Exponential and logarithmic functions

IB Mathematics Applications and Interpretation SL· 6 min read

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

📘 Definition

Exponential function

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

A function where the independent variable $x$ is the exponent of a constant base $a$ . Domain is all real numbers, range is all positive real numbers.

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 the exponential function with base $a$ . By definition, $lo g_{a} x = y$ if and only if $a^{y} = x$ . Domain is all positive real numbers, range is all real numbers.

Example:

\log_2 8 = 3 $, s in ce$ 2^3 = 8$

Exponential and logarithmic functions are inverses, so their graphs are reflections of each other across the line $y = x$ . This definition is the foundation for all conversions and equation solving.

📐 Worked Example

Convert $5^{3} = 125$ to logarithmic form, and convert $lo g_{4} 16 = 2$ to exponential form.

1
Recall the inverse relationship: $a^{y} = x ⟺ lo g_{a} x = y$
2
For $5^{3} = 125$ , we have $a = 5$ , $y = 3$ , $x = 125$ , so
3
$\log_5 125 = 3$
4
For $lo g_{4} 16 = 2$ , we have $a = 4$ , $x = 16$ , $y = 2$ , so
5
$4^2 = 16$

Exam tip:

Always check your conversion by substituting back to confirm you placed the base, exponent, and argument correctly.

2. Laws of Logarithms★★☆☆☆⏱ 20 min

All logarithm laws are derived directly from the laws of exponents. These rules let you simplify complex logarithmic expressions and solve equations.

Law Name	Logarithm Form	Exponent Equivalent
Product Rule	$lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$	$a^{m} a^{n} = a^{m + n}$
Quotient Rule	$lo g_{a} (\frac{x}{y}) = lo g_{a} x - lo g_{a} y$	$\frac{a ^{m}}{a ^{n}} = a^{m - n}$
Power Rule	$lo g_{a} (x^{n}) = n lo g_{a} x$	$(a^{m})^{n} = a^{mn}$
Change of Base	$lo g_{a} b = \frac{l n b}{l n a} = \frac{l o g _{10} b}{l o g _{10} a}$	N/A

📐 Worked Example

Simplify $2 ln 3 + ln 4 - ln 12$

1
Apply the power rule first to the first term:
2
$2 \ln 3 = \ln 3^2 = \ln 9$
3
Combine the first two terms with the product rule:
4
$\ln 9 + \ln 4 = \ln(9 \times 4) = \ln 36$
5
Apply the quotient rule for the final simplification:
6
$\ln 36 - \ln 12 = \ln\left(\frac{36}{12}\right) = \ln 3$

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

When the variable is in the exponent, we use logarithms to 'bring down' the exponent and solve. For logarithmic equations, we convert to exponential form after combining terms, then always check for extraneous solutions.

📐 Worked Example

Solve $lo g_{2} (x + 2) + lo g_{2} x = 3$

1
Combine the logs using the product rule:
2
$\log_2 [x(x+2)] = 3$
3
Convert from logarithmic to exponential form:
4
$x(x+2) = 2^3 = 8$
5
Rearrange into standard quadratic form:
6
$x^2 + 2x - 8 = 0$
7
Factor and solve the quadratic:
8
$(x+4)(x-2) = 0 \implies x = -4 \text{ or } x = 2$
9
Check solutions: $x = - 4$ gives undefined logs (negative arguments), so it is extraneous. $x = 2$ has positive arguments, so it is valid.
10
Final solution: $x = 2$

📐 Worked Example

Solve $3^{2 x - 1} = 24$ for $x$ , give your answer to 3 significant figures.

1
Take the natural logarithm of both sides:
2
$\ln\left(3^{2x -1}\right) = \ln 24$
3
Apply the power rule to bring down the exponent:
4
$(2x -1) \ln 3 = \ln 24$
5
Rearrange to isolate the term with $x$ :
6
$2x - 1 = \frac{\ln 24}{\ln 3} \approx 2.893$
7
Solve for $x$ :
8
$2x \approx 3.893 \implies x \approx 1.95$

4. Exponential Growth and Decay Modeling★★★☆☆⏱ 20 min

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📘 Definition

Exponential growth/decay model

A(t) = A_0 e^{kt}

Standard continuous model where $A_{0}$ is the initial quantity at $t = 0$ , $k$ is the constant growth/decay rate, $k > 0$ for growth and $k < 0$ for decay.

