Algebra 2 · F.BF.B.5 · F.LE.A.4

Laws of logarithms

Multiplying powers of the same base means adding their exponents, and a logarithm is an exponent. So what is the logarithm of a product, a quotient or a power?

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Definition of a logarithmAlgebra 2

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Change of baseAlgebra 2 Expanding and condensing logarithmsAlgebra 2 Solving by taking a logarithm of both sidesAlgebra 2 Log-log transforms and power modelsStatistics Solving exponential equations by matching basesAlgebra 2 Solving logarithmic equationsAlgebra 2 Derivatives of logarithms and logarithmic differentiationCalculus 1 +4 more

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Logarithms11

▴ 1 above Definition of a logarithm Expanding and condensing logarithms

Laws of logarithms

Solving by taking a logarithm of both sides Solving exponential equations by matching bases ▾ Show all 11 (5 more)

Why this exists

Multiplying is harder than adding. Adding two long numbers is a column of single-digit sums; multiplying them is a grid of products that then has to be added up as well. Raising a number to a high power is harder again. Before electronic calculators, people did long multiplications by looking up the logarithms of the two numbers in printed tables, adding them, and looking the sum up backwards. The slide rule does the same thing with lengths. Both rest on one fact: the logarithm of a product is the sum of the logarithms.

That fact still matters now that calculators do the arithmetic. It is what lets a logarithm be broken into simpler pieces, what lets an unknown exponent be brought down to where it can be solved for, and what turns curving exponential data into a straight line. All of it comes from three laws, and each of the three is a law of exponents read through the definition of a logarithm.

The lesson is taught over two days. The first takes the product and quotient laws, the second the power law and what it makes possible.

Definition

Let b be an allowed base (b > 0 and b \neq 1), let M and N be positive numbers, and let p be any real number. The laws of logarithms are these three.

The product law: the logarithm of a product is the sum of the logarithms of the factors.

\log_b(MN) = \log_b M + \log_b N

The quotient law: the logarithm of a quotient is the logarithm of the top minus the logarithm of the bottom.

\log_b \frac{M}{N} = \log_b M - \log_b N

The power law: the logarithm of a power is the exponent times the logarithm of the base of the power.

\log_b\left(M^p\right) = p \log_b M

All the logarithms in one law have the same base b. Each law can be read in either direction: left to right it breaks one logarithm into simpler ones, right to left it gathers several into one.

Intuition

A logarithm is an exponent. Write 4 and 8 as powers of 2: 4 = 2^2 and 8 = 2^3. Their product is

\begin{aligned} 4 \cdot 8 &= 2^2 \cdot 2^3 \\ &= 2^{2+3} = 2^5 = 32 \end{aligned}

because powers with the same base multiply by adding exponents. The exponents are the logarithms, \log_2 4 = 2 and \log_2 8 = 3, and they added to give \log_2 32 = 5. Multiplying the numbers added their logarithms.

The picture makes this a matter of lengths. Put each power of 2 on a line at a distance from 1 equal to its logarithm: 1 at 0, 2 at 1, 4 at 2, 8 at 3, and so on.

Figure 1Each number sits at its logarithm. Lengths 2 and 3 laid end to end make 5, and the numbers 4 and 8 multiply to 32.

To multiply 4 by 8, start at 4 and step on by the length that 8 sits from 1, which is 3. You land at 32. On this ruler, multiplying is laying lengths end to end, which is adding. Dividing is the reverse: stepping back from 32 by the length of 8 lands at 4, so \displaystyle \frac{32}{8} = 4 and \log_2 32 - \log_2 8 = \log_2 4. And squaring 8 is laying the length of 8 down twice, so \log_2\left(8^2\right) = 2 \log_2 8 = 6, which is where 64 sits. The ruler also works between the ticks, since every positive number has a logarithm; a slide rule is two such rulers sliding against each other.

