Exponential and logarithmic functions

IB Mathematics Applications and Interpretation HL· 35 min read

1. Core Definitions & Inverse Relationship★★☆☆☆⏱ 15 min

📘 Definition

Exponential Function

f(x) = a^x, \quad a>0, a \neq 1

A function where the independent variable $x$ appears as the exponent of a constant base $a$ . All real numbers are in the domain, and output values are always positive.

Example:

$f (x) = 2^{x}$ , $f (x) = e^{x}$

📘 Definition

Logarithmic Function

f(x) = \log_a x, \quad a>0, a \neq 1

The inverse of an exponential function with base $a$ . By definition, $lo g_{a} x = y$ if and only if $a^{y} = x$ . Only positive input values are defined.

Example:

$lo g_{2} 8 = 3$ because $2^{3} = 8$

Since exponential and logarithmic functions are inverses, their graphs are reflections of each other over the line $y = x$ . This inverse relationship is the foundation for all algebraic manipulation involving these functions.

📐 Worked Example

Convert $5^{3} = 125$ to logarithmic form, and convert $lo g_{4} (\frac{1}{16}) = - 2$ to exponential form.

1
Recall the core inverse definition: $a^{y} = x ⟺ lo g_{a} x = y$
2
For $5^{3} = 125$ , we identify $a = 5$ , $y = 3$ , $x = 125$ . Substituting into the definition gives:
3
$\log_5 125 = 3$
4
For $lo g_{4} (\frac{1}{16}) = - 2$ , identify $a = 4$ , $y = - 2$ , $x = \frac{1}{16}$ . Substituting gives:
5
$4^{-2} = \frac{1}{16}$

Exam tip:

Always check your conversion by substituting back into the original form to avoid mixing up the base and exponent.

2. Laws of Logarithms★★☆☆☆⏱ 20 min

All logarithm laws derive directly from exponent laws, because logarithms are inverses of exponents. The core rules for any valid base $a$ are:

Product rule: $lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$
Quotient rule: $lo g_{a} (\frac{x}{y}) = lo g_{a} x - lo g_{a} y$
Power rule: $lo g_{a} x^{n} = n lo g_{a} x$
Change of base rule: $lo g_{a} b = \frac{l n b}{l n a} = \frac{l o g _{c} b}{l o g _{c} a}$ for any positive $c \neq = 1$

📐 Worked Example

Simplify $2 ln 3 + ln 4 - ln 12$ into a single logarithm.

1
Apply the power rule to the first term, bringing the 2 out front into the exponent:
2
$2 \ln 3 = \ln 3^2 = \ln 9$
3
Apply the product rule to the first two terms, combining the sum of logs into the log of a product:
4
$\ln 9 + \ln 4 = \ln (9 \times 4) = \ln 36$
5
Apply the quotient rule, combining the difference of logs into the log of a quotient:
6
$\ln 36 - \ln 12 = \ln \left(\frac{36}{12}\right) = \ln 3$
7
The final simplified form is $ln 3$ .

3. Graphs and Key Properties★★★☆☆⏱ 20 min

Exponential and logarithmic functions have consistent shapes and key features for bases $a > 1$ (the most common case in IB exams). The table below summarizes core properties:

Feature	Exponential $f (x) = a^{x}, a > 1$	Logarithmic $f (x) = lo g_{a} x, a > 1$
Domain	$x \in R$	$x > 0$
Range	$y > 0$	$y \in R$
Asymptote	Horizontal: $y = 0$	Vertical: $x = 0$
Intercept	$y$ -intercept at $(0, 1)$	$x$ -intercept at $(1, 0)$
Monotonicity	Strictly increasing	Strictly increasing

📐 Worked Example

Sketch the graph of $f (x) = 2^{x - 1} + 3$ , label the asymptote and $y$ -intercept.

1
This is a transformation of the base $f (x) = 2^{x}$ graph: shifted 1 unit right, 3 units up.
2
The base graph has a horizontal asymptote at $y = 0$ . Shifting up 3 units moves the asymptote to:
3
$y = 3$
4
Calculate the $y$ -intercept by substituting $x = 0$ :
5
$f(0) = 2^{-1} + 3 = \frac{1}{2} + 3 = 3.5$
6
The graph is strictly increasing, approaches $y = 3$ as $x \to - \infty$ , and increases to infinity as $x \to + \infty$ , with $y$ -intercept at $(0, 3.5)$ .

Exam tip:

Exponential functions always have horizontal asymptotes; logarithmic functions always have vertical asymptotes. Mixing these up is a very common exam error.

