Properties of Logarithms

Learning goals

Read every rule as an exponent law run backwards
Apply the product, quotient and power rules with their domains
Change base with $\log_b M = \tfrac{\log_a M}{\log_a b}$
Expand and condense logarithmic expressions using all three rules
Use $\log_{10}$ to count the digits of a huge power

The mirror behind every rule

Everything here rests on the definition you already have. For a base $b$ with $b > 0$ and $b \ne 1$ , and for $M > 0$ ,

\log_b M = x \quad\text{means exactly}\quad b^x = M.

A logarithm is an exponent. It is the exponent you must put on $b$ to reach $M$ . Two consequences follow immediately and get used constantly: $b^{\log_b M} = M$ for every $M > 0$ , and $\log_b (b^x) = x$ for every real $x$ . Those two say that raising $b$ to a power and taking a base- $b$ logarithm are inverse operations, each undoing the other.

Here is what that gives you with actual numbers. Since $8 = 2^3$ and $4 = 2^2$ , their product is $8 \cdot 4 = 2^{3+2} = 2^5 = 32$ . Multiplying the numbers matched adding their exponents, $3$ and $2$ . But $3$ and $2$ are also the base-two logarithms of $8$ and $4$ , so that same addition is $\log_2 8 + \log_2 4 = 3 + 2 = 5 = \log_2 32$ . Multiplying two numbers matched adding their logarithms. That one example is the whole idea of this lesson. The three rules below just say it for any base and any numbers, not only $2$ , $8$ , and $4$ .

To see why it always works, recall the exponent laws, which hold for every real exponent whenever the base is positive:

b^x \cdot b^y = b^{x+y}, \qquad \frac{b^x}{b^y} = b^{x-y}, \qquad \left(b^x\right)^p = b^{xp}.

Each of the three logarithm rules is one of these laws, read backwards. That gives us a single proof strategy, and we will use it three times without variation:

Name the logarithms. Put $x = \log_b M$ and $y = \log_b N$ .
Translate to exponents. The definition says $b^x = M$ and $b^y = N$ .
Apply the matching exponent law to combine them.
Translate back by taking $\log_b$ of both sides.

The exponent laws and the logarithm rules line up exactly, one for one:

Exponent law	Logarithm rule it becomes
$b^x \cdot b^y = b^{x+y}$	$\log_b(MN) = \log_b M + \log_b N$
$\dfrac{b^x}{b^y} = b^{x-y}$	$\log_b\!\left(\dfrac{M}{N}\right) = \log_b M - \log_b N$
$\left(b^x\right)^p = b^{xp}$	$\log_b\!\left(M^p\right) = p \log_b M$

Multiplication on the inside becomes addition on the outside, division becomes subtraction, and an exponent becomes a multiplier. One level of difficulty is stripped away every time.

The product rule

For $b > 0$ with $b \ne 1$ , and for $M > 0$ and $N > 0$ :

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

The product rule#

Let $x = \log_b M$ and $y = \log_b N$ . Both of these logarithms exist precisely because $M$ and $N$ are positive, and by the definition of a logarithm the two statements say exactly that $b^x = M$ and $b^y = N$ .

Multiply those two equations together. The left sides give $MN$ and the right sides give $b^x \cdot b^y$ , so $MN = b^x \cdot b^y$ . The exponent law for a product of powers with the same base collapses the right side into a single power, $b^x \cdot b^y = b^{x+y}$ , and therefore

MN = b^{\,x+y}.

Read that last line back through the definition of a logarithm. It says that the exponent you must put on $b$ to reach $MN$ is $x + y$ , which is to say $\log_b(MN) = x + y$ . (The left side is legitimate: $M$ and $N$ are positive, so $MN$ is positive too.) Substituting back what $x$ and $y$ stood for,

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

The condition $M > 0$ and $N > 0$ is not decorative, and it is stronger than requiring the product to be positive. Take $M = N = -2$ . Their product is $4$ , so the left side $\log_b 4$ is a perfectly good number, but $\log_b(-2)$ does not exist, so the right side is not merely wrong, it is meaningless. The rule needs each factor positive, not just their product.

