| Primary source | OpenStax College Algebra 2e, Section 6.5: "Logarithmic Properties" |
| Direct link | https://openstax.org/books/college-algebra-2e/pages/6-5-logarithmic-properties |
| Supplementary | OpenStax Precalculus 2e, Section 4.5: "Logarithmic Properties" |
| Supplementary link | https://openstax.org/books/precalculus-2e/pages/4-5-logarithmic-properties |
College Algebra 2e Section 6.5 begins on page 610 of the PDF.
Every logarithm rule is an exponent rule in disguise.
If you write $M = b^x$ (so $x = \log_b M$) and $N = b^y$ (so $y = \log_b N$), then:
\[ MN = b^x \cdot b^y = b^{x+y} \]
In logarithmic language, the exponent of $MN$ base $b$ is $x + y = \log_b M + \log_b N$. That is the product rule.
Every other rule works the same way: translate to exponent language, apply an exponent rule, translate back. The log rules are not new facts. They are the exponent rules viewed from a different angle.
The practical payoff is large: logarithms convert multiplication to addition, exponentiation to multiplication, and roots to division. Before calculators, this was how engineers and scientists did complex arithmetic using slide rules and log tables. Today, the same transformations appear in calculus (the derivative of $\ln x$), in statistics (log-likelihoods), and in every formula involving exponential growth or decay.
If the definition of logarithm ($\log_b M = x$ means $b^x = M$) is unfamiliar, review exponential functions first.
Let $M, N > 0$, $b > 0$, $b \neq 1$, and let $p$ be any real number.
| Rule | Logarithm form | Corresponding exponent rule |
|---|---|---|
| Product rule | $\log_b(MN) = \log_b M + \log_b N$ | $b^{x+y} = b^x \cdot b^y$ |
| Quotient rule | $\log_b\!\left(\dfrac{M}{N}\right) = \log_b M - \log_b N$ | $b^{x-y} = b^x / b^y$ |
| Power rule | $\log_b(M^p) = p \log_b M$ | $(b^x)^p = b^{px}$ |
| Change of base | $\log_b M = \dfrac{\log M}{\log b} = \dfrac{\ln M}{\ln b}$ | (bridge formula) |
| Special values | $\log_b(b) = 1$, $\log_b(1) = 0$ | $b^1 = b$, $b^0 = 1$ |
| Inverse property | $\log_b(b^x) = x$ and $b^{\log_b x} = x$ | Logarithm and exponential are inverses |
$\log_b M = x$ means exactly $b^x = M$.
The logarithm base $b$ of $M$ is the exponent you put on $b$ to get $M$.
| Logarithm statement | Equivalent exponential statement |
|---|---|
| $\log_3 9 = 2$ | $3^2 = 9$ |
| $\log_5 125 = 3$ | $5^3 = 125$ |
| $\log_2 \frac{1}{4} = -2$ | $2^{-2} = \frac{1}{4}$ |
| $\ln e = 1$ | $e^1 = e$ |
| $\log 1000 = 3$ | $10^3 = 1000$ |
Common logarithm: $\log M$ (base 10 implied when no base is written). Natural logarithm: $\ln M$ (base $e \approx 2.718$).
\[ \log_b(MN) = \log_b M + \log_b N \]
Derivation. Let $x = \log_b M$ and $y = \log_b N$, so $M = b^x$ and $N = b^y$.
\[ MN = b^x \cdot b^y = b^{x+y} \]
Converting back: $\log_b(MN) = x + y = \log_b M + \log_b N$.
Example 1. Expand $\log_3(27x)$.
\[ \log_3 27 + \log_3 x = 3 + \log_3 x \]
(since $3^3 = 27$)
Example 2. Condense $\log 5 + \log 8$.
\[ \log(5 \cdot 8) = \log 40 \]
Common Error -- Adding Inside the Logarithm. $\log_b(M + N) \neq \log_b M + \log_b N$. The product rule applies to multiplication inside. There is no rule that simplifies a logarithm of a sum. Test with numbers: $\log(10 + 10) = \log 20 \approx 1.3$, but $\log 10 + \log 10 = 1 + 1 = 2$. Not equal.
\[ \log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N \]
Derivation. Using the same substitution: $\dfrac{M}{N} = \dfrac{b^x}{b^y} = b^{x-y}$, so $\log_b\!\left(\dfrac{M}{N}\right) = x - y = \log_b M - \log_b N$.
