IB Math AA HL Exponentials & Logarithms — Complete Cheatsheet

Exponentials and Logarithms underpin some of the highest-leverage questions in IB Mathematics Analysis & Approaches HL. The core skills — index laws, log laws, change of base, hidden-quadratic substitutions — appear on every Paper 1, and Section B questions routinely reach 18 marks by combining logs with sequences, calculus, or graph transformations. The HL extension over SL adds the discriminant approach to log domains, two-intersection-point parameter problems, and the "express in terms of $p$ and $q$" identities that test whether you really understand log laws or just memorised the four rules.

The single biggest source of dropped marks is inventing log laws that don't exist — most commonly $\ln(a + b) = \ln a + \ln b$. The second is forgetting to check the domain of the original equation after solving. This cheatsheet condenses every formula and trap from SL 1.5–1.6 and AHL 1.7–1.8 into one page you can revise from. For the printable PDF, full notes, the tutorial booklet, and the marked-up solutions, scroll to the bottom.

§1 — Laws, Definitions & Key Results SL 1.5–1.6

The eight index laws

Trick: convert to the same base first. Once bases match, exponents add, subtract, or multiply.

What is a logarithm?

"To what power must I raise the base to get this number?"

$$\log_a x = y \;\Leftrightarrow\; a^y = x \qquad (a > 0,\; a \neq 1,\; x > 0).$$

Special notations: $\ln x = \log_e x$ (base $e \approx 2.718$); $\log x = \log_{10} x$.

Instant results: $\log_a 1 = 0$; $\log_a a = 1$; $\log_a(a^n) = n$; $a^{\log_a x} = x$; $\ln e = 1$; $e^{\ln x} = x$; $\ln(e^x) = x$.

The four laws of logarithms

Trick: apply the power rule first to bring down any exponents, then combine using product/quotient rules.

§2 — Solving Strategies SL 1.5–1.6, AHL 1.7–1.8

Same base — the simplest method

Both sides can be written with the same base. Equate exponents directly.

Example: Solve $27^{x+1} = 9^{2x-3}$. Write both as powers of 3: $3^{3(x+1)} = 3^{2(2x-3)}$, so $3x + 3 = 4x - 6 \Rightarrow x = 9$.

Example (simplify): $\dfrac{4^{x+1} \cdot 8^{x-1}}{16^x} = \dfrac{2^{2x+2} \cdot 2^{3x-3}}{2^{4x}} = \dfrac{2^{5x-1}}{2^{4x}} = 2^{x-1}$.

Taking logarithms — when bases differ

Equation has $a^x = k$ or $a^f = b^g$ with different bases. Take $\ln$ both sides; the exponent comes down as a multiplier.

Example: Solve $2^{3x+1} = 3^{x-2}$ (exact form). $(3x + 1) \ln 2 = (x - 2) \ln 3$, expand and collect: $x(3 \ln 2 - \ln 3) = -2 \ln 3 - \ln 2$, so $x = \dfrac{\ln 18}{\ln(3/8)}$.

Hidden quadratic — the $u = a^x$ substitution

Equation contains $a^{2x}$ and $a^x$ together: $\alpha\,a^{2x} + \beta\,a^x + \gamma = 0$. Let $u = a^x$, so $a^{2x} = u^2$. Solve the quadratic in $u$, back-substitute, and discard $u \leq 0$ since $a^x > 0$ always.

Example: Solve $4^x - 5 \cdot 2^x + 4 = 0$. Let $u = 2^x$, $4^x = u^2$: $u^2 - 5u + 4 = 0 \Rightarrow (u - 1)(u - 4) = 0$. So $2^x = 1 \Rightarrow x = 0$, or $2^x = 4 \Rightarrow x = 2$.

Example with shifted exponent: $9^x - 4 \cdot 3^{x+1} + 27 = 0$. First expand: $4 \cdot 3^{x+1} = 12 \cdot 3^x$. Let $u = 3^x$: $u^2 - 12u + 27 = 0 \Rightarrow (u - 3)(u - 9) = 0 \Rightarrow x = 1$ or $x = 2$.

