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Solution

We are given: $\left( 2^{\frac{1}{\log_x 4}} \right) \left( 2^{\frac{1}{\log_x 16}} \right) \left( 2^{\frac{1}{\log_x 256}} \right) \cdots = 2$ Using the identity $a^{\frac{1}{\log_b a}} = b$ : $4 = x^2,\; 16 = x^4,\; 256 = x^8$ So, $2^{\frac{1}{\log_x 4}} = 2^{\frac{1}{\log_x x^2}} = 2^{1/2} = 2^{\frac12}$ $2^{\frac{1}{\log_x 16}} = 2^{\frac{1}{\log_x x^4}} = 2^{1/4} = 2^{\frac14}$ $2^{\frac{1}{\log_x 256}} = 2^{\frac{1}{\log_x x^8}} = 2^{1/8} = 2^{\frac18}$ Thus the infinite product is: $2^{\frac12} \cdot 2^{\frac14} \cdot 2^{\frac18} \cdot 2^{\frac1{16}} \cdots$ Combine the exponents: $2^{\left( \frac12 + \frac14 + \frac18 + \frac1{16} + \cdots \right)}$ The geometric series: $\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots = 1$ Therefore, the product equals: $2^1 = 2$ Hence the value of $x$ is: $\boxed{\frac12}$