Results

The three laws. For an allowed base b, positive M and N, and any real p:

\log_b(MN) = \log_b M + \log_b N

\log_b \frac{M}{N} = \log_b M - \log_b N

\log_b\left(M^p\right) = p \log_b M

Many factors. Using the product law again and again, for any number of positive factors,

\begin{aligned} &\log_b(M_1 M_2 \cdots M_k) \\ &\quad = \log_b M_1 + \log_b M_2 + \cdots + \log_b M_k \end{aligned}

A reciprocal. Putting M = 1 in the quotient law, and using \log_b 1 = 0,

\log_b \frac{1}{N} = -\log_b N

A root. A root is a power with a unit fraction exponent, \displaystyle \sqrt[n]{M} = M^{\frac{1}{n}}, so the power law gives

\log_b \sqrt[n]{M} = \frac{1}{n} \log_b M

These sit beside the facts that come straight from the definition: \log_b 1 = 0, \log_b b = 1, \log_b\left(b^y\right) = y and b^{\log_b x} = x.

Why it is true

Each law is a law of exponents in disguise. The proofs all start the same way: give the logarithms names, and switch to exponential form.

Let m = \log_b M and n = \log_b N. These exist because M and N are positive. In exponential form they say

b^m = M \qquad b^n = N

The product law. Multiply the two equations. Powers with the same base multiply by adding exponents (the product of powers law), so

MN = b^m \cdot b^n = b^{m+n}

In logarithmic form this says the exponent that turns b into MN is m + n:

\log_b(MN) = m + n = \log_b M + \log_b N

The quotient law. Divide instead. Powers with the same base divide by subtracting exponents (the quotient of powers law), so

\frac{M}{N} = \frac{b^m}{b^n} = b^{m-n}

and in logarithmic form

\log_b \frac{M}{N} = m - n = \log_b M - \log_b N

The power law. Raise b^m = M to the power p. A power of a power multiplies the exponents (the power of a power law, which holds for every real p because b is positive), so

M^p = \left(b^m\right)^p = b^{mp}

In logarithmic form,

\log_b\left(M^p\right) = mp = p \log_b M

In each proof the only step that is not a rewrite between the two forms is a law of exponents. Products of powers add exponents, so logarithms of products add; quotients of powers subtract exponents, so logarithms of quotients subtract; powers of powers multiply exponents, so the logarithm of a power is a multiple. \blacksquare

Worked examples

Examples 1 to 5 belong to the first day: the product and quotient laws, used on numbers, on known values and on expressions. Examples 6 to 9 belong to the second: the power law, and what it does for numbers too large to write out.

Example 1

Two logarithms into one, and one into two

(a) Find \log_2(8 \cdot 32) in two ways: by the product law, and by working out the product first.

(b) Find \log_6 4 + \log_6 9.

Solution

(a) By the product law, \log_2(8 \cdot 32) = \log_2 8 + \log_2 32. Since 2^3 = 8 and 2^5 = 32, these are 3 and 5, so

\begin{aligned} \log_2(8 \cdot 32) &= \log_2 8 + \log_2 32 \\ &= 3 + 5 = 8 \end{aligned}

The other way: 8 \cdot 32 = 256, and 2^8 = 256, so \log_2 256 = 8. The two ways agree, as the law says they must.

(b) Neither logarithm is a nice number on its own: no whole power of 6 is 4 or 9. Read the product law from right to left instead. Both logarithms have base 6, so they combine into the logarithm of the product:

\begin{aligned} \log_6 4 + \log_6 9 &= \log_6(4 \cdot 9) \\ &= \log_6 36 = 2 \end{aligned}

because 6^2 = 36. Two awkward logarithms made one simple one.

Example 2

The quotient law on numbers

Find each value.

(a) \log_3 162 - \log_3 2 (b) \log 5000 - \log 5 (c) \log_2 3 - \log_2 24

Solution

In each, both logarithms have the same base, so the quotient law gathers the difference into one logarithm, of the first argument divided by the second.

(a) \displaystyle \log_3 162 - \log_3 2 = \log_3 \frac{162}{2} = \log_3 81 = 4, since 3^4 = 81.

(b) The common logarithm has base 10 in both terms, so \displaystyle \log 5000 - \log 5 = \log \frac{5000}{5} = \log 1000 = 3, since 10^3 = 1000.

