# Laws of Logarithms

> IB Mathematics: Applications and Interpretation HL · Number and Algebra
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u1-laws-of-logarithms/

This module covers the core algebraic laws for operating on logarithms, including product, quotient, power rules, and the change of base formula. You will learn to simplify expressions and prepare for solving logarithmic and exponential equations.

**Prerequisites:** [Definition of logarithms and their relationship to exponents](https://www.owlsprep.com/study/ib-math-ai-hl-u1-introduction-to-logarithms/)

## Learning objectives

- Recall and apply the three core laws of logarithms correctly
- Simplify logarithmic expressions using combinations of logarithm laws
- Use the change of base formula to evaluate logarithms with any base
- Recognize common errors when manipulating logarithmic expressions

## Product and Quotient Laws of Logarithms

**Product Law of Logarithms** — The logarithm of a product of two positive numbers equals the sum of the logarithms of the individual numbers, for the same base. Valid for $x>0, y>0, b>0, b \neq 1$.

*Notation:* \log_b(xy) = \log_b x + \log_b y

**Worked example:** Simplify $\log_2 8 + \log_2 4$

1. Check conditions: 8 and 4 are positive, same base 2, so product rule applies.
2. Apply the product law:
3. $$\log_2(8 \times 4) = \log_2 32$$
4. Evaluate by rewriting in exponential form: $2^5 = 32$, so $\log_2 32 = 5$.
5. Final result: $5$

**Quotient Law of Logarithms** — The logarithm of a quotient of two positive numbers equals the difference of the logarithm of the numerator and the logarithm of the denominator, for the same base. Valid for $x>0, y>0, b>0, b \neq 1$.

*Notation:* \log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y

**Worked example:** Simplify $\log_3 36 - \log_3 4$

1. Check conditions: 36 and 4 are positive, same base 3, so quotient rule applies.
2. Apply the quotient law:
3. $$\log_3\left(\frac{36}{4}\right) = \log_3 9$$
4. Evaluate: $3^2 = 9$, so $\log_3 9 = 2$.

> **Exam tip:** Always confirm all logarithm arguments are positive before applying any law. Negative arguments lead to undefined values.

## Power Law of Logarithms

The power law extends logarithm operations to expressions where the argument of the logarithm is raised to a power. This is the most commonly used law for solving exponential equations later in the course.

**Power Law of Logarithms** — The logarithm of a positive number raised to a power $n$ equals $n$ multiplied by the logarithm of the original number. Valid for $x>0, b>0, b \neq 1, n \in \mathbb{R}$.

*Notation:* \log_b(x^n) = n\log_b x

**Worked example:** Simplify $2\log_5 10 - \log_5 4$

1. Apply the power law to the first term first:
2. $$2\log_5 10 = \log_5(10^2) = \log_5 100$$
3. Substitute back to get: $\log_5 100 - \log_5 4$
4. Apply the quotient law:
5. $$\log_5\left(\frac{100}{4}\right) = \log_5 25$$
6. Evaluate: $5^2 = 25$, so the simplified result is $2$.

**Check your understanding**

Test your understanding of the power law

1. What is the expanded form of $\log(10x^3)$?

   - A) $3\log 10 + \log x$
   - B) $\log 10 + 3\log x$
   - C) $3(\log 10 + \log x)$
   - D) $\log 10 + \log(3x)$

   *Why:* First apply the product rule: $\log(10x^3) = \log 10 + \log x^3$. Then apply the power rule to $\log x^3$ to get $\log 10 + 3\log x$.

## Change of Base Formula

IB exams often ask you to evaluate logarithms with bases that are not 10 or $e$, which most calculators cannot compute directly. The change of base formula rewrites any logarithm in terms of a calculator-friendly base.

**Change of Base Formula** — Any logarithm can be rewritten as the ratio of two logarithms with a new chosen base $c$. The most common choices for $c$ are 10 (common log) or $e$ (natural log) for calculator use. Valid for $a>0, b>0, b \neq 1, c>0, c \neq 1$.

*Notation:* \log_b a = \frac{\log_c a}{\log_c b}

**Worked example:** Evaluate $\log_7 25$ to 3 decimal places

1. Apply change of base with natural log:
2. $$\log_7 25 = \frac{\ln 25}{\ln 7}$$
3. Calculate numerator and denominator:
4. $$\ln 25 \approx 3.2189, \quad \ln 7 \approx 1.9459$$
5. Divide the values: $\frac{3.2189}{1.9459} \approx 1.654$
6. The same result is obtained using base 10: $\frac{\log_{10} 25}{\log_{10} 7} \approx 1.654$

> **Exam tip:** If your calculator only outputs one type of log, remember that change of base works with both common and natural log, you will get the same result either way.

