A swimming club treasurer uses the formula \(D = 8^{\frac{2}{3}} \times 4^{-\frac{1}{2}}\) to work out the size, in litres, of a chemical dose for the pool....
Question 1 Report
A swimming club treasurer uses the formula \(D = 8^{\frac{2}{3}} \times 4^{-\frac{1}{2}}\) to work out the size, in litres, of a chemical dose for the pool.
- Work out the value of \(8^{\frac{2}{3}}\). (2)
- Work out the value of \(4^{-\frac{1}{2}}\). (2)
- Hence find the value of \(D\). (1)
Answer Details
Fractional indices combine a root with a power: for a fraction \(\frac{p}{q}\), the index law \(x^{\frac{p}{q}}=(\sqrt[q]{x})^{p}\) applies, and a negative index means "take the reciprocal".
(a) The denominator of the index gives the root and the numerator gives the power:
\[8^{\frac{2}{3}}=(\sqrt[3]{8})^{2}=2^{2}=4\] [2 marks](b) A negative index means reciprocal, so \(4^{-\frac{1}{2}}=\dfrac{1}{4^{\frac{1}{2}}}\). Since \(4^{\frac{1}{2}}=\sqrt{4}=2\):
\[4^{-\frac{1}{2}}=\dfrac{1}{2}\] [2 marks](c) Multiplying the two results from parts (a) and (b):
\[D=4 \times \dfrac{1}{2}=2 \text{ litres}\] [1 mark]Taking the root before the power (here, the cube root of 8 rather than \(8^{2}=64\) first) keeps the numbers small and avoids unnecessary arithmetic with large values.
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