Here is what the product rule looks like as a picture. Mark the numbers $1$ through $10$ on a ruler, but place each number $n$ at a distance proportional to $\log_{10} n$ from the left end instead of at its own value. On that ruler, distances are logarithms, so adding two logarithms means laying two distances end to end.

On a logarithmic scale the distance from 1 to a number n is proportional to the base-10 logarithm of n. Laying the distance for 2 and the distance for 3 end to end lands exactly on 6, because log 2 + log 3 = log 6. Sliding one such scale along another is how a slide rule multiplies.

The second bar is the distance from $1$ to $3$ , but it has been shifted so it starts at $2$ . It ends at $6$ . Multiplying by $3$ always moves you the same distance to the right on this scale, no matter where you start, and that fixed distance is $\log_{10} 3$ . A slide rule is exactly two of these scales, and sliding one against the other adds the distances for you.

The quotient rule

For $b > 0$ with $b \ne 1$ , and for $M > 0$ and $N > 0$ :

\log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N.

The quotient rule#

This is the product-rule proof with one symbol changed. Let $x = \log_b M$ and $y = \log_b N$ , so $b^x = M$ and $b^y = N$ . Divide the two equations instead of multiplying them:

\frac{M}{N} = \frac{b^x}{b^y} = b^{\,x-y},

using the exponent law for a quotient of powers with the same base. ( $N \ne 0$ since $N > 0$ , so the division is legal, and $M/N > 0$ , so its logarithm exists.) Reading $M/N = b^{x-y}$ back through the definition gives $\log_b(M/N) = x - y$ , which is

\log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N.

One special case is worth keeping in your head. Put $M = 1$ and use $\log_b 1 = 0$ :

\log_b\!\left(\frac{1}{N}\right) = \log_b 1 - \log_b N = -\log_b N.

Taking a reciprocal on the inside flips the sign on the outside. That single fact converts every division into a subtraction and every “one over” into a minus sign.

Check your understanding

Which single logarithm is equal to $\log_7 45 - \log_7 9$ ?

\log_7 36

\log_7 5

\log_7 405

\dfrac{\log_7 45}{\log_7 9}

The power rule

For $b > 0$ with $b \ne 1$ , for $M > 0$ , and for any real number $p$ :

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

The power rule#

Same pattern once more. Let $x = \log_b M$ , which exists because $M > 0$ , so $b^x = M$ . Raise both sides to the power $p$ :

M^p = \left(b^x\right)^p = b^{\,px},

using the exponent law for a power of a power, which holds for every real exponent because $b > 0$ . Since $b^{px}$ is positive, $M^p$ is positive and its logarithm exists. Reading the line back through the definition gives $\log_b\!\left(M^p\right) = px$ , and substituting $x = \log_b M$ gives the rule:

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

Notice what the power rule does: an exponent trapped inside a logarithm walks out to the front as a plain multiplier. That is the property that made logarithms indispensable, because it turns the brutal operation of raising to a power into ordinary multiplication.

The domain condition here is the one students lose. The rule requires $M > 0$ . Consider the familiar-looking claim

\log_b\!\left(x^2\right) = 2 \log_b x.

This is true for every $x > 0$ , and it is false for every $x < 0$ , for a reason that is easy to miss: when $x$ is negative, $x^2$ is still positive. The left side is therefore a perfectly good number, but $\log_b x$ does not exist for a negative $x$ , so the right side is not a number at all. At $x = 0$ neither side exists. So the equation holds exactly when $x > 0$ , no more and no less.