Logarithmic equations — four steps

1. Collect logs into a single $\log$ on each side using the four laws.
2. Exponentiate to remove the logs.
3. Solve the algebraic equation that remains.
4. Check the domain — every original log argument must be $> 0$.

Example: Solve $\log_2 x + \log_2(x - 2) = 3$. Combine: $\log_2[x(x - 2)] = 3 \Rightarrow x(x - 2) = 8 \Rightarrow x^2 - 2x - 8 = 0 \Rightarrow (x - 4)(x + 2) = 0$. Domain: $x > 0$ and $x - 2 > 0$, so reject $x = -2$. Final answer: $x = 4$ (R1 mark for the domain check).

Mixed-base log equations — use $\ln$

Convert all logs to $\ln$ via $\log_a x = \dfrac{\ln x}{\ln a}$. Let $u = \ln x$, collect, and solve linearly. On Paper 2 you can also graph both sides on the GDC and read off the intersection.

Example (2025 Nov TZ3 P1): $3 \log_5(10x) - \log_4 x = 1$. Convert: $\dfrac{3(\ln 10 + \ln x)}{\ln 5} - \dfrac{\ln x}{\ln 4} = 1$. Let $u = \ln x$, collect, solve: $u \approx -2.880 \Rightarrow x = e^{-2.880} \approx 0.0561$.

§3 — Graphs, Transformations & Applications SL 1.5–1.6, AHL 1.7–1.8

Graphs of $y = a^x$ and $y = \log_a x$

These two graphs are reflections of each other in $y = x$ — they are inverses.

Asymptotes after transformation

For exp: vertical shifts move the asymptote; horizontal shifts do NOT. For log: horizontal shifts move the asymptote; vertical shifts do NOT.

Example: $y = 2^{x-3} + 5$. Asymptote: $y = 5$; range: $y > 5$; domain: $\mathbb{R}$; $y$-int at $x = 0$ is $y = 2^{-3} + 5 = 5.125$.

Example: $y = \ln(x + 4) - 2$. Asymptote: $x = -4$; domain: $x > -4$; $x$-int from $\ln(x + 4) = 2 \Rightarrow x = e^2 - 4$.

Exponential growth and decay

Any percentage growth/decay, half-life, doubling time, compound interest, or population modelling. Standard models:

• $P(t) = P_0 \cdot a^t$  ($a > 1$: growth; $0 < a < 1$: decay)
• $P(t) = P_0 \cdot e^{kt}$  ($k > 0$: growth; $k < 0$: decay)
• At $t = 0$: $P = P_0$.

Half-life: $a^{T_{1/2}} = \tfrac{1}{2}$  $\Rightarrow$  $T_{1/2} = \dfrac{\ln 2}{|\ln a|} = \dfrac{\ln 2}{|k|}$.

Doubling time: $T_2 = \dfrac{\ln 2}{\ln a} = \dfrac{\ln 2}{k}$.

Decibel scale (2024 P2): $L = 10 \log_{10}(I \times 10^{12})$; doubling $I$ adds $10 \log_{10} 2 \approx 3$ dB.

Logarithms in sequences

Key result: if $a, ar, ar^2$ are in GP, then $\ln a, \ln a + \ln r, \ln a + 2 \ln r$ are in AP with common difference $d = \ln r$. The reverse also holds — logs in AP $\Rightarrow$ originals in GP.

Example (2024 Nov P2): $S_n = \sum 5(\log_3 c)^r$. This is GP with ratio $r = \log_3 c$; converges iff $|\log_3 c| < 1$, giving $\tfrac{1}{3} < c < 3$, $c \neq 1$.

§4 — Advanced Exam Insights AHL 1.7–1.8

"Express in terms of $p$ and $q$"

Given $\log a = p$ and $\log b = q$, find $\log$ of an expression involving $a$ and $b$. Step 1: factorise the argument into prime powers of $a$ and $b$ only. Step 2: apply product + power rules. For a different base, apply change of base first.

