Rational Exponents (AI HL)
A rational exponent is just a root and a power written as one number, so \(8^{2/3}\) means "cube root of 8, then square it", not two separate operations you have to guess the order of. This page covers the index laws you need for fractional and negative powers, with worked examples and the mistakes that cost the most marks. It's part of the broader Exponential & Logarithmic Models topic.
11 questions on this sub-topic.
The index laws you need
Covered under IB syllabus reference AHL1.10: simplifying expressions, both numerically and algebraically, involving rational exponents. These aren't a separate formula-booklet entry - they follow directly from the integer index laws you already know, just extended to fractional and negative powers.
Fractional powers
\(a^{1/n} = \sqrt[n]{a}\), \(a^{m/n} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}\)
The denominator \(n\) is the root, the numerator \(m\) is the power. Take the root first to keep the numbers small before you raise to the power.
Negative powers
\(a^{-m/n} = \dfrac{1}{a^{m/n}}\)
A negative exponent means "reciprocal", not "negative answer". Combine with the fractional-power rule above once the sign is dealt with.
Need the wider syllabus context for exponential and logarithmic work? See Exponential & Logarithmic Models.
Worked examples
Simplify each expression without a calculator.
(a) \(8^{2/3}\)
(b) \(16^{-3/4}\)
(c) \(x^{1/2} \cdot x^{3/2}\)
Worked solution
(a) \(8^{2/3} = (\!\sqrt[3]{8})^2 = 2^2 = 4\) A1
(b) \(16^{-3/4} = \dfrac{1}{(\!\sqrt[4]{16})^3} = \dfrac{1}{2^3}\) M1
\( = \dfrac{1}{8}\) A1
(c) \(x^{1/2} \cdot x^{3/2} = x^{1/2+3/2} = x^2\) A1
Solve \(e^{2x} - 5e^{x} + 6 = 0.\)
(a)(i) Give the exact answer with \(x<0.9\).
(a)(ii) Give the exact answer with \(x>0.9.\)
Worked solution
\(y^2-5y+6=0.\) M1
\((y-2)(y-3)=0\Rightarrow y=2,3.\) A1
\(x=\ln2.\) A1 \(x=\ln3.\) A1
Simplify each expression without a calculator.
(a) 82/3
(b) 16−3/4
(c) \(x^{1/2} \cdot x^{3/2}\)
Worked solution
(a) \(8^{2/3} = (\!\sqrt[3]{8})^2 = 2^2 = 4\) A1
(b) \(16^{-3/4} = \dfrac{1}{(\!\sqrt[4]{16})^3} = \dfrac{1}{2^3}\) M1
\( = \dfrac{1}{8}\) A1
(c) \(x^{1/2} \cdot x^{3/2} = x^{1/2+3/2} = x^2\) A1
Common mistakes
- Treating a negative exponent as a negative value. \(16^{-3/4}\) is not \(-16^{3/4}\) - the minus sign in the exponent means "take the reciprocal", so the answer for a positive base is always positive.
- Forgetting a common denominator when adding fractional exponents. Multiplying \(x^{1/2}\) by \(x^{3/2}\) means adding \(\tfrac{1}{2}+\tfrac{3}{2}\), not multiplying the powers or the bases - this only goes wrong when the fractions don't already share a denominator.
- Rooting last instead of first. Computing \(a^{m/n}\) as \(\sqrt[n]{a^m}\) means raising to a large power before rooting, which produces huge numbers to simplify by hand; \((\sqrt[n]{a})^m\) keeps the arithmetic manageable.
Ready to practise properly?
11 rational-exponent questions, marked instantly like the real exam.
Quick answers
What is a rational (fractional) exponent?
\(a^{1/n}\) means the \(n\)th root of \(a\), and \(a^{m/n}\) means that root raised to the power \(m\): \((\sqrt[n]{a})^m\), which is the same as \(\sqrt[n]{a^m}\).
How do you simplify a negative rational exponent?
Take the reciprocal of the positive-exponent version: \(a^{-m/n} = \dfrac{1}{a^{m/n}}\). The negative sign flips the fraction, it does not make the value negative.