The repair is an absolute value. For any $x \ne 0$ we have $x^2 = |x|^2$ with $|x| > 0$ , so the power rule applies to $|x|$ and gives the identity that is true for every nonzero $x$ :

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

Check it on $x = -3$ : the left side is $\log_b 9$ , and the right side is $2\log_b 3 = \log_b 9$ . They agree, and neither one asks you to take the logarithm of a negative number.

Check your understanding

For which values of $x$ is the statement $\log_5\!\left(x^2\right) = 2\log_5 x$ true?

A every real

x

B every

x \ne 0

C every

x > 0

D no value of

x

Expanding and condensing

Two skills come out of the three rules, and they are the same skill run in opposite directions.

Expanding takes one logarithm of a complicated expression and breaks it into a sum and difference of simpler logarithms. Work from the outside in: split the top and bottom of a fraction first (quotient rule). Then split the products (product rule), and then bring exponents down to the front (power rule). Roots are exponents in disguise, so $\sqrt{y} = y^{1/2}$ before you start.

Condensing runs the film backwards, aiming for one logarithm with no coefficient left outside it. The product and quotient rules combine two bare logarithms, $\log_b M \pm \log_b N$ ; they say nothing about a coefficient in front. So the first step is always the power rule: send every coefficient back up as an exponent, turning each term into a bare logarithm. Only then can the product and quotient rules add everything into a single numerator and subtract the rest into a single denominator.

Throughout, assume every variable inside a logarithm is positive. Textbooks say this so often that they stop saying it out loud, but you now know why they must.

Worked example 1 Evaluate $\log_2 6 + \log_2 \tfrac{2}{3}$ without a calculator

Neither logarithm is a whole number on its own, so evaluating them separately is hopeless. Combine them first with the product rule.

\log_2 6 + \log_2 \frac{2}{3} = \log_2\!\left(6 \cdot \frac{2}{3}\right) = \log_2 4.

Now the question is what exponent turns $2$ into $4$ , and the answer is $2$ , since $2^2 = 4$ :

\log_2 6 + \log_2 \frac{2}{3} = 2.

The ugly numbers were never the point. The rule let them cancel before we ever had to face them.

Worked example 2 Expand $\log_5\!\left(\dfrac{25x^3}{\sqrt{y}}\right)$ for $x, y > 0$

Start with the fraction, using the quotient rule to split numerator from denominator:

\log_5\!\left(\frac{25x^3}{\sqrt{y}}\right) = \log_5\!\left(25x^3\right) - \log_5 \sqrt{y}.

The numerator is a product, so the product rule splits it again, and the denominator is a root, which is the power $y^{1/2}$ :

= \log_5 25 + \log_5\!\left(x^3\right) - \log_5\!\left(y^{1/2}\right).

Now the power rule brings both exponents to the front, and $\log_5 25 = 2$ because $5^2 = 25$ :

\log_5\!\left(\frac{25x^3}{\sqrt{y}}\right) = 2 + 3\log_5 x - \frac{1}{2}\log_5 y.

Every logarithm left in the answer is as simple as it can be, which is what “expand” asks for.

Worked example 3 Condense $\tfrac{1}{2}\log_b x + 3\log_b y - 2\log_b z$ into one logarithm

We want a single logarithm with no coefficient in front, and the product and quotient rules only combine bare logarithms. So clear the coefficients first by running the power rule backwards:

\frac{1}{2}\log_b x + 3\log_b y - 2\log_b z = \log_b\!\left(x^{1/2}\right) + \log_b\!\left(y^3\right) - \log_b\!\left(z^2\right).

The two logarithms being added combine into a product, and the one being subtracted goes into the denominator:

= \log_b\!\left(\frac{x^{1/2} \cdot y^3}{z^2}\right) = \log_b\!\left(\frac{y^3\sqrt{x}}{z^2}\right).

A single logarithm, as requested. Reversing the steps expands it back to where we started, which is a good way to check your work.

Check your understanding

Condense $3\log_b x - \log_b y$ into a single logarithm.