*Calculator:* allowed

## Combining Multiple Logarithm Laws

Exam questions almost always require combining multiple laws to simplify an expression or solve for an unknown. A good general approach is to apply the power law first to any coefficients, then combine terms with product/quotient rules.

**Worked example:** Express $3\log_a 2 + \frac{1}{2}\log_a 25 - \log_a 10$ as a single logarithm

1. Apply the power law to all terms with coefficients:
2. $$3\log_a 2 = \log_a 2^3 = \log_a 8, \quad \frac{1}{2}\log_a 25 = \log_a 25^{1/2} = \log_a 5$$
3. Substitute back: $\log_a 8 + \log_a 5 - \log_a 10$
4. Apply the product law to the first two terms:
5. $$\log_a (8 \times 5) - \log_a 10 = \log_a 40 - \log_a 10$$
6. Apply the quotient law to combine the remaining terms:
7. $$\log_a \left(\frac{40}{10}\right) = \log_a 4$$
8. Final result: $\log_a 4$

**Check your understanding**

Check your ability to combine multiple laws

1. Which of the following is equivalent to $\ln\left(\frac{e^2 x^3}{y}\right)$?

   - A) $2 + 3\ln x - \ln y$
   - B) $2e + 3\ln x - \ln y$
   - C) $2 + 3\ln\left(\frac{x}{y}\right)$
   - D) $\frac{2 + 3\ln x}{\ln y}$

   *Why:* Expand step-by-step: $\ln(e^2 x^3) - \ln y = \ln e^2 + \ln x^3 - \ln y = 2\ln e + 3\ln x - \ln y = 2 + 3\ln x - \ln y$, since $\ln e = 1$.

## Common pitfalls

- **Wrong:** Claiming $\log_b(x+y) = \log_b x + \log_b y$
  - Why it fails: The product law applies to products of $x$ and $y$, not sums of $x$ and $y$. There is no general rule for the logarithm of a sum.
  - Correct: Use $\log_b(xy) = \log_b x + \log_b y$ for products, leave $\log_b(x+y)$ unmodified unless you can factor the argument.
- **Wrong:** Claiming $\frac{\log_b x}{\log_b y} = \log_b\left(\frac{x}{y}\right)$
  - Why it fails: The quotient rule is often confused with change of base. The quotient rule gives a difference of logs, not a ratio.
  - Correct: $\log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y$, and $\frac{\log_b x}{\log_b y} = \log_y x$ by change of base.
- **Wrong:** Claiming $(\log_b x)^n = n\log_b x$
  - Why it fails: The power law applies to a power of the argument of the logarithm, not a power of the entire logarithm.
  - Correct: $\log_b(x^n) = n\log_b x$; $(\log_b x)^n$ cannot be simplified with the power law.
- **Wrong:** Accepting solutions that make any original logarithm argument negative
  - Why it fails: Logarithms are only defined for positive arguments, even if the simplified expression is defined for a negative value, the original expression is undefined.
  - Correct: Always check all original logarithm arguments are positive after solving for an unknown, and discard any invalid solutions.

## Cheatsheet

| Law Name | Formula | Restrictions |
| --- | --- | --- |
| Product Rule | $\log_b(xy) = \log_b x + \log_b y$ | $x>0, y>0, b>0, b\neq1$ |
| Quotient Rule | $\log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y$ | $x>0, y>0, b>0, b\neq1$ |
| Power Rule | $\log_b(x^n) = n\log_b x$ | $x>0, b>0, b\neq1$ |
| Change of Base | $\log_b a = \frac{\ln a}{\ln b} = \frac{\log_{10} a}{\log_{10} b}$ | $a>0, b>0, b\neq1$ |

## What's next

Mastering logarithm laws is a foundational skill for almost all subsequent topics in IB AI HL. These laws are used constantly to solve exponential equations, model exponential growth and decay, transform non-linear data into linear form for regression, and work with logarithmic functions in calculus. You will immediately apply these laws to solve exponential and logarithmic equations, which form the basis of almost all modeling problems you will encounter throughout the course.

- [Solving exponential equations with logarithms](https://www.owlsprep.com/study/ib-math-ai-hl-u1-solving-exponential-equations-with-logarithms/)

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