IB Math AI HL Exponents & Logarithms — Complete Cheatsheet
Every exponent and logarithm skill for IB Mathematics Applications & Interpretation HL — index & log laws, negative & fractional indices, the natural log, change of base, and the log scales behind decibels, the Richter scale & pH. Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Exponents and logarithms are the engine room of IB Mathematics Applications & Interpretation HL. AI treats them less as abstract algebra and more as a modelling toolkit: an exponent packages anything that grows or decays by a fixed factor — money, populations, radioactive samples — while a logarithm is the inverse key that pulls the unknown back down out of the exponent and tames ranges spanning billions onto a readable scale. Because a calculator is always allowed, the marks live in choosing the right method and reading the answer correctly — not in grinding the arithmetic by hand.
This cheatsheet condenses the whole of SL 1.5 and AHL 1.9 — the index laws, negative and fractional indices, solving exponential equations, the laws of logarithms, the natural log, change of base, and the real-world log scales behind decibels, the Richter scale and pH — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — The laws of exponents SL 1.5
An exponent is repeated multiplication, and every manipulation below reduces to one idea: same base, combine the indices. These are the fluency skills the rest of the topic is built on.
Product:$a^{m}\times a^{n}=a^{m+n}$ — same base, add the indices
Quotient:$\dfrac{a^{m}}{a^{n}}=a^{m-n}$ — same base, subtract the indices
Power of a power:$(a^{m})^{n}=a^{mn}$ — multiply the indices
Over a product:$(ab)^{n}=a^{n}b^{n}$ and $\left(\dfrac{a}{b}\right)^{n}=\dfrac{a^{n}}{b^{n}}$
Zero index:$a^{0}=1$ for any $a\neq 0$
TrickWhen the bases already match you never need the calculator: add the top indices and subtract the bottom one, e.g. $\dfrac{2^{7}\times2^{3}}{2^{4}}=2^{7+3-4}=2^{6}=64$.
TrapThese laws only apply to a common base. $2^{3}\times5^{2}$ cannot be merged into a single power, and $a^{m}+a^{n}$ never simplifies by adding indices — the laws are about multiplying, not adding.
§2 — Negative & fractional indices SL 1.5
Extending the index laws to negative and fractional powers is where marks are quietly won or lost — the rules are short, but the sign and the order matter.
Negative:$a^{-n}=\dfrac{1}{a^{n}}$ — a negative index means reciprocal, giving a positive fraction
Unit fraction:$a^{1/n}=\sqrt[n]{a}$ — the $n$-th root
General fraction:$a^{m/n}=\left(\sqrt[n]{a}\right)^{m}=\sqrt[n]{a^{m}}$ — root first, then power
TrickRead the fraction as "denominator = root, numerator = power", and take the root first to keep the numbers small: $27^{2/3}=\left(\sqrt[3]{27}\right)^{2}=3^{2}=9$.
TrapA negative index is not a negative number: $2^{-3}=\tfrac{1}{8}$, never $-8$. Flip the base to make the index positive, then evaluate the fraction.
Same-base equation:$b^{ax+c}=b^{k}\ \Rightarrow\ ax+c=k$ — equal bases force equal indices
NoteType the model in one line on the GDC as A*b^t and use the power key: a factor $b=1.05$ is 5% growth per step, $b=0.92$ is 8% decay. Never multiply the base by $t$.
TrapSame-base equating only works once both sides are a power of the same base. Rewrite first: $8^{x}=2^{x+5}\Rightarrow(2^{3})^{x}=2^{x+5}\Rightarrow3x=x+5\Rightarrow x=2.5$. If no common base exists, take logs (§7).
§4 — Logarithms: the inverse of an exponent SL 1.5
A logarithm answers the question an exponent hides: to what power must I raise the base to get this number? That single idea is what lets you pull an unknown down out of an exponent.
Definition:$x=\log_b a \iff b^{x}=a$, valid for $b\gt 0,\ b\neq 1,\ a\gt 0$
Common log:$\log x=\log_{10}x$ — base 10, the scale behind Richter, decibels and pH
Natural log:$\ln x=\log_{e}x$, where $e\approx 2.71828$ — the base of continuous growth
Inverse pair:$\log_b(b^{x})=x$ and $b^{\log_b x}=x$ — logs and powers undo each other
TrickRead every logarithm aloud as a power: $\log_2 32=5$ because $2^{5}=32$, and $\log_{10}1000=3$ because $10^{3}=1000$. Your GDC evaluates any $\log$ or $\ln$ directly.
NoteBecause the argument $a$ must be positive, the $\log$ of $0$ or of a negative number is undefined — a solution that forces a negative argument must be rejected.
§5 — The laws of logarithms AHL 1.9
Because logs turn powers into products and products into sums, three laws let you compress or expand any logarithmic expression — and it is the power law that finally lets you solve for an exponent.
Product:$\log_b(xy)=\log_b x+\log_b y$
Quotient:$\log_b\!\dfrac{x}{y}=\log_b x-\log_b y$
Power:$\log_b x^{n}=n\log_b x$ — brings the exponent down as a multiplier
Special values:$\log_b 1=0$ and $\log_b b=1$
TrickCombine into a single log, then ask "what power of the base is the argument?": $\log_5 100-\log_5 4=\log_5 25=2$, because $5^{2}=25$.
