16. For which value of $k$ is $\sqrt{2016}+\sqrt{56}$ equal to $14^{k}$ ?
A $\frac{1}{2}$
B $\frac{3}{4}$
C $\frac{5}{4}$
D $\frac{3}{2}$
E $\frac{5}{2}$
SOLUTION
D

We have already seen (in the solution to question 12) that $2016=2^{5} \times 3^{2} \times 7$. It follows that $2016=\left(2^{4} \times 3^{2}\right) \times 2 \times 7=\left(2^{2} \times 3\right)^{2} \times 2 \times 7=12^{2} \times 14$. Therefore $\sqrt{2016}=12 \sqrt{14}$.
Also, $\sqrt{56}=\sqrt{2^{2} \times 14}=2 \sqrt{14}$. Hence $\sqrt{2016}+\sqrt{56}=12 \sqrt{14}+2 \sqrt{14}=14 \sqrt{14}$.
Now, $14 \sqrt{14}=14^{1} \times 14^{1 / 2}=14^{1+1 / 2}=14^{3 / 2}$. Therefore $k=\frac{3}{2}$.