IB Maths AA SL · Unit 2: Functions
IB Maths AA SL Exponential and Logarithmic Functions Questions
Exam-style IB Maths AA SL exponential and logarithmic functions questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 15 questions
- Paper 1: 8
- Paper 2: 7
- 4 easy
- 5 medium
- 4 hard
- 2 very hard
- 3 worked examples
Practise Exponential and Logarithmic Functions questions →
AA SL formula booklet
What's examined in AA SL exponential and logarithmic functions
The question bank covers these exponential and logarithmic functions question types (number of questions in brackets):
- Solving Exp Log Equations (7)
- Modelling and Graphing Functions (5)
- Log Laws and Function Properties (3)
Exponential and Logarithmic Functions worked examples
Worked example 1: Solving basic exponential equations · easy
Solve $e^{3x - 1} = 20$ algebraically. Give your exact answer in terms of natural logarithms.
1. Take the natural logarithm of both sides: $\ln(e^{3x - 1}) = \ln 20$.
2. Simplify via inverse property: $3x - 1 = \ln 20$.
3. Add $1$ to both sides: $3x = \ln 20 + 1$.
4. Divide by $3$: $\mathbf{x = \frac{\ln 20 + 1}{3}}$.
Examiner tip: Do not approximate $\ln 20$ with a decimal when a question specifically asks for an exact algebraic answer. Doing so will cost you the final accuracy mark.
Worked example 2: Logarithmic laws combination · medium
Solve $\log_2 x + \log_2(x - 6) = 4$ algebraically.
1. Apply the product rule: $\log_2(x(x - 6)) = 4$.
2. Convert to exponential form: $x(x - 6) = 16$.
3. Expand and rearrange: $x^2 - 6x - 16 = 0$.
4. Factorise: $(x - 8)(x + 2) = 0 \implies x = 8$ or $x = -2$.
5. Reject $x = -2$ since $\log_2(-2)$ is undefined. Valid solution: $\mathbf{x = 8}$.
Examiner tip: Always check final roots against the domain of the original logarithms; arguments must be strictly positive. Leaving $x = -2$ as a final answer will lose reasoning marks.
Worked example 3: Hidden quadratics in exponentials · hard
Solve $e^{2x} - 8e^x + 15 = 0$ algebraically. Give exact answers in logarithmic form.
1. Rewrite using index laws: $(e^x)^2 - 8(e^x) + 15 = 0$.
2. Substitute $u = e^x$: $u^2 - 8u + 15 = 0$.
3. Factorise: $(u - 5)(u - 3) = 0 \implies u = 5$ or $u = 3$.
4. Substitute back: $e^x = 5$ or $e^x = 3$.
5. Take natural logs: $\mathbf{x = \ln 5}$ and $\mathbf{x = \ln 3}$.
Examiner tip: Using a substitution $u = e^x$ makes the quadratic structure obvious and drastically reduces factorisation errors.
FAQ
How many IB Maths AA SL exponential and logarithmic functions questions are there?
There are 15 exam-style exponential and logarithmic functions questions in the AA SL question bank (Paper 1: 8 · Paper 2: 7), graded 4 easy, 5 medium, 4 hard, 2 very hard. Every question has a full IB-style mark scheme (M, A and R marks).
Is exponential and logarithmic functions on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 8 · Paper 2: 7. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA SL Unit 2 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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