4. Solving Exponential and Logarithmic Equations★★★☆☆⏱ 25 min

✓ Calculator OK

To solve equations involving exponential functions with the variable in the exponent, we take logarithms of both sides to bring down the exponent, then isolate the variable. For logarithmic equations, we use logarithm laws to combine terms, then convert to exponential form to solve.

📐 Worked Example

Solve $3^{2 x} = 5^{x + 1}$ for $x$ , give your answer to 3 significant figures.

1
Take the natural logarithm of both sides to bring down exponents:
2
$\ln \left(3^{2x}\right) = \ln \left(5^{x+1}\right)$
3
Apply the power rule to both sides:
4
$2x \ln 3 = (x+1) \ln 5$
5
Expand the right-hand side and collect terms with $x$ on the left:
6
$2x \ln 3 - x \ln 5 = \ln 5 \implies x (2 \ln 3 - \ln 5) = \ln 5$
7
Isolate $x$ and calculate the numerical value:
8
$x = \frac{\ln 5}{2 \ln 3 - \ln 5} \approx \frac{1.609}{2.197 - 1.609} \approx 2.74$
9
Final answer to 3 significant figures is $x \approx 2.74$ .

✓ Quick check

Test your understanding with this quick question:

What is the solution to $lo g_{2} (x + 1) = 3$ ?
- A) 2
- B) 5
- C) 7
- D) 8
Reveal answer
C) 7 —
Use the definition of logarithm: $lo g_{2} (x + 1) = 3 ⟺ 2^{3} = x + 1 ⟹ 8 = x + 1 ⟹ x = 7$ . Don't forget to subtract 1 after calculating $2^{3}$ !

5. Common Pitfalls

Wrong move:

Claiming $ln (x + y) = ln x + ln y$

Why:

The product rule only applies to products of terms, not sums. There is no general rule for the logarithm of a sum.

Correct move:

$ln (x y) = ln x + ln y$ , $ln (x + y)$ cannot be simplified further.

Wrong move:

Accepting solutions that result in negative arguments for logarithms

Why:

Logarithms of non-positive numbers are undefined, so extraneous solutions are very common with logarithmic equations.

Correct move:

Always substitute all solutions back into the original equation to check all arguments are positive, and reject any invalid solutions.

Wrong move:

Stating $f (x) = e^{x} + 2$ has a vertical asymptote at $x = 2$

Why:

Confusing the type and position of asymptotes for transformed exponential functions.

Correct move:

Exponential functions have horizontal asymptotes. $f (x) = e^{x} + 2$ has a horizontal asymptote at $y = 2$ .

Wrong move:

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

Why:

Misapplying the quotient rule for logarithms. The quotient rule applies to the logarithm of a quotient, not a quotient of logarithms.

Correct move:

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

Wrong move:

When solving $a^{x} = b$ , writing $x = \frac{l o g b}{a}$

Why:

Common algebraic error when isolating the variable $x$ .

Correct move:

If $a^{x} = b$ , then $x = lo g_{a} b = \frac{l n b}{l n a}$ , there is no division by the base $a$ .

6. Quick Reference Cheatsheet

Rule/Relationship	Formula	Key Graph Feature
Core Inverse Relationship	$a^{y} = x ⟺ lo g_{a} x = y$	Exponential: Horizontal asymptote
Product Rule	$lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$	Logarithm: Vertical asymptote
Quotient Rule	$lo g_{a} (x / y) = lo g_{a} x - lo g_{a} y$	Inverses reflect over $y = x$
Power Rule	$lo g_{a} x^{n} = n lo g_{a} x$	Domain of log: $x > 0$
Change of Base Rule	$lo g_{a} b = \frac{l n b}{l n a}$	Range of exp: $y > 0$

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.

2023 · 1
Solve exponential growth equation
2022 · 2
Simplify logarithmic expression
2021 · 1
Sketch transformed log graph

Going deeper

IB SyllabusIB AI HL Syllabus Topic 2: FunctionsOfficial syllabus reference for this subtopic

What's Next

Exponential and logarithmic functions are the foundation for nearly all continuous modeling topics in IB AI HL. You will use these functions to model real-world scenarios like population growth, radioactive decay, compound interest, and cooling processes. Mastery of algebraic manipulation and graph properties is critical for success on both paper 1 and paper 2 exams, and these concepts will also be extended when you study derivatives and integrals of transcendental functions later in the course. Building a strong understanding now will make more complex application and calculus topics much easier to master.