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Question
group activity: exponent relay challenge
work in groups of 3 - 4 students. solve each problem in order as a relay. each group member takes a turn solving one problem, shows their work, then passes it to the next member. your group cannot move to the next problem until the current one is correct. the first group to finish all problems correctly wins!
problem work/steps final answer
- simplify 81^(1/2)
- simplify 8^(2/3)
3 simplify 32^(1/5) / 2^(1/5)
4 simplify 27^(-2/3)
- simplify 9^(1/2) + 10^(1/2)
reflection:
after completing the relay, discuss as a group:
- what does a fractional exponent mean in your own words?
- which problem was most challenging for your group and why?
- how do negative fractional exponents change your approach?
Step1: Recall fractional - exponent rule
For \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\). For \(81^{\frac{1}{2}}\), \(a = 81\), \(m = 1\), \(n=2\). So \(81^{\frac{1}{2}}=\sqrt{81}=9\).
Step2: Apply fractional - exponent rule
For \(8^{\frac{2}{3}}\), first find the cube - root of 8, \(\sqrt[3]{8}=2\), then square it. \(2^{2}=4\).
Step3: Use exponent rules for division
\(a^{m}\div a^{n}=a^{m - n}\). For \(32^{\frac{1}{5}}\div2^{\frac{1}{5}}=(32\div2)^{\frac{1}{5}}\). Since \(32\div2 = 16\), then \((32\div2)^{\frac{1}{5}}=16^{\frac{1}{5}}=\sqrt[5]{16}\).
Step4: Apply negative - exponent rule
\(a^{-m}=\frac{1}{a^{m}}\). For \(27^{-\frac{2}{3}}\), first find \(27^{\frac{2}{3}}\). \(\sqrt[3]{27}=3\), then \(3^{2}=9\). So \(27^{-\frac{2}{3}}=\frac{1}{9}\).
Step5: Simplify square - roots
\(9^{\frac{1}{2}}=\sqrt{9}=3\), \(10^{\frac{1}{2}}=\sqrt{10}\). So \(9^{\frac{1}{2}}+10^{\frac{1}{2}}=3 + \sqrt{10}\).
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