(c) Here the larger argument is on the bottom: \displaystyle \log_2 3 - \log_2 24 = \log_2 \frac{3}{24} = \log_2 \frac{1}{8} = -3, since \displaystyle 2^{-3} = \frac{1}{8}. A quotient below 1 gives a negative logarithm, which matches the subtraction of a larger logarithm from a smaller one.

Example 3

Building new logarithms from two known ones

Given \log 2 \approx 0.30103 and \log 3 \approx 0.47712, find \log 6, \log 1.5 and \log 20 without a calculator.

Solution

Write each argument as a product or quotient of 2, 3 and 10, whose common logarithms are known (\log 10 = 1 because 10^1 = 10).

Since 6 = 2 \cdot 3, the product law gives

\begin{aligned} \log 6 &= \log 2 + \log 3 \\ &\approx 0.30103 + 0.47712 = 0.77815 \end{aligned}

Since \displaystyle 1.5 = \frac{3}{2}, the quotient law gives

\begin{aligned} \log 1.5 &= \log 3 - \log 2 \\ &\approx 0.47712 - 0.30103 = 0.17609 \end{aligned}

Since 20 = 2 \cdot 10, the product law gives

\begin{aligned} \log 20 &= \log 2 + \log 10 \\ &\approx 0.30103 + 1 = 1.30103 \end{aligned}

Each agrees with a calculator to all five places shown. A table of logarithms can be filled in the same way: once the logarithms of the primes are known, those of every other whole number follow by addition.

Example 4

Laws on unknown numbers

Suppose \log_b M = 3.2 and \log_b N = 1.7 for some allowed base b and positive numbers M and N. Find \log_b(MN), \displaystyle \log_b \frac{M}{N}, \displaystyle \log_b \frac{N}{M} and \log_b(bM).

Solution

Nothing about b, M or N is needed beyond the two logarithms, because each law expresses the new logarithm through the old ones.

By the product law, \log_b(MN) = 3.2 + 1.7 = 4.9.

By the quotient law, \displaystyle \log_b \frac{M}{N} = 3.2 - 1.7 = 1.5, and \displaystyle \log_b \frac{N}{M} = 1.7 - 3.2 = -1.5. Swapping top and bottom changes the sign, since \displaystyle \frac{N}{M} is the reciprocal of \displaystyle \frac{M}{N}.

By the product law and \log_b b = 1, \log_b(bM) = \log_b b + \log_b M = 1 + 3.2 = 4.2. Multiplying by the base adds exactly 1 to the logarithm.

Example 5

The laws on expressions

(a) Write \log_7(5x) as a sum of logarithms.

(b) Write \displaystyle \log_4 \frac{y}{16} as a difference, and simplify.

Solution

(a) The argument is the product of 5 and x. For x > 0 both factors are positive, and the product law gives

\log_7(5x) = \log_7 5 + \log_7 x

(b) For y > 0, the quotient law gives \displaystyle \log_4 \frac{y}{16} = \log_4 y - \log_4 16 = \log_4 y - 2, since 4^2 = 16.

\begin{aligned} \ln x + \ln(x + 2) &= \ln\big(x(x + 2)\big) \\ &= \ln\left(x^2 + 2x\right) \end{aligned}

The law needs both factors positive, x > 0 and x + 2 > 0, which together say x > 0. For those x the two sides are equal. The single logarithm on the right is defined for more values: at x = -3, for instance, x^2 + 2x = 3 is positive, but \ln(-3) on the left has no real value. So the equation holds for x > 0 and only there.

Example 6

The power law on numbers

Find each value.

(a) \log_5\left(25^4\right) (b) \displaystyle \log_4 \sqrt[3]{16} (c) \displaystyle \log_3 \frac{1}{9^4}

Solution

(a) By the power law, the exponent 4 comes out in front: \log_5\left(25^4\right) = 4 \log_5 25 = 4 \cdot 2 = 8. As a check, 25 = 5^2, so 25^4 = 5^8.