Example 3. Expand $\ln\!\left(\dfrac{e^3}{x}\right)$.
\[ \ln e^3 - \ln x = 3 - \ln x \]
Example 4. Condense $\ln 12 - \ln 4$.
\[ \ln\!\left(\frac{12}{4}\right) = \ln 3 \]
Common Error -- Subtracting Inside the Logarithm. $\log_b M - \log_b N = \log_b\!\left(\dfrac{M}{N}\right)$, not $\log_b(M - N)$. The subtraction of logarithms corresponds to division inside, not subtraction inside.
\[ \log_b(M^p) = p \log_b M \]
Derivation. With $M = b^x$: $M^p = (b^x)^p = b^{px}$, so $\log_b(M^p) = px = p \log_b M$.
Example 5. Expand $\log_2(x^5)$.
\[ 5 \log_2 x \]
Example 6. Condense $\frac{1}{2}\ln x$.
\[ \ln x^{1/2} = \ln\sqrt{x} \]
Example 7. Expand $\log\!\sqrt[3]{x^2}$.
Write as $\log(x^{2/3})$, then apply the power rule: \[ \frac{2}{3}\log x \]
Calculators typically have $\log$ (base 10) and $\ln$ (base $e$) buttons. To evaluate $\log_b M$ for any other base $b$, use:
\[ \log_b M = \frac{\log M}{\log b} = \frac{\ln M}{\ln b} \]
Derivation. Let $x = \log_b M$, so $M = b^x$. Take the natural log of both sides: \[ \ln M = x \ln b \implies x = \frac{\ln M}{\ln b} \]
Example 8. Evaluate $\log_5 50$.
\[ \log_5 50 = \frac{\ln 50}{\ln 5} = \frac{3.912}{1.609} \approx 2.431 \]
Check: $5^{2.431} \approx 50$. $\checkmark$
Method 1 (convert to exponential form): If the equation has one logarithm, convert and solve.
Example 9. Solve $\log_2(3x + 1) = 4$.
Convert: $3x + 1 = 2^4 = 16$.
Solve: $3x = 15 \implies x = 5$.
Check: $\log_2(3 \cdot 5 + 1) = \log_2 16 = 4$. $\checkmark$
Method 2 (condense, then convert): If the equation has multiple logarithms on one side, condense using the rules, then convert.
Example 10. Solve $\log x + \log(x - 3) = 1$.
Condense the left side (product rule): $\log(x(x-3)) = 1$.
Convert (base 10): $x(x-3) = 10^1 = 10$.
Expand and rearrange: $x^2 - 3x - 10 = 0$.
Factor: $(x-5)(x+2) = 0$, so $x = 5$ or $x = -2$.
Check both solutions in the original equation.
$x = 5$: $\log 5 + \log 2 = \log 10 = 1$. $\checkmark$
$x = -2$: $\log(-2)$ is undefined (logarithm of a negative number does not exist in the reals). Discard.
Solution: $x = 5$ only.
Always check solutions in logarithmic equations. The condensing step can introduce solutions that make an argument of the original logarithm negative. These are extraneous solutions and must be discarded.
Method 3 (equate arguments): If both sides of the equation have a single logarithm with the same base, equate the arguments.
If $\log_b A = \log_b B$ (with $A, B > 0$), then $A = B$.
Example 11. Solve $\ln(x + 4) = \ln(3x - 2)$.
Since the logarithms have the same base: $x + 4 = 3x - 2$, so $6 = 2x$, thus $x = 3$.
Check: $\ln(7) = \ln(7)$. $\checkmark$
These are the two directions of applying the rules.
Expanding: Write a single complicated logarithm as a sum and difference of simpler ones (useful for solving equations and for differentiation in calculus).
Condensing: Write a sum and difference of logarithms as a single logarithm (useful for solving logarithmic equations).
Example 12. Expand $\ln\!\left(\dfrac{x^3 \sqrt{y}}{z^4}\right)$.
\[ \ln(x^3) + \ln(\sqrt{y}) - \ln(z^4) = 3\ln x + \frac{1}{2}\ln y - 4\ln z \]
Example 13. Condense $2\log_3 x - \log_3(x+1) + \frac{1}{3}\log_3(x+2)$.
Apply power rule to each term, then combine: \[ \log_3(x^2) - \log_3(x+1) + \log_3((x+2)^{1/3}) \] \[ = \log_3\!\left(\frac{x^2 (x+2)^{1/3}}{x+1}\right) \]
| Error | Example | Correction |
|---|---|---|
| Logarithm of a sum | $\log(M+N) = \log M + \log N$ | No such rule; this is false |
| Quotient inside vs. subtraction | $\log M - \log N = \log(M-N)$ | Correct form: $\log(M/N)$ |
| Forgetting to check for extraneous solutions | $x = -2$ accepted without checking | Must verify arguments are positive |
| Wrong change-of-base order | $\log_5 50 = \frac{\ln 5}{\ln 50}$ | Numerator is what you want: $\frac{\ln 50}{\ln 5}$ |
| Pulling coefficients out of log sums | $\log(a + b) = \log a + \log b$ (repeated) | Only for products: $\log(ab) = \log a + \log b$ |
Problem 1. Evaluate: $\log_4 64$.
