# Logarithms

Canonical: https://duckyhelper.com/learn/algebra-2/logarithms/
Updated: 2026-10-01

A logarithm answers the question "what exponent?": \(\log_b(x) = y\) means \(b^y = x\). So \(\log_2(32) = 5\) because \(2^5 = 32\). The rules turn products into sums, quotients into differences and powers into multipliers. Solve an exponential equation by taking a log of both sides. Solve a log equation by rewriting it in exponential form, then check that every log input is positive.

## The key idea

Logs and exponents are inverses. Every log statement is an exponent statement read backwards:

$$
\log_b(x) = y \iff b^y = x \qquad (b > 0,\ b \ne 1,\ x > 0)
$$

**The same fact in both forms**

| Exponential form | Log form |
| --- | --- |
| \(2^5 = 32\) | \(\log_2(32) = 5\) |
| \(10^{-2} = 0.01\) | \(\log(0.01) = -2\) |
| \(e^0 = 1\) | \(\ln(1) = 0\) |

\(\log\) with no base means base 10, and \(\ln\) means base \(e \approx 2.718\). The three rules come straight from the exponent rules:

$$
\log_b(MN) = \log_b M + \log_b N, \quad \log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N, \quad \log_b(M^p) = p\log_b M
$$

Change of base lets any calculator find any log:

$$
\log_b(x) = \frac{\log x}{\log b} = \frac{\ln x}{\ln b}
$$

## Worked examples

**Example 1: evaluate a log by hand**

Problem: Find \(\log_4(8)\) without a calculator.

1. Call the answer y. By definition, 4 to the y is 8.

   $$
   4^y = 8
   $$
2. Write both sides as powers of 2.

   $$
   (2^2)^y = 2^3
   $$
3. Same base, so the exponents must match.

   $$
   2y = 3
   $$
4. Divide by 2.

   $$
   y = \frac{3}{2}
   $$

Answer: \(\log_4(8) = \frac{3}{2}\)

**Example 2: condense into one log**

Problem: Write \(2\log_3(x) + \log_3(5) - \log_3(x + 1)\) as a single logarithm.

1. Power rule first: the 2 becomes an exponent.

   $$
   2\log_3(x) + \log_3(5) - \log_3(x + 1) = \log_3(x^2) + \log_3(5) - \log_3(x + 1)
   $$
2. Product rule: a sum of logs is the log of a product.

   $$
   \log_3(x^2) + \log_3(5) - \log_3(x + 1) = \log_3(5x^2) - \log_3(x + 1)
   $$
3. Quotient rule: a difference of logs is the log of a quotient.

   $$
   \log_3(5x^2) - \log_3(x + 1) = \log_3\left(\frac{5x^2}{x + 1}\right)
   $$

Answer: \(\log_3\left(\frac{5x^2}{x + 1}\right)\)

**Example 3: solve an exponential equation**

Problem: Solve \(5^{2x - 1} = 40\). Round to three decimal places.

1. 40 is not a power of 5, so take the log of both sides and use the power rule to bring the exponent down.

   $$
   (2x - 1)\log(5) = \log(40)
   $$
2. Divide both sides by \(\log 5\).

   $$
   2x - 1 = \frac{\log(40)}{\log(5)}
   $$
3. Add 1 and divide by 2.

   $$
   x = \frac{1}{2}\left(1 + \frac{\log(40)}{\log(5)}\right)
   $$
4. On a calculator, \(\frac{\log 40}{\log 5} \approx 2.292\), so \(x \approx \frac{3.292}{2}\).

Answer: \(x \approx 1.646\)

**Example 4 (test-hard): a log equation with an extraneous answer**

Problem: Solve \(\log_2(x) + \log_2(x - 2) = 3\).

1. Product rule: combine into one log.

   $$
   \log_2(x(x - 2)) = 3
   $$
2. Rewrite in exponential form: \(2^3 = 8\).

   $$
   x(x - 2) = 8
   $$
3. Expand and set to zero.

   $$
   x^2 - 2x - 8 = 0
   $$
4. Factor.

   $$
   (x - 4)(x + 2) = 0
   $$
5. So \(x = 4\) or \(x = -2\). But \(\log_2(-2)\) is not defined, so \(-2\) is extraneous. Check 4: \(\log_2 4 + \log_2 2 = 2 + 1 = 3\).

Answer: \(x = 4\)

## Common mistakes

- **Splitting a log of a sum.** \(\log(a + b)\) is not \(\log a + \log b\). Fix: the product rule is for \(\log(ab)\) only.
- **Canceling logs in a fraction.** \(\frac{\log 40}{\log 5}\) is not \(\log 8\). Fix: divide the two log values; only \(\log 40 - \log 5\) equals \(\log 8\).
- **Moving a coefficient the wrong way.** \(2\log_3 x = \log_3(x^2)\), not \(\log_3(2x)\). Fix: the number in front becomes an exponent on the input.
- **Keeping answers that make a log input negative or zero.** Fix: after solving, put each answer into every log in the original equation.
- **Mixing up the base and the answer.** \(\log_2 32 = 5\) means \(2^5 = 32\), not \(5^2\) or \(32^2\). Fix: say "2 to what power is 32?"