(b) A cube root is the power \displaystyle \frac{1}{3}, so

\begin{aligned} \log_4 \sqrt[3]{16} &= \log_4\left(16^{\frac{1}{3}}\right) = \frac{1}{3} \log_4 16 \\ &= \frac{1}{3} \cdot 2 = \frac{2}{3} \end{aligned}

As a check, 16 = 4^2, so \displaystyle \sqrt[3]{16} = \left(4^2\right)^{\frac{1}{3}} = 4^{\frac{2}{3}}.

\begin{aligned} \log_3 \frac{1}{9^4} &= -\log_3\left(9^4\right) = -4 \log_3 9 \\ &= -4 \cdot 2 = -8 \end{aligned}

As a check, 9^4 = 3^8, and \displaystyle \frac{1}{3^8} = 3^{-8}.

Example 7

More logarithms from the same two

Given \log 2 \approx 0.30103 and \log 3 \approx 0.47712, find \log 8, \displaystyle \log \sqrt{3} and \log 72.

Solution

Since 8 = 2^3, the power law gives \log 8 = 3 \log 2 \approx 3(0.30103) = 0.90309.

Since \displaystyle \sqrt{3} = 3^{\frac{1}{2}}, the power law gives \displaystyle \log \sqrt{3} = \frac{1}{2} \log 3 \approx \frac{0.47712}{2} = 0.23856.

Since 72 = 8 \cdot 9 = 2^3 \cdot 3^2, the product law and then the power law give

\begin{aligned} \log 72 &= \log\left(2^3\right) + \log\left(3^2\right) \\ &= 3 \log 2 + 2 \log 3 \\ &\approx 0.90309 + 0.95424 = 1.85733 \end{aligned}

Every whole number built from the primes 2 and 3 has a logarithm that is a combination of these two.

Example 8

The power law on expressions

(a) Write \log_b\left(x^5\right) without the power. (b) Write 3 \ln t as a single logarithm. (c) Write \displaystyle \log_2 \sqrt{w} as a multiple of \log_2 w.

Solution

(a) For x > 0, the power law gives \log_b\left(x^5\right) = 5 \log_b x.

(b) Read the power law from right to left: a number in front of a logarithm becomes an exponent on its argument. For t > 0, 3 \ln t = \ln\left(t^3\right).

The condition matters in (a). If x were negative, x^5 would be negative and neither side would be defined; for an even power such as x^2 the left side is defined for negative x but the right is not, a case taken up under Boundaries.

Example 9

How many digits?

The number 6^{50} is far too long for a calculator display. Using \log 6 \approx 0.77815 from Example 3, find how many digits it has.

Solution

A whole number has 39 digits exactly when it is at least 10^{38} and less than 10^{39}; in general, the number of digits is fixed by which two powers of 10 it lies between. The common logarithm reads that off, because \log is increasing and \log\left(10^k\right) = k.

By the power law,

\log\left(6^{50}\right) = 50 \log 6 \approx 50(0.77815) = 38.9075

So 6^{50} = 10^{38.9075\ldots}, which lies between 10^{38} and 10^{39}. The number has 39 digits.

The logarithm says more. Since 10^{38.9075} = 10^{0.9075} \cdot 10^{38} by the product of powers law, and 10^{0.9075} \approx 8.08, the number is about 8.08 \times 10^{38}. Written out in full, 6^{50} has 39 digits and begins 808.

Boundaries

Every argument must be positive, and every base allowed. The laws are proved for positive M and N, because only then do \log_b M and \log_b N exist. A product can be positive when its factors are not: (-4)(-2) = 8, and \log_2 8 = 3, but \log_2(-4) and \log_2(-2) have no real value, so the product law has nothing to say there.

With variables, the two sides may have different domains. In Example 5, the single logarithm \ln\left(x^2 + 2x\right) is defined for x < -2 as well, while \ln x + \ln(x + 2) is not. An identity between logarithms holds where every logarithm in it is defined. This is exactly why solving logarithmic equations needs every answer checked against the original equation.

Even powers and negative bases of powers. The power law needs M > 0. For x \neq 0, x^2 = |x|^2, and |x| is positive, so the correct statement for every nonzero x is

\log_b\left(x^2\right) = 2 \log_b |x|

Without the absolute value it fails at x = -3, where the left side is \log_b 9 and the right side does not exist.

