Using the base-change formula $\log_{a^n} b = \frac{1}{n} \log_a b$, we can rewrite the terms with a common base of 2: $\log_2 x + \log_{2^2} x + \log_{2^4} x = \frac{7}{4}$. This simplifies to $\log_2 x + \frac{1}{2}\log_2 x + \frac{1}{4}\log_2 x = \frac{7}{4}$. Factoring out $\log_2 x$, we get $(1 + \frac{1}{2} + \frac{1}{4})\log_2 x = \frac{7}{4} \Rightarrow (\frac{7}{4})\log_2 x = \frac{7}{4}$. This implies $\log_2 x = 1$, so $x = 2^1 = 2$. Option 1 (x=4) gives $\frac{7}{2}$, Option 2 (x=1) gives $0$, and Option 4 (x=16) gives $7$.