## Quick methods

> **Tip: Same base? Match the exponents**
>
> If both sides can be written as powers of the same base, like \(4^y = 8\) as \(2^{2y} = 2^3\), set the exponents equal. No calculator needed. When the numbers are not related powers, take logs instead.

> **Note: Change of base on any calculator**
>
> Most calculators only have \(\log\) and \(\ln\). For \(\log_2(10)\), type \(\log(10) \div \log(2) \approx 3.32\). Either log works as long as you use the same one on top and bottom.

## Practice

**5 practice questions**

1. What is \(\log_2\left(\frac{1}{64}\right)\)?
   A. \(-6\)
   B. \(6\)
   C. \(-32\)
   D. \(\frac{1}{6}\)

   Answer: \(-6\). \(2^6 = 64\), so \(2^{-6} = \frac{1}{64}\) and the log is \(-6\). A fraction less than 1 always gives a negative log when the base is more than 1.

2. Which expression is equal to \(\log(x) + \log(5) - \log(2)\) for \(x > 0\)?
   A. \(\log\left(\frac{5x}{2}\right)\)
   B. \(\log(5x - 2)\)
   C. \(\log(x + 3)\)
   D. \(\log(10x)\)

   Answer: \(\log\left(\frac{5x}{2}\right)\). Adding logs multiplies the inputs and subtracting divides: \(\log\left(\frac{x \cdot 5}{2}\right)\). \(\log(5x - 2)\) treats subtracting logs as subtracting inputs, and \(\log(10x)\) multiplies by 2 instead of dividing.

3. Solve \(3^x = 20\). Round to the nearest hundredth.
   A. 0.37
   B. 1.30
   C. 2.73
   D. 6.67

   Answer: 2.73. \(x = \frac{\log 20}{\log 3} \approx \frac{1.301}{0.477} \approx 2.73\). 0.37 divides the logs upside down, 1.30 is just \(\log 20\), and 6.67 is \(20 \div 3\). Check: \(3^{2.73} \approx 20\).

4. Solve \(\log_3(x + 1) = 4\).
   A. \(11\)
   B. \(63\)
   C. \(80\)
   D. \(82\)

   Answer: \(80\). Exponential form: \(x + 1 = 3^4 = 81\), so \(x = 80\). 63 uses \(4^3\) instead of \(3^4\), and 11 uses \(3 \cdot 4\).

5. Solve \(\log_5(x) + \log_5(x - 4) = 1\).

   Answer: 5. Combine: \(\log_5(x(x - 4)) = 1\), so \(x^2 - 4x = 5\), which factors as \((x - 5)(x + 1) = 0\). \(x = -1\) makes \(\log_5(x)\) undefined, so only \(x = 5\) works.

## Frequently asked questions

### What is the difference between log and ln?

Both are logarithms with different bases. \(\log\) with no base written usually means base 10, the common log. \(\ln\) is base \(e\), about 2.718, the natural log. They follow the same rules, and either one works for change of base.

### Why can't you take the log of a negative number?

A log asks what power of a positive base gives the input. A positive base raised to any real power is always positive, so no real exponent gives 0 or a negative number. That is why answers that make a log input zero or negative get thrown out.

### How do I use the change of base formula?

Divide the log of the input by the log of the base, using the same kind of log on top and bottom: \(\log_2(10) = \frac{\log 10}{\log 2} \approx 3.32\). Check by raising: \(2^{3.32}\) is about 10.

### Is log(a + b) equal to log a + log b?

No. The product rule says \(\log(ab) = \log a + \log b\). There is no rule that splits a log of a sum, so leave \(\log(a + b)\) as it is. For example, \(\log(10 + 10) = \log 20 \approx 1.30\), but \(\log 10 + \log 10 = 2\).

## Related

- [Inverse functions](https://duckyhelper.com/learn/algebra-2/inverse-functions/)
- [Exponent rules and how to use them](https://duckyhelper.com/learn/algebra-1/exponent-rules/)
- [Exponential growth and decay on the SAT](https://duckyhelper.com/learn/sat-math/exponential-functions/)
- [Sequences and series](https://duckyhelper.com/learn/algebra-2/sequences-and-series/)
- [Algebra 2 study guides](https://duckyhelper.com/learn/algebra-2/)

## Try asking Ducky

- "Can you explain what a log actually means with a simple example?"
- "I got x = -2 and x = 4. Why is only 4 right?"
- "Give me five log rule problems and check my answers."

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