There is no law for a sum or a difference. The laws turn products, quotients and powers of arguments into sums, differences and multiples of logarithms. Nothing turns \log_b(M + N) into an expression in \log_b M and \log_b N, because there is no law of exponents for b^m + b^n. For example \log_2(8 + 8) = \log_2 16 = 4, while \log_2 8 + \log_2 8 = 6, which is \log_2(8 \cdot 8).

Nor for products of logarithms. The power law moves an exponent on the argument. A power of the whole logarithm is something else: (\log_2 8)^2 = 3^2 = 9, while \log_2\left(8^2\right) = 2 \log_2 8 = 6. Likewise \log_b M \cdot \log_b N is not the logarithm of any simple combination of M and N.

One base at a time. \log_2 8 + \log_3 9 cannot be combined by the product law, because the two logarithms are exponents on different bases. Moving between bases is the job of Change of base.

Where this goes next

Expanding and condensing logarithms applies the laws several times in a row, breaking a logarithm of a product of powers into its simplest pieces, or gathering many logarithms into one. The power law is the key move in taking a logarithm of both sides: in an equation such as 5^x = 30, the logarithm of 5^x is x \log 5, and the unknown is no longer in an exponent. The quotient and power laws together give Change of base, which turns a logarithm in any base into common or natural logarithms. Gathering two logarithms into one is the first step in solving logarithmic equations, where the domain conditions above decide which answers survive.

The product law is also why Logarithmic scales work: multiplying a quantity by 10 adds 1 to its common logarithm, so equal steps on the scale are equal factors in the quantity. Taking logarithms of data turns an exponential pattern into a straight line in semi-log plots, and a power law into a line in log-log plots. In calculus the laws make logarithmic differentiation possible: a product of many factors becomes a sum before it is differentiated.

Exercises

Exercises 1 to 14 go with the first day (Examples 1 to 5), Exercises 15 to 25 with the second (Examples 6 to 9), and the reasoning group draws on both. In Exercises 7 to 10 and 19 to 21, \log_b 2 = p and \log_b 5 = q for some allowed base b.

Day 1: combine and find the value

\log_3 6 + \log_3 13.5
\log_4 2 + \log_4 32
\log_2 96 - \log_2 3
\log 40 + \log 25
\log_5 10 - \log_5 250
\ln(2e) - \ln 2

Day 1: in terms of p and q

\log_b 10
\log_b 2.5
\log_b 0.4
\log_b(10b)

Day 1: one step with expressions

Use one law once. Say for which values of the variable the two sides are equal.

Write \log_6(7y) as a sum.
Write \displaystyle \ln \frac{x}{4} as a difference.
Write \log_2 x + \log_2 9 as a single logarithm.
Write \log_5(x + 3) - \log_5 x as a single logarithm.

Day 2: the power law

\log_2\left(32^3\right)
\log_{16}\left(2^{12}\right)
\displaystyle \log_5 \sqrt[4]{125}
\log\left(0.01^5\right)
\log_b 8, in terms of p
\log_b 50, in terms of p and q
\displaystyle \log_b \sqrt{5}, in terms of q
Write 4 \log_3 x as a single logarithm.
Write \displaystyle \frac{1}{2} \ln y as a single logarithm.

Day 2: large numbers

Use \log 2 \approx 0.30103 and \log 3 \approx 0.47712.

How many digits does 2^{64} have?
Which is larger, 2^{40} or 3^{25}? Decide by comparing their common logarithms.

Reasoning

Prove the quotient law a second way, from the product law and the power law, by writing \displaystyle \frac{M}{N} = M \cdot N^{-1}.
Work out \log_2(8 + 8) and \log_2 8 + \log_2 8. Which logarithm is the second one really equal to, and why can no law turn \log_b(M + N) into logarithms of M and N?
Explain why \log_3\left(x^2\right) = 2 \log_3 x fails at x = -3, and give a statement that holds for every nonzero x.
Show that \displaystyle \log_b \frac{1}{N} = -\log_b N for positive N in two ways: once with the quotient law, once with the power law.

