Index Laws (AA SL)
Index laws let you simplify and combine powers without a calculator - multiplying, dividing, raising a power to another power, and handling zero, negative and fractional exponents. They're the toolkit every later algebra topic leans on, so getting them automatic now saves marks everywhere else. It's part of the broader Logs & Exponents topic.
11 questions on this sub-topic.
The key rules
Covered under IB syllabus reference SL1.5 (laws of exponents with integer exponents) and SL1.7 (laws of exponents with rational exponents, covering fractional indices). These aren't formula-booklet entries - they're assumed prior knowledge you're expected to apply fluently without looking anything up.
Product & quotient rules
\(a^m \times a^n = a^{m+n}\)
\(a^m \div a^n = a^{m-n}\)
Add exponents when multiplying same-base powers, subtract when dividing. Also \((a^m)^n = a^{mn}\) when raising a power to a power.
Zero, negative & fractional indices
\(a^0 = 1, \quad a^{-n} = \dfrac{1}{a^n}, \quad a^{m/n} = \sqrt[n]{a^{m}}\)
Any non-zero base to the power 0 is 1. A negative exponent flips to a reciprocal. A fractional exponent is a root - the denominator is the root, the numerator is the power.
Need logarithm laws or the full Logs & Exponents overview? See Logs & Exponents.
Worked examples
Write \(\dfrac{\sqrt[3]{x^{2}}\cdot x}{x^{1/2}}\) as a single power of \(x\).
Worked solution
\(\sqrt[3]{x^2}=x^{2/3}\) and \(\sqrt{x}=x^{1/2}\). M1
\(x^{2/3}\cdot x^{1}\div x^{1/2}=x^{2/3+1-1/2}.\) A1
\(\tfrac23+1-\tfrac12=\tfrac76\), so \(=x^{7/6}.\) A1
Evaluate \(\dfrac{16^{3/4}\times2^{-1}}{8^{2/3}}\), giving your answer as an integer.
Worked solution
\(16=2^4\) and \(8=2^3.\) M1
\(16^{3/4}=(2^4)^{3/4}=2^3=8\) and \(8^{2/3}=(2^3)^{2/3}=2^2=4.\) A1
\(2^3\times2^{-1}=2^2=4.\) M1
\(\dfrac{4}{4}=1.\) A1
Evaluate.
(a) \(27^{2/3}\)
(b) \(5^{0}+5^{-1}\)
Worked solution
(a) \(27^{1/3}=3\Rightarrow27^{2/3}=3^2\) M1
\(=9.\) A1
(b) \(5^0+5^{-1}=1+\tfrac15=\tfrac65.\) A1
Evaluate.
(a) \(8^{-2/3}\)
(b) \(\left(\tfrac{1}{16}\right)^{1/2}\)
Worked solution
(a) \(8^{1/3}=2\Rightarrow8^{2/3}=4\), and the negative index inverts: \(8^{-2/3}\) M1
\(=\tfrac14.\) A1
(b) \(\left(\tfrac1{16}\right)^{1/2}=\sqrt{\tfrac1{16}}\) M1
\(=\tfrac14.\) A1
Common mistakes
- Adding exponents when the bases don't match. \(a^m\times b^n\) can't be simplified by adding exponents unless \(a=b\) - rewrite one base in terms of the other first, as with \(16=2^4\) or \(8=2^3\).
- Mishandling a negative exponent's sign. \(a^{-n}\) means \(\dfrac{1}{a^n}\), not \(-a^n\) - the negative flips the term to a reciprocal, it doesn't make the value negative.
- Applying the power rule instead of the product rule (or vice versa). \((a^m)^n\) multiplies the exponents, but \(a^m \times a^n\) adds them - mixing the two up is one of the most common slips under exam pressure.
Ready to practise properly?
11 index-laws questions, marked instantly like the real exam.
Quick answers
What are the main index laws I need for IB AA SL?
The product rule \(a^m \times a^n = a^{m+n}\), the quotient rule \(a^m \div a^n = a^{m-n}\), and the power rule \((a^m)^n = a^{mn}\), together with \(a^0 = 1\) and \(a^{-n} = \tfrac{1}{a^n}\).
What does a fractional index like \(a^{1/n}\) mean?
\(a^{1/n}\) is the \(n\)th root of \(a\), and \(a^{m/n}\) is the \(n\)th root of \(a\) raised to the power \(m\) (or equivalently the \(n\)th root of \(a^m\)).