For which value of \(k\) is \(\sqrt{2016}+\sqrt{56}\) equal to \(14^k\)?

A)\(\:\dfrac{1}{2} \qquad\) B) \(\dfrac{3}{4} \qquad\) C) \(\dfrac{5}{4} \qquad\) D) \(\dfrac{3}{2} \qquad\) E) \(\dfrac{5}{2}\)

If we factorise \(2016\) into its prime factors, we find \(2016 = 2^5\times 3^2 \times 7\).

Doing the same to \(56\), we find \(56 = 2^3\times 7\). Thus

\[\begin{align*} \sqrt{2016}+\sqrt{56} &= \sqrt{2^5\times 3^2 \times 7}+\sqrt{2^3\times 7} \\ &= \sqrt{(2^4\times 3^2) \times 14}+\sqrt{2^2\times 14}\\ &= 2^2 \times 3 \times \sqrt{14}+2\sqrt{14}= 14\sqrt{14}\\ &= 14^{3/2}, \end{align*}\]so \(k = \dfrac{3}{2}\) and the answer is (D).