Simplification of logarithmic expressions

IB Mathematics: Applications and Interpretation SL· 37 min read

1. Core Logarithm Laws for Simplification★★☆☆☆⏱ 10 min

All logarithmic simplification relies on four core laws, which are directly derived from matching laws of exponents. Since logarithms are the inverse of exponential functions, every index rule maps directly to a log law.

📘 Definition

Core Logarithm Laws

For base $b > 0, b \neq = 1$ , $x > 0$ , $y > 0$ , real $k$

Product Rule: $lo g_{b} (x y) = lo g_{b} x + lo g_{b} y$ \ 2. Quotient Rule: $lo g_{b} (\frac{x}{y}) = lo g_{b} x - lo g_{b} y$ \ 3. Power Rule: $lo g_{b} (x^{k}) = k lo g_{b} x$ \ 4. Change of Base: $lo g_{b} a = \frac{l o g _{c} a}{l o g _{c} b}$ for any $c > 0, c \neq = 1$

Example:

$lo g_{2} (8 \times 4) = lo g_{2} 8 + lo g_{2} 4 = 3 + 2 = 5$

📐 Worked Example

Simplify $lo g_{3} 12 + lo g_{3} 9 - lo g_{3} 4$

1
Combine the first two terms using the product rule:
2
$\log_3 (12 \times 9) - \log_3 4 = \log_3 108 - \log_3 4$
3
Combine the two terms using the quotient rule:
4
$\log_3 \left(\frac{108}{4}\right) = \log_3 27$
5
Simplify the final logarithm, since $27 = 3^{3}$ :
6
$\log_3 27 = 3$

Exam tip:

Always check that all arguments are positive at the end of simplification, most IB questions will state variables are positive so you don't need to adjust for negative domains.

2. Simplifying Logarithms with Different Bases★★★☆☆⏱ 15 min

When simplifying expressions with logarithms of different bases, you first need to convert all terms to the same base using the change of base law. The most common choices for a shared base are 10 (common log) or $e$ (natural log), both are accepted in IB exams.

📐 Worked Example

Simplify $lo g_{4} 15 - lo g_{16} 25$ , writing your answer in terms of $lo g_{2} 3$ and $lo g_{2} 5$ .

1
Rewrite both logs with base 2 using change of base:
2
$\log_4 15 = \frac{\log_2 15}{\log_2 4} = \frac{\log_2 (3 \times 5)}{2} = \frac{\log_2 3 + \log_2 5}{2}$
3
Rewrite the second term similarly:
4
$\log_{16} 25 = \frac{\log_2 25}{\log_2 16} = \frac{2\log_2 5}{4} = \frac{\log_2 5}{2}$
5
Subtract the second term from the first and simplify:
6
$\frac{\log_2 3 + \log_2 5}{2} - \frac{\log_2 5}{2} = \frac{1}{2}\log_2 3$

3. Simplifying Natural Logarithmic Expressions★★★☆☆⏱ 12 min

Natural logarithms (base $e$ ) follow exactly the same laws as any other logarithm, but they have special simplification properties that come from the definition of base $e$ . These are frequently used in IB AI SL when working with exponential growth and decay.

📘 Definition

Special Properties of Natural Logarithms

$ln e = 1$ \ 2. $ln (e^{k}) = k$ for any real $k$ \ 3. $e^{l n k} = k$ for any $k > 0$

📐 Worked Example

Simplify $ln (e^{3} x^{2}) - ln (e x) + ln x$ for $x > 0$ .

1
Expand the first term using the product rule:
2
$\ln e^3 + \ln x^2 - \ln(e x) + \ln x$
3
Apply the natural log property and power rule:
4
$3 + 2\ln x - (\ln e + \ln x) + \ln x$
5
Simplify $ln e = 1$ , expand and collect like terms:
6
$3 + 2\ln x - 1 - \ln x + \ln x = 2 + 2\ln x$

4. Common Pitfalls

Wrong move:

Claiming $lo g_{b} (x + y) = lo g_{b} x + lo g_{b} y$

Why:

The product rule applies to products of arguments, not sums of arguments. There is no general rule for the log of a sum.

Correct move:

Only use the product rule for products: $lo g_{b} (x y) = lo g_{b} x + lo g_{b} y$ . $lo g_{b} (x + y)$ cannot be simplified further.

Wrong move:

Claiming $\frac{l o g _{b} x}{l o g _{b} y} = lo g_{b} (\frac{x}{y})$

Why:

The quotient rule is confused with the change of base rule. The quotient rule gives a difference of logs, not a ratio.

Correct move:

$lo g_{b} (\frac{x}{y}) = lo g_{b} x - lo g_{b} y$ , and $\frac{l o g _{b} x}{l o g _{b} y} = lo g_{y} x$ by change of base.

Wrong move:

Claiming $lo g_{b} (x^{k}) = (lo g_{b} x)^{k}$

Why:

The power rule misinterpreted: the exponent becomes a multiplier outside the log, not an exponent on the entire log term.

Correct move:

The power rule is $lo g_{b} (x^{k}) = k lo g_{b} x$ , always written as a multiplier.

Wrong move:

Simplifying $lo g_{b} x^{2} = 2 lo g_{b} x$ for all real $x$

Why:

$lo g_{b} x^{2}$ is defined for $x \neq = 0$ , but $2 lo g_{b} x$ is only defined for $x > 0$ , so this simplification is incomplete.

Correct move:

Write $lo g_{b} x^{2} = 2 lo g_{b} ∣ x ∣$ when simplifying for all non-zero $x$ .

5. Quick Reference Cheatsheet

Law Name	Formula	Notes
Product Rule	$lo g_{b} (x y) = lo g_{b} x + lo g_{b} y$	$x > 0, y > 0$
Quotient Rule	$lo g_{b} (\frac{x}{y}) = lo g_{b} x - lo g_{b} y$	$x > 0, y > 0$
Power Rule	$lo g_{b} (x^{k}) = k lo g_{b} x$	Valid for any real $k$
Change of Base	$lo g_{b} a = \frac{l o g _{c} a}{l o g _{c} b}$	Use $c = e$ or $c = 10$
Natural Log Property	$ln e^{k} = k$	$ln e = 1$
Inverse Property	$b^{l o g_{b} a} = a$	Requires $a > 0$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · Paper 1
Simplify 3-term log expression
2023 · Paper 1
Combine logs of different bases
2021 · Paper 2
Simplify natural log expression

What's Next

Simplifying logarithmic expressions is a mandatory foundation for solving both logarithmic and exponential equations, which are heavily tested across both IB AI SL papers. These skills also appear repeatedly in context for exponential growth and decay problems, logarithmic regression, and applied science questions like pH calculation. Mastering simplification now eliminates avoidable errors in later, higher-weighted exam questions, and builds the algebraic fluency needed for full marks on extended response questions.