Example (2025 May TZ3 P1 Q1): $\log_{10} 2 = p$, $\log_{10} 3 = q$. (a) $\log_{10} 24 = \log_{10}(2^3 \cdot 3) = 3p + q$. (b) $\log_3 8 = \dfrac{\log_{10}(2^3)}{\log_{10} 3} = \dfrac{3p}{q}$.

Power inside the base

$\log_{a^k} x = \dfrac{\log_a x}{k}$. A power inside the base divides the result. Proof: $\log_{a^k} x = \dfrac{\ln x}{\ln a^k} = \dfrac{\ln x}{k \ln a} = \dfrac{1}{k} \log_a x$.

Examples: $\log_{1000} a = \dfrac{\log_{10} a}{3}$; $\log_{27} 9 = \dfrac{\log_3 9}{\log_3 27} = \dfrac{2}{3}$; $\log_{16} 8 = \dfrac{3}{4}$.

Domain of a logarithmic function — discriminant method

"Find values of $k$ such that $\log_a(Ax^2 + Bx + C)$ is defined for all $x \in \mathbb{R}$." For the log to be defined everywhere, the argument must be strictly positive everywhere: require $A > 0$ AND $\Delta = B^2 - 4AC < 0$. Then solve the resulting inequality in $k$.

Example (2024 Nov P2 Q5): $h(x) = \log_{10}(4x^2 - rx + r - 1)$. Need $4x^2 - rx + r - 1 > 0$ always. Leading coeff $= 4 > 0$ ✓. Discriminant: $r^2 - 16(r - 1) < 0 \Rightarrow r^2 - 16r + 16 < 0 \Rightarrow r = 8 \pm 4\sqrt{3}$. Final answer: $8 - 4\sqrt{3} < r < 8 + 4\sqrt{3}$.

Two distinct intersection points of log graphs

Two log/exp functions intersect at two distinct points; find conditions on a parameter. Step 1: set $f(x) = g(x)$ and use log laws to reach a single log on each side. Step 2: equate arguments to get a quadratic in $x$. Step 3: two distinct roots $\Rightarrow \Delta > 0$. Step 4: check both roots are in the domain of the original functions.

Example (2023 Nov TZ1 P1 Q10): $f = \ln(2x - 9)$, $g = 2 \ln x - \ln d$, intersect at 2 points. Equate $\Rightarrow 2x - 9 = x^2/d \Rightarrow x^2 - 2dx + 9d = 0$. Two roots: $\Delta = 4d^2 - 36d > 0 \Rightarrow d > 9$ (since $d > 0$). For $d = 10$: $x = 10 \pm \sqrt{10}$; check $10 - \sqrt{10} > 4.5$ ✓; $q - p = 2\sqrt{10}$.

Logarithms in series — the classic 18-mark question

Series with $\ln x$ terms. Asked: GP or AP? Find the sum and find $x$. For GP, equate ratios: $r = u_2 / u_1 = u_3 / u_2$. For AP, equate differences. Key move: divide through by $\ln x$ (valid since $x > 1$, so $\ln x \neq 0$).

Example (2022 May TZ1 P1 Q10): $\ln x + p \ln x + \tfrac{1}{3} \ln x + \cdots$, $x > 1$.
GP case: ratio constant $\Rightarrow p = \dfrac{1}{3p} \Rightarrow p = \pm \dfrac{1}{\sqrt{3}}$; converges since $|r| = \dfrac{1}{\sqrt{3}} < 1$. For $p > 0$, $S_\infty = 3 + \sqrt{3}$ gives $\ln x = 2 \Rightarrow x = e^2$.
AP case: $p - 1 = \dfrac{1}{3} - p \Rightarrow p = \dfrac{2}{3}$; common difference $d = -\dfrac{1}{3} \ln x$. With $S_n = \ln(x^{-3})$ you obtain $n = 9$.

Worked Example — Hidden Quadratic with Domain Check

Common Student Questions