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) \(\dfrac{5}{2}\)
Could we write \(2016\) as the product of its prime factors? Could we do the same for \(56\)?
How could we now simplify the square roots?
