A swimming club's chemical dosing rate is modelled by \(D = k^{\frac{3}{2}}\), where \(k\) is a positive constant satisfying \(k^2 = 2401\). A second club u...

This question chains several index-law rules together: finding a base from a squared value, applying fractional and negative indices to that base, and then writing a large number as a product of prime factors.

(a) Since \(k\) is stated to be positive and \(k^{2}=2401\), taking the positive square root gives \(k=\sqrt{2401}=49\). [1 mark]

(b) Using \(k=49\) in \(D=k^{\frac{3}{2}}\), the fractional index means "square root, then cube" (or "cube, then square root" - both give the same result):

(c) A negative index means take the reciprocal of the positive-index value:

(d) Multiplying the two results:

(e) Since \(k=49=7^{2}\) and \(k^{2}=2401\), it follows that \(2401=(7^{2})^{2}=7^{4}\); dividing 2401 repeatedly by 7 confirms this: \(2401\div7=343\), \(343\div7=49\), \(49\div7=7\), \(7\div7=1\), four divisions by 7 in total. [2 marks]

(f) Writing both numbers in terms of the same prime: \(2401=7^{4}\) and \(343=7^{3}\). Since \(7^{3}\) divides exactly into \(7^{4}\), with \(2401\div343=7\) confirming this, \(343\) is indeed a factor of \(2401\), and since it is the larger of the two numbers being compared, it is also the highest common factor. The safety officer's statement is correct. [2 marks]

Recognising that \(2401\) is a power of \(49\) (itself a power of 7) early on is what keeps every later part as clean whole-number arithmetic rather than needing a calculator for large roots.