Answers

4. \log_3(6 \cdot 13.5) = \log_3 81 = 4.
3. \log_4(2 \cdot 32) = \log_4 64 = 3, since 4^3 = 64.
5. \displaystyle \log_2 \frac{96}{3} = \log_2 32 = 5.
3. \log(40 \cdot 25) = \log 1000 = 3.
-2. \displaystyle \log_5 \frac{10}{250} = \log_5 \frac{1}{25} = -2, since \displaystyle 5^{-2} = \frac{1}{25}.
1. \displaystyle \ln \frac{2e}{2} = \ln e = 1.
p + q. 10 = 2 \cdot 5, so \log_b 10 = \log_b 2 + \log_b 5.
q - p. \displaystyle 2.5 = \frac{5}{2}, so \log_b 2.5 = \log_b 5 - \log_b 2.
p - q. \displaystyle 0.4 = \frac{2}{5}, so \log_b 0.4 = \log_b 2 - \log_b 5.
p + q + 1. \log_b(10b) = \log_b 10 + \log_b b, and \log_b b = 1.
\log_6 7 + \log_6 y, for y > 0.
\ln x - \ln 4, for x > 0.
\log_2(9x), for x > 0.
\displaystyle \log_5 \frac{x + 3}{x}, for x > 0. The left side needs x + 3 > 0 and x > 0, so x > 0. The single logarithm is also defined for x < -3, where the left side is not.
15. 3 \log_2 32 = 3 \cdot 5 = 15.
3. \displaystyle 12 \log_{16} 2 = 12 \cdot \frac{1}{4} = 3, since \displaystyle 16^{\frac{1}{4}} = 2.
\displaystyle \frac{3}{4}. \displaystyle \log_5\left(125^{\frac{1}{4}}\right) = \frac{1}{4} \log_5 125 = \frac{1}{4} \cdot 3.
-10. 5 \log 0.01 = 5 \cdot (-2) = -10.
3p. \log_b\left(2^3\right) = 3 \log_b 2.
p + 2q. \log_b(2 \cdot 5^2) = \log_b 2 + 2 \log_b 5.
\displaystyle \frac{q}{2}. \displaystyle \log_b\left(5^{\frac{1}{2}}\right) = \frac{1}{2} \log_b 5.
\log_3\left(x^4\right), for x > 0.
\displaystyle \ln \sqrt{y}, for y > 0. The \displaystyle \frac{1}{2} becomes the exponent: \displaystyle \ln\left(y^{\frac{1}{2}}\right).
20 digits. \log\left(2^{64}\right) = 64 \log 2 \approx 19.26592, so 2^{64} lies between 10^{19} and 10^{20}.
2^{40} is larger. \log\left(2^{40}\right) \approx 40(0.30103) = 12.0412 and \log\left(3^{25}\right) \approx 25(0.47712) = 11.928. The common logarithm is increasing, so the number with the larger logarithm is larger.
By the product law, \log_b\left(M \cdot N^{-1}\right) = \log_b M + \log_b\left(N^{-1}\right). By the power law with p = -1, \log_b\left(N^{-1}\right) = -\log_b N, so the sum is \log_b M - \log_b N. Both laws apply because M and N^{-1} are positive.
\log_2(8 + 8) = \log_2 16 = 4, while \log_2 8 + \log_2 8 = 6 = \log_2 64, the logarithm of 8 \cdot 8. A law for \log_b(M + N) would need a law of exponents for b^m + b^n, and a sum of two powers is not a single power of b with an exponent built from m and n.
At x = -3 the left side is \log_3 9 = 2, but \log_3(-3) has no real value, so the right side does not exist. The power law needs a positive argument. Since x^2 = |x|^2, the true statement is \log_3\left(x^2\right) = 2 \log_3 |x| for every x \neq 0.
Quotient law: \displaystyle \log_b \frac{1}{N} = \log_b 1 - \log_b N = 0 - \log_b N. Power law: \displaystyle \frac{1}{N} = N^{-1}, so \log_b\left(N^{-1}\right) = -1 \cdot \log_b N.