$4^x = 64 = 4^3$, so $x = 3$.
$\log_4 64 = 3$.
Problem 2. Expand: $\log\!\left(\dfrac{100x^2}{y}\right)$.
$\log 100 + \log x^2 - \log y = 2 + 2\log x - \log y$
Problem 3. Condense: $3\ln x + \ln 5 - 2\ln y$.
$\ln(x^3) + \ln 5 - \ln(y^2) = \ln\!\left(\dfrac{5x^3}{y^2}\right)$
Problem 4. Solve: $\log_3(x - 1) = 2$.
$x - 1 = 3^2 = 9 \implies x = 10$.
Check: $\log_3(9) = 2$. $\checkmark$
Problem 5. Solve: $\ln x + \ln(x + 2) = \ln 8$.
Condense left: $\ln(x(x+2)) = \ln 8$.
Equate arguments: $x^2 + 2x = 8$.
$x^2 + 2x - 8 = 0 \implies (x+4)(x-2) = 0$.
$x = -4$ or $x = 2$.
Check $x = -4$: $\ln(-4)$ undefined. Discard.
Check $x = 2$: $\ln 2 + \ln 4 = \ln 8$. $\checkmark$
Solution: $x = 2$.
Problem 6. Use the change of base formula to evaluate $\log_7 200$ to four decimal places.
$\log_7 200 = \dfrac{\ln 200}{\ln 7} = \dfrac{5.2983}{1.9459} \approx 2.7228$
Check: $7^{2.7228} \approx 200$. $\checkmark$
Problem 7. Show that $\log_b a = \dfrac{1}{\log_a b}$.
Apply the change-of-base formula to both sides:
$\log_b a = \dfrac{\ln a}{\ln b}$ and $\log_a b = \dfrac{\ln b}{\ln a}$.
Therefore $\dfrac{1}{\log_a b} = \dfrac{1}{\ln b / \ln a} = \dfrac{\ln a}{\ln b} = \log_b a$. $\checkmark$
Problem 8. Solve: $2^{x+1} = 3^x$. Express the answer exactly and as a decimal.
Take $\ln$ of both sides: \[ (x+1)\ln 2 = x \ln 3 \] \[ x \ln 2 + \ln 2 = x \ln 3 \] \[ \ln 2 = x(\ln 3 - \ln 2) = x \ln\frac{3}{2} \] \[ x = \frac{\ln 2}{\ln(3/2)} \approx \frac{0.6931}{0.4055} \approx 1.709 \]
Problem 9. The half-life of a substance is 12 years. Starting from 100 grams, how long until 10 grams remain?
The decay formula is $A(t) = 100 \cdot \left(\dfrac{1}{2}\right)^{t/12}$.
Set $A(t) = 10$: \[ \left(\frac{1}{2}\right)^{t/12} = \frac{1}{10} \]
Take $\ln$ of both sides: \[ \frac{t}{12} \ln\!\left(\frac{1}{2}\right) = \ln\!\left(\frac{1}{10}\right) \] \[ \frac{t}{12} \cdot (-\ln 2) = -\ln 10 \] \[ t = \frac{12 \ln 10}{\ln 2} = 12 \log_2 10 \approx \frac{12 \cdot 2.3026}{0.6931} \approx 39.9 \text{ years} \]
A logarithm is an exponent counter.
$\log_b M$ asks: "How many times do I multiply $b$ by itself to get $M$?"
From that perspective, the rules are obvious:
This is why logarithms converted multiplication into addition for slide-rule users: multiplying two numbers was replaced by reading off their "counts" from a table, adding the counts, and reading back the result. The rules you are learning made that process possible.
For students who find math intimidating: The one fact to anchor is the definition: $\log_b M = x$ means $b^x = M$. If you can convert between logarithm form and exponential form, you can evaluate any logarithm and check any answer. The three rules are just exponent rules seen from a new angle; you already know them.
For students interested in proof: The product rule can be proved rigorously once you have defined $\ln x = \displaystyle\int_1^x \frac{1}{t}\,dt$ (the integral definition). Then $\ln(MN) = \ln M + \ln N$ follows from the additive property of integrals over adjacent intervals. This is the calculus-based approach, which avoids circular reasoning.
For students interested in careers: Log scales appear in decibels (acoustics), the Richter scale (seismology), pH (chemistry), star magnitude (astronomy), and the entropy formula in information theory. Every time you see a scale that compresses a huge range of values into a small readable range, you are looking at a logarithm.