I}{I _{0}}$ and use the first law to split off the 2.

Hint 2
(c) Going from 70 dB to 90 dB adds 20 dB. Each factor of 10 in intensity adds $10 lo g_{10} 10 = 10$ dB.

Full answer
(a) New level $= 10 lo g_{10} (2 \times \frac{I}{I _{0}}) = 10 lo g_{10} 2 + 10 lo g_{10} \frac{I}{I _{0}} = 10 lo g_{10} 2 + L$ . The increase is $10 lo g_{10} 2 \approx 3.0$ dB, and $L$ does not appear in it, so it is the same from any starting level. (b) About 73 dB, not 140 dB. (c) $90 - 70 = 20$ dB, and $20 = 10 lo g_{10} 100$ . So the intensity must be 100 times bigger: 100 machines. Check: $70 + 10 lo g_{10} 100 = 70 + 20 = 90$ . ✓ This is why the decibel scale uses a logarithm. Multiplying the intensity by the same factor always adds the same number of decibels.

The bigger idea

The three laws of logarithms are the three index laws, read in reverse. The index laws say what happens to the indices when you multiply, divide or raise powers. A logarithm is an index, so the logarithm laws say the same thing in the other direction. This is what an inverse function does. It undoes the original, so its rules mirror the original's rules.

The laws changed how science was done. Napier published his first table in 1614, and Briggs's base-10 tables followed. Kepler saw Napier's work a few years later and welcomed it. His great planetary tables of 1627, the Rudolphine Tables, included tables of logarithms. In 1620 Edmund Gunter marked a logarithmic scale on a ruler and added lengths with a pair of dividers. Around 1622 William Oughtred slid two such scales against each other. That was the first slide rule: the explorer above, made of wood. Slide rules stayed in daily use until electronic calculators replaced them in the 1970s.

On a logarithmic scale, equal steps mean equal ratios, not equal differences. That is why decibels, pH and earthquake magnitudes use one. The next lecture uses the third law for its most important job: bringing an unknown down from the index, to solve equations such as $3^{x} = 17$ .

Key concepts in this lesson

The logarithm of a product is the sum of the logarithms, $lo g_{a} (A B) = lo g_{a} A + lo g_{a} B$ , because multiplying powers adds their indices.
The logarithm of a quotient is the difference of the logarithms, $lo g_{a} \frac{A}{B} = lo g_{a} A - lo g_{a} B$ , because dividing powers subtracts their indices.
The logarithm of a power brings the power to the front, $lo g_{a} (A^{n}) = n lo g_{a} A$ , because a power of a power multiplies the indices.
For any allowed base, $lo g_{a} a = 1$ and $lo g_{a} 1 = 0$ , and so $lo g_{a} \frac{1}{B} = - lo g_{a} B$ .
There is no law for the logarithm of a sum or a difference, or for a product, quotient or power of whole logarithms, and one counterexample with numbers shows it.
Any law you forget can be rebuilt in three lines by naming the logarithms as indices and using the index law.