Given that nonzero reals $a,b,c,d$ satisfy $a^b=c^d$ and $\frac{a}{2c}=\frac{b}{d}=2$, compute $\frac{1}{c}$.


[i]2021 CCA Math Bonanza Lightning Round #2.2[/i]
1. Given the equations:
   \[
   a^b = c^d \quad \text{and} \quad \frac{a}{2c} = \frac{b}{d} = 2
   \]
   From \(\frac{a}{2c} = 2\), we can express \(a\) in terms of \(c\):
   \[
   a = 2 \cdot 2c = 4c
   \]
   From \(\frac{b}{d} = 2\), we can express \(b\) in terms of \(d\):
   \[
   b = 2d
   \]

2. Substitute \(a = 4c\) and \(b = 2d\) into the first equation \(a^b = c^d\):
   \[
   (4c)^{2d} = c^d
   \]

3. Simplify the equation:
   \[
   (4c)^{2d} = c^d
   \]
   Taking the \(d\)-th root on both sides:
   \[
   (4c)^{2d} = c^d \implies (4c)^2 = c
   \]

4. Solve for \(c\):
   \[
   16c^2 = c
   \]
   Divide both sides by \(c\) (assuming \(c \neq 0\)):
   \[
   16c = 1 \implies c = \frac{1}{16}
   \]

5. Compute \(\frac{1}{c}\):
   \[
   \frac{1}{c} = \frac{1}{\frac{1}{16}} = 16
   \]

The final answer is \(\boxed{16}\)