Example:

Population growth, radioactive decay, cooling

Most applied exponential problems in IB AI SL follow this standard form. The first step is always to use given information to find the unknown constant $k$ , then use the model to answer the question.

📐 Worked Example

A bacterial population grows exponentially. At $t = 0$ there are 100 bacteria, after 3 hours there are 800 bacteria. Find $k$ in the model $A (t) = 100 e^{k t}$ , then find the population after 5 hours.

1
Substitute the known values ( $t = 3, A = 800$ ) into the model:
2
$800 = 100 e^{3k}$
3
Divide both sides by 100 to isolate the exponential:
4
$8 = e^{3k}$
5
Take natural log of both sides:
6
$\ln 8 = 3k$
7
Solve for $k$ :
8
$k = \frac{\ln 8}{3} = \ln 2 \approx 0.693$
9
Calculate population at $t = 5$ :
10
$A(5) = 100 e^{5 \ln 2} = 100 (e^{\ln 2})^5 = 100 \times 2^5 = 3200$

Exam tip:

Always check that your final population or quantity makes sense: growth should give a larger number than the initial, decay should give a smaller number.

5. Common Pitfalls

Wrong move:

Forgetting to check for extraneous solutions when solving logarithmic equations

Why:

Quadratic equations often produce negative solutions that make log arguments negative, which are undefined

Correct move:

Always substitute every solution back into the original equation to confirm all log arguments are positive

Wrong move:

Writing $lo g_{a} x^{n} = (lo g_{a} x)^{n}$ instead of $n lo g_{a} x$

Why:

Confusion between the exponent of the argument and the exponent of the entire logarithm

Correct move:

Remember the power rule brings the exponent of the argument out front as a multiplier

Wrong move:

Splitting $lo g_{a} (x + y)$ into $lo g_{a} x + lo g_{a} y$

Why:

Confusing the product rule for logarithms with addition inside the log

Correct move:

The product rule only applies to products inside the log, not sums. $lo g_{a} (x + y)$ cannot be simplified in general.

Wrong move:

Taking the logarithm of both sides of $a^{x} + b = c$ before isolating $a^{x}$

Why:

Logarithms do not distribute over addition, so early log application leads to wrong simplification

Correct move:

Always rearrange to isolate the exponential term on one side before taking logarithms

Wrong move:

Confusing the domain of exponential and logarithmic functions

Why:

Mixing up which function can accept negative inputs

Correct move:

Exponentials accept any real $x$ and output positive values; logarithms only accept positive $x$ and output any real value

6. Quick Reference Cheatsheet

Concept	Key Formula	Note
Exponential ↔ Log conversion	$a^{y} = x ⟺ lo g_{a} x = y$	Inverse relationship
Product Rule	$lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$	$x > 0, y > 0$
Quotient Rule	$lo g_{a} (x / y) = lo g_{a} x - lo g_{a} y$	$x > 0, y > 0$
Power Rule	$lo g_{a} (x^{n}) = n lo g_{a} x$	$x > 0$
Change of Base	$lo g_{a} b = \frac{l n b}{l n a}$	Works for any positive base
Continuous Model	$A (t) = A_{0} e^{k t}$	$k > 0$ growth, $k < 0$ decay
Domain Check	$lo g_{a} x$ requires $x > 0$	Always check for extraneous solutions

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 · 1
Exponential growth model problem
2023 · 2
Solve quadratic logarithmic equation

What's Next

Exponential and logarithmic functions are one of the most frequently tested topics in IB AI SL, appearing in both pure and applied questions across both papers. They form the foundation for modeling all proportional growth and decay problems, which are core to applications in finance, biology, and physics. Mastering the rules here will make it much easier to tackle applied problems later in the course, where you will use these functions to model real-world phenomena and interpret results in context.