Convert all logarithms to a common base, which is base 2.

We use the change of base formula: $\log_b a = \frac{\log_c a}{\log_c b}$.

For $\log_4 x$:

$$\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}$$

For $\log_{16} x$:

$$\log_{16} x = \frac{\log_2 x}{\log_2 16} = \frac{\log_2 x}{4}$$

$$\frac{4y}{4} + \frac{2y}{4} + \frac{y}{4} = \frac{21}{4}$$ $$\frac{4y + 2y + y}{4} = \frac{21}{4}$$ $$\frac{7y}{4} = \frac{21}{4}$$

Multiply both sides by 4:

$$7y = 21$$

Divide by 7:

$$y = \frac{21}{7}$$ $$y = 3$$

$$\log_2 x = 3$$

By the definition of a logarithm, if $\log_b a = c$, then $b^c = a$.

$$x = 2^3$$ $$x = 8$$

The final answer is 8.