Properties of Logarithms

Once you know that a logarithm is an exponent, the log rules stop being a list to memorize and start being things you already believe. There are exactly three: the product rule, the quotient rule, and the power rule. Each one is an exponent rule wearing different clothes.

These three rules let you expand one complicated log into simple pieces, or condense a pile of logs into a single one. Both directions show up constantly — expanding when you're solving log equations, condensing when you're cleaning up an answer. Master the three moves and every "simplify this log expression" problem is mechanical.

The three rules

Product rule: $\log_b (xy) = \log_b x + \log_b y$ . A product inside the log becomes a sum of logs.

Quotient rule: $\log_b \dfrac{x}{y} = \log_b x - \log_b y$ . A quotient inside the log becomes a difference of logs — top minus bottom, in that order.

Power rule: $\log_b x^n = n \log_b x$ . An exponent inside the log moves out front as a multiplier. This is the rule you'll lean on most, because it turns "the variable is stuck in an exponent" into "the variable is out front where I can solve for it."

Why the rules work

Every log rule is an exponent rule in disguise, because logs are exponents. You know that $b^m \cdot b^n = b^{m+n}$ — when you multiply powers, the exponents add. Since $\log_b x$ and $\log_b y$ are the exponents of $x$ and $y$ , multiplying $x$ and $y$ must add those exponents. That's the product rule.

Try it with real numbers: $\log_2 32 = 5$ . But $32 = 8 \cdot 4$ , and $\log_2 8 + \log_2 4 = 3 + 2 = 5$ . Same answer both ways. The quotient rule mirrors $\dfrac{b^m}{b^n} = b^{m-n}$ , and the power rule mirrors $(b^m)^n = b^{mn}$ .

What the rules do NOT say

The most common log mistake in Algebra 2 is inventing a rule for $\log_b(x + y)$ . There isn't one. The product rule turns a product inside the log into a sum of logs — it says nothing about a sum inside the log. $\log_b(x + y)$ cannot be split, rewritten, or simplified with these properties. It just stays as it is.

Quick sanity check: $\log_2(8 + 8) = \log_2 16 = 4$ , but $\log_2 8 + \log_2 8 = 3 + 3 = 6$ . Not equal — so $\log_b(x+y)$ and $\log_b x + \log_b y$ are genuinely different things. Related traps: $\dfrac{\log_b x}{\log_b y}$ is not $\log_b \dfrac{x}{y}$ , and $(\log_b x)^n$ is not $n \log_b x$ . The rules only apply to products, quotients, and powers inside the log.

Worked examples

Example 1: the product rule with numbers

Show that $\log_2 (8 \cdot 4) = \log_2 8 + \log_2 4$ .

Evaluate the left side first

\log_2 (8 \cdot 4) = \log_2 32

2^5 = 32

, so

\log_2 32 = 5

Now the right side

\log_2 8 + \log_2 4 = 3 + 2

Both sides equal

5

✓

5 = 5

Answer: Both sides equal $5$ — the product inside became a sum outside.

Example 2: the power rule

Evaluate $\log_2 4^3$ two ways.

Way 1 — compute the inside first

4^3 = 64, \quad \log_2 64 = 6

Way 2 — move the exponent out front

\log_2 4^3 = 3 \log_2 4

Since

\log_2 4 = 2

3 \log_2 4 = 3 \cdot 2 = 6

Same answer both ways ✓

Answer: $\log_2 4^3 = 6$

Example 3: expand a log expression

Fully expand $\log_b \dfrac{x^2 y}{z}$ .

Quotient rule: top minus bottom

\log_b \dfrac{x^2 y}{z} = \log_b (x^2 y) - \log_b z

Product rule on the first log

\log_b (x^2 y) = \log_b x^2 + \log_b y

Power rule moves the

2

out front

\log_b x^2 = 2\log_b x

Put it all together

2\log_b x + \log_b y - \log_b z

Answer: $\log_b \dfrac{x^2 y}{z} = 2\log_b x + \log_b y - \log_b z$

Example 4: condense to a single log

Write $2\log 5 + \log 4$ as one log, then evaluate it.

Power rule in reverse

2\log 5 = \log 5^2 = \log 25

Product rule in reverse

\log 25 + \log 4 = \log(25 \cdot 4)

Multiply inside

\log 100

Base

10

: since

10^2 = 100

\log 100 = 2

Answer: $2\log 5 + \log 4 = \log 100 = 2$

Try one yourself

Common questions

Do I have to memorize the log rules?

You should know all three cold, but memorize them alongside the exponent rules they come from: product $\leftrightarrow$ $b^m b^n = b^{m+n}$ , quotient $\leftrightarrow$ $\dfrac{b^m}{b^n} = b^{m-n}$ , power $\leftrightarrow$ $(b^m)^n = b^{mn}$ . If you forget one, test it with easy numbers like base $2$ and powers of $2$ .

Can I simplify

\log_b(x + y)

No. None of the three properties applies to a sum (or difference) inside a log. Check with numbers: $\log_2 16 = 4$ , but $\log_2 8 + \log_2 8 = 6$ . If the inside is a sum you can't factor, the expression is already as simple as it gets.

Do the rules work for common logs and natural logs?

Yes. The rules hold for any valid base, so $\ln(xy) = \ln x + \ln y$ and $\log x^n = n\log x$ work exactly the same way. The base never changes within a rule — every log in the equation keeps the base you started with.

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