TrapThe laws act on products and quotients, never on sums: $\log(x+y)\neq\log x+\log y$. And a quotient becomes $\log x-\log y$ — subtract the logs, do not divide them.
§6 — The natural logarithm & continuous change AHL 1.9
When growth or decay is continuous rather than in discrete steps, the model is written with base $e$, and $\ln$ is the tool that unlocks the time variable.
Continuous growth:$V=V_0e^{rt}$ — $V_0$ is the initial value, $r$ the continuous rate
Solve for time:$e^{rt}=m\ \Rightarrow\ rt=\ln m\ \Rightarrow\ t=\dfrac{\ln m}{r}$
Half-life decay:$N=N_0e^{-kt}$ with $\tfrac12=e^{-kh}\ \Rightarrow\ k=\dfrac{\ln 2}{h}$
TrickTo undo a base-$e$ exponential, isolate $e^{rt}$ first, then take $\ln$ of both sides — $\ln$ and $e$ cancel, leaving $rt$ on its own to divide out.
TrapThe decay constant is $k=\dfrac{\ln 2}{h}\approx\dfrac{0.693}{h}$, not $\tfrac{0.5}{h}$. A longer half-life gives a smaller $k$; the natural log is essential.
§7 — Solving exponential equations with logs AHL 1.9
When the two sides share no common base, take logs of both sides and use the power law to bring the unknown down. Any base works, so read $\ln$ or $\log$ straight off the GDC.
Solve $a^{x}=b$:$x=\dfrac{\log b}{\log a}=\dfrac{\ln b}{\ln a}$ — keep $\log b$ on top
Change of base:$\log_b a=\dfrac{\ln a}{\ln b}=\dfrac{\log_{10}a}{\log_{10}b}$ — rewrite any base as $\ln$ or $\log$
With a shift (cooling):$T=R+A\,b^{t}\ \Rightarrow\ b^{t}=\dfrac{T-R}{A}\ \Rightarrow\ t=\dfrac{\ln\!\left((T-R)/A\right)}{\ln b}$
TrickChange of base means you never need a "$\log_b$" button: to find $\log_{1.05}3$, just type $\dfrac{\ln 3}{\ln 1.05}$. Give the answer to 3 s.f. unless told otherwise.
TrapFor a compound-growth "how many whole years until it first exceeds a target", $n=\dfrac{\log(T/P)}{\log(1+i)}$ — then round UP to the next whole year, never to 3 s.f. With a cooling model, forgetting to subtract $R$ before taking logs is the other classic slip.
§8 — Log scales: Richter, decibels & pH AHL 1.9
Some real-world quantities span factors of billions, so science measures them on a logarithmic scale where each equal step multiplies the quantity by a fixed factor. This is the flagship AI application of logarithms.
A logarithmic scale: the numbers on top jump ×10 each equal step, while their positions — the base-10 logarithms below — rise by just 1. That is how a log scale squeezes billions into a short, readable ruler.
Decibels:$L=10\log_{10}\!\dfrac{I}{I_0}$ dB, with threshold $I_0=10^{-12}\,\text{W m}^{-2}$
Richter:a rise of $\Delta M$ in magnitude means the amplitude grows by a factor $10^{\Delta M}$
Acidity:$\text{pH}=-\log_{10}[H^{+}]$ — one pH unit is a tenfold change in $[H^{+}]$
TrickWrite the defining log equation first, then work outwards: for decibels, form the ratio $I/I_0$, take $\log_{10}$, then multiply by 10. To go the other way, undo each step in turn.
TrapA difference on a log scale is a ratio in reality, not a sum: $+2$ on the Richter scale is $\times100$ amplitude, and $+10$ dB is $\times10$ intensity. Never drop the $I_0$ or the factor of $10$ in the decibel formula.
§9 — Synthesis: linearising models with logs AHL 1.9
This is where Exponents & Logarithms meets Non-Linear Modelling & Regression. A curved exponential or power model is no use for a straight-line fit — until you take logs, which linearise it so least-squares regression applies.
TrickThe plotted axis is a logarithm, so reverse it to get a real value back: a semi-log prediction is $y=e^{\,mx+c}$ (base $e$) or $y=10^{\,mx+c}$ (base 10). Match the exponential base to the log you took.
TrapThe gradient and intercept are logarithms of the constants, not the constants themselves. Antilog them ($e^{\ldots}$ for $\ln$, $10^{\ldots}$ for $\log_{10}$) — reporting the raw gradient as $b$ or $k$ is the standard mistake.
Rewrite each side as one base, then equate the indices
No common base — take logs instead
"Evaluate / simplify $\log_b a$"
Ask "what power of $b$ gives $a$?"; combine logs first
$\log(x+y)\neq\log x+\log y$
"Solve $a^{x}=b$"
$x=\dfrac{\ln b}{\ln a}$ (any base)
Keep $\log b$ on top; give 3 s.f.