\log_b(3x - y)

\log_b\!\left(\dfrac{x^3}{y}\right)

\log_b\!\left(\dfrac{3x}{y}\right)

3\log_b\!\left(\dfrac{x}{y}\right)

Worked example 4 How a log table multiplies, using $\log_{10} 2 \approx 0.3010$ and $\log_{10} 3 \approx 0.4771$

Those two entries are enough to compute the logarithm of any number built from $2$ s and $3$ s. That is exactly how a four-figure table was used for three centuries.

Take $\log_{10} 12$ . Factor the inside first: $12 = 2^2 \cdot 3$ . Now the product rule splits the factors and the power rule brings the exponent down:

\log_{10} 12 = 2\log_{10} 2 + \log_{10} 3 \approx 2(0.3010) + 0.4771 = 1.0791.

Take $\log_{10} 1.5$ . Write it as a quotient, $1.5 = \tfrac{3}{2}$ , and subtract:

\log_{10} 1.5 = \log_{10} 3 - \log_{10} 2 \approx 0.4771 - 0.3010 = 0.1761.

Now put the two together to see the actual trick. Since $\log_{10}(MN) = \log_{10}M + \log_{10}N$ , multiplying $12 \times 1.5$ is the same as adding their logarithms:

\log_{10} 12 + \log_{10} 1.5 \approx 1.0791 + 0.1761 = 1.2552.

A table lets you look that sum back up to find the number it belongs to. Because $\log_{10} 18 \approx 1.2553$ , the closest entry is $18$ , which you can check directly: $12 \times 1.5 = 18$ . The small gap between $1.2552$ and $1.2553$ is just the rounding in our four-decimal entries; a real table, carrying more decimal places, would land on $18$ even more precisely. Two table lookups and an addition replaced a multiplication; two lookups and a subtraction replaced a division.

Change of base

Every logarithm in the lesson so far has had a friendly base. Real questions are not so polite: a population that triples, an interest rate, a half-life, all produce logarithms in bases nobody has tabulated. Worse, a calculator offers you only a couple of bases, not the one you happen to need.

The fix is the change-of-base formula. For $b, a > 0$ with $b \ne 1$ and $a \ne 1$ , and for $M > 0$ :

\log_b M = \frac{\log_a M}{\log_a b}.

Any base you can compute in will do for $a$ . The formula converts the base you want into the base you have.

Here is why it works, in three lines. Let $x = \log_b M$ , so $b^x = M$ . Take a base- $a$ logarithm of both sides and use the power rule on the left:

\log_a\!\left(b^x\right) = x\log_a b = \log_a M.

Solve for $x$ by dividing by $\log_a b$ , and since $x = \log_b M$ , that gives the formula. The division is always legal, because $\log_a b = 0$ would mean $a^0 = b$ , that is $b = 1$ , and $b = 1$ was excluded as a base from the start. The full derivation, with every legality check spelled out, is in where the change-of-base formula comes from, at the bottom of this lesson.

One special case is worth knowing: swapping the base and the argument inverts the value, so $\log_2 8 = 3$ sits alongside $\log_8 2 = \tfrac{1}{3}$ . It falls straight out of the formula. The one line of algebra is in why swapping the base and the argument inverts the value, at the bottom of this lesson.

Worked example 5 Evaluate $\log_8 32$ exactly, and $\log_3 50$ to three decimal places

For $\log_8 32$ , notice that both $8$ and $32$ are powers of $2$ , so change to base $2$ , where both logarithms are exact:

\log_8 32 = \frac{\log_2 32}{\log_2 8} = \frac{5}{3}.

Check it against the definition: $8^{5/3} = \left(2^3\right)^{5/3} = 2^5 = 32$ . Correct.

For $\log_3 50$ there is no such luck, so change to base $10$ , which is what a calculator’s log key gives you:

\log_3 50 = \frac{\log_{10} 50}{\log_{10} 3} \approx \frac{1.6990}{0.4771} \approx 3.561.

A quick sanity check keeps you honest: $3^3 = 27$ and $3^4 = 81$ , and $50$ sits between them, so the answer had to land between $3$ and $4$ .

Check your understanding

Rewrite $\log_6 90$ using base-10 logarithms.

\dfrac{\log_{10} 6}{\log_{10} 90}

\log_{10} 90 - \log_{10} 6

\dfrac{\log_{10} 90}{\log_{10} 6}

\log_{10} 90 \cdot \log_{10} 6

What the rules buy you

The power rule does something no amount of arithmetic can: it measures a number too large to write down. How many digits does $2^{40}$ have? Multiplying it out is possible but tedious, and for $2^{4000}$ it would be hopeless. Logarithms answer the question in one line.

The idea is that a positive integer $N$ has $d$ digits exactly when $10^{d-1} \le N < 10^d$ . (A three-digit number runs from $100$ to $999$ , which is $10^2 \le N < 10^3$ .) So the digit count is read off from where $\log_{10} N$ falls between consecutive whole numbers.

Worked example 6 How many digits does $2^{40}$ have? Use $\log_{10} 2 \approx 0.30103$

Take the base-10 logarithm and let the power rule pull the exponent down:

\log_{10}\!\left(2^{40}\right) = 40 \log_{10} 2 \approx 40(0.30103) = 12.0412.

That value sits between $12$ and $13$ . Because $10^t$ grows as $t$ grows, this pins the number itself:

10^{12} < 2^{40} < 10^{13}.

A number that is at least $10^{12}$ but below $10^{13}$ has exactly $13$ digits. So $2^{40}$ has $13$ digits, and indeed $2^{40} = 1{,}099{,}511{,}627{,}776$ .

The same line handles $2^{4000}$ , where $4000(0.30103) = 1204.12$ tells you the answer has $1205$ digits, without writing a single one of them.

That is the shape of every application in the next lessons. Whenever an unknown is stuck in an exponent, take a logarithm and the power rule drags it down to where ordinary algebra can reach it.

Common mistakes

Splitting a sum. $\log_b(M + N)$ is not $\log_b M + \log_b N$ . The product rule is about a product inside the logarithm, never a sum. Test it: $\log_2(4 + 4) = \log_2 8 = 3$ , while $\log_2 4 + \log_2 4 = 2 + 2 = 4$ .
Confusing a quotient of logs with the log of a quotient. $\dfrac{\log_b M}{\log_b N}$ is not $\log_b M - \log_b N$ . The first is a change of base and equals $\log_N M$ ; the second is the quotient rule. They measure different things and must not be substituted for each other.
Forgetting the domain conditions. $\log_b(MN) = \log_b M + \log_b N$ needs $M > 0$ and $N > 0$ , not just $MN > 0$ : with $M = N = -2$ the left side is $\log_b 4$ but the right side does not exist. And $\log_b\!\left(x^2\right) = 2\log_b x$ is false for every $x < 0$ , because the left side exists there but $\log_b x$ does not; the identity valid for all $x \ne 0$ is $\log_b\!\left(x^2\right) = 2\log_b |x|$ .
Moving an exponent that is not inside. $(\log_b M)^2$ is not $2\log_b M$ . The power rule only applies when the exponent is on the argument, inside the logarithm. When in doubt, write the parentheses.
Condensing before clearing coefficients. $2\log_b x + \log_b y$ is not $\log_b(2xy)$ . Send the $2$ up as an exponent first, giving $\log_b\!\left(x^2 y\right)$ .