"Half-life / continuous decay"
$N=N_0e^{-kt}$ with $k=\dfrac{\ln 2}{h}$
$k$ is a rate ($\ln 2/h$), not $0.5/h$
"Whole years to first exceed"
$n=\dfrac{\log(T/P)}{\log(1+i)}$, then round up
Round UP, never to 3 s.f.
"Decibels / Richter / pH"
Write the defining log law; form the ratio, then $\log_{10}$
Include $I_0$; a dB gap is a ratio, not a sum
"Linearise / semi-log plot"
$\ln y=\ln a+(\ln b)x$; regress $\ln y$ on $x$
Gradient and intercept are logs — antilog them
Worked Example — AI HL-Style Sound Levels
Question (AI HL Paper 2 style — 6 marks)
The loudness of a sound of intensity $I$ (in $\text{W m}^{-2}$) is measured in decibels by $L=10\log_{10}\!\dfrac{I}{I_0}$, where $I_0=10^{-12}\,\text{W m}^{-2}$ is the threshold of hearing. (a) A machine produces sound of intensity $I=3.5\times10^{-4}\,\text{W m}^{-2}$. Find its sound level $L$, in dB. (b) A library has a sound level of $40$ dB. Find the intensity of sound in the library. (c) Hence find how many times more intense the machine is than the library.
Solution
(a) Form the ratio, then take the log: $\dfrac{I}{I_0}=\dfrac{3.5\times10^{-4}}{10^{-12}}=3.5\times10^{8}$. So $L=10\log_{10}(3.5\times10^{8})=85.4$ dB (3 s.f.). (M1)(A1)
(b) Reverse the formula: $40=10\log_{10}\dfrac{I}{I_0}\Rightarrow\log_{10}\dfrac{I}{I_0}=4\Rightarrow\dfrac{I}{I_0}=10^{4}$. Hence $I=10^{4}\times10^{-12}=10^{-8}\,\text{W m}^{-2}$. (M1)(A1)
(c) The ratio of intensities is $\dfrac{3.5\times10^{-4}}{10^{-8}}=3.5\times10^{4}=35\,000$ times. (M1)(A1)(Check: the levels differ by $85.4-40=45.4$ dB, and $10^{45.4/10}=10^{4.54}\approx3.5\times10^{4}$.)
Examiner's note: always form the ratio $I/I_0$ first — dropping the threshold $I_0=10^{-12}$ or the factor of $10$ are the two standard slips. Keep answers to 3 s.f. and carry full calculator accuracy between parts. Remember that a difference in decibels is a multiplicative ratio of intensities, which is exactly why part (c) lands on a clean power of ten.
Common Student Questions
What's the difference between $\log$ and $\ln$, and does it matter which I use?
$\log$ on its own means $\log_{10}$ (base 10); $\ln$ means $\log_e$ (base $e\approx2.718$). For solving $a^{x}=b$ either one works, because change of base gives the identical answer $x=\frac{\log b}{\log a}=\frac{\ln b}{\ln a}$. Use $\ln$ for continuous-growth models written with $e$, and $\log_{10}$ for log scales such as decibels, Richter and pH. Both keys are on your GDC.
Why isn't $\log(x+y)$ equal to $\log x+\log y$?
Because the log laws come from the index laws, which are about multiplying, not adding. $\log x+\log y$ equals $\log(xy)$ — a product inside — whereas $\log(x+y)$ cannot be simplified at all. Splitting the log of a sum is the single most common error on this topic: there is simply no law for it.
How do I solve $a^{x}=b$ when $b$ isn't a neat power of $a$?
Take logs of both sides and use the power law: $\log(a^{x})=\log b\Rightarrow x\log a=\log b\Rightarrow x=\frac{\log b}{\log a}$. Any base works, so type $\frac{\ln b}{\ln a}$ (or $\frac{\log b}{\log a}$) straight into the GDC. Keep $\log b$ on top — flipping the fraction is a classic slip — and round to 3 s.f.
For "how many years until it first exceeds…", why round up instead of to 3 s.f.?
Because $n$ counts whole compounding periods. Solving $(1+i)^{n}=T/P$ gives a decimal such as $n=13.4$, but after 13 full years the value has not yet reached the target, so you need the 14th year. "First exceeds" always rounds up to the next whole period; rounding down, or to 3 s.f., loses the mark.
On a scale like decibels or Richter, what does "a difference of 2" actually mean?
A difference on a logarithmic scale is a ratio, not an addition. Each Richter unit multiplies the amplitude by 10, so $+2$ means $10^{2}=\times100$. Each $+10$ dB multiplies sound intensity by 10, and each pH unit changes $[H^{+}]$ tenfold. That compression is the whole point of a log scale: it turns multiplying by huge factors into simply adding small numbers.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every index and logarithm law in full, with worked negative, fractional and change-of-base examples and the exact GDC keystrokes.
Tutorial booklet — 25+ AI HL-style questions from index-law fluency to full modelling applications (continuous decay, compound growth, decibels and pH).
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations for every exponential and logarithmic method.
Predicted-paper questions — the exact exponent, logarithm and log-scale question types most likely in the next session.