Key takeaways

A logarithm is an exponent, so every rule here is an exponent law read backwards: $\log_b M = x$ means $b^x = M$ .
Product rule: $\log_b(MN) = \log_b M + \log_b N$ , valid when $M > 0$ and $N > 0$ .
Quotient rule: $\log_b\!\left(\dfrac{M}{N}\right) = \log_b M - \log_b N$ , valid when $M > 0$ and $N > 0$ . In particular $\log_b\!\left(\tfrac{1}{N}\right) = -\log_b N$ .
Power rule: $\log_b\!\left(M^p\right) = p\log_b M$ for $M > 0$ and any real $p$ . For $x < 0$ use $\log_b\!\left(x^2\right) = 2\log_b |x|$ .
Change of base: $\log_b M = \dfrac{\log_a M}{\log_a b}$ for any valid base $a$ . Dividing is safe because $\log_a b \ne 0$ whenever $b \ne 1$ .
To expand, split quotients, then products, then bring exponents down. To condense, send coefficients up as exponents first, then combine.
The rules turn multiplication into addition and powers into multiplication, which is why $\log_{10}$ can count the digits of $2^{40}$ without anyone multiplying it out.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Logarithms (Khan Academy)
Use the properties of logarithms (free textbook) (OpenStax)
Logarithmic properties (free textbook) (OpenStax)
Logarithm functions (Paul's Online Math Notes)
Working with exponents and logarithms (Math is Fun)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

Where the change-of-base formula comes from

The lesson states the formula and uses it; this is the derivation, which rests on nothing but the definition and the power rule you already have.

The change-of-base formula#

Let $x = \log_b M$ , which exists because $M > 0$ , and translate it with the definition: $b^x = M$ .

Two equal positive numbers have equal logarithms in any valid base, so we may apply $\log_a$ to both sides of $b^x = M$ :

\log_a\!\left(b^x\right) = \log_a M.

The power rule applies to the left side, since $b > 0$ , and pulls the exponent to the front: $x \log_a b = \log_a M$ .

To solve for $x$ we want to divide by $\log_a b$ , which is only legal if that number is not zero. It is not zero, and here is why: $\log_a b = 0$ would mean $a^0 = b$ , that is $b = 1$ , and $b = 1$ was excluded as a base from the start. So the division is safe, and

x = \frac{\log_a M}{\log_a b},

which, since $x = \log_b M$ , is the formula.

Why swapping the base and the argument inverts the value

The change-of-base formula lets you pick any legal base for $a$ . Nothing stops you picking $M$ itself, the very number you are taking the logarithm of, provided $M > 0$ and $M \ne 1$ so that $M$ is a legal base. Since $\log_M M = 1$ , the numerator collapses:

\log_b M = \frac{\log_M M}{\log_M b} = \frac{1}{\log_M b}.

So $\log_b M$ and $\log_M b$ are reciprocals of each other. Read it as a statement about exponents and it is almost obvious: if it takes the power $3$ to get from $2$ to $8$ , it must take the power $\tfrac{1}{3}$ to get from $8$ back to $2$ , because $8^{1/3} = 2$ . The identity is the same fact, written in logarithms.

A bit of history (optional)

This lesson handed you $\log_{10} 2 \approx 0.3010$ and invited you to use it. Somebody had to compute that number first, and there was nowhere to look it up.

Henry Briggs, an English professor of geometry, set himself the whole table. He would calculate the base-10 logarithm of thirty thousand numbers, to fourteen decimal places. The result appeared in 1624. He had no calculator and no base-10 table to lean on. He had a single rule with any real leverage in it, and it is the power rule of this lesson.

Watch what he does with it, beginning from the one value he already knows, $\log_{10} 10 = 1$ . Taking a square root halves a logarithm, so the square root of $10$ delivers a second entry immediately. The square root of that delivers a third. He continues for fifty-four rounds, each root halving the logarithm and dragging the number itself closer to $1$ . From that ladder of roots, recombined by the product rule, every remaining entry in the table could be assembled.

The arithmetic occupied him for years, all of it on paper, to more decimal places than your calculator will show you. His choice of base is a big reason a bare log key still means base $10$ today. And the numbers you have been quoting, the $0.3010$ and the $0.4771$ , are the ones he computed.