We are given the equation $\frac{\sqrt{4^{3x+1}}}{\sqrt{4^{2x+1}}} = 2$ and we need to solve for $x$.

1. Problem Description

We are given the equation $\frac{\sqrt{4^{3x+1}}}{\sqrt{4^{2x+1}}} = 2$ and we need to solve for $x$.

2. Solution Steps

First, we simplify the equation.

$\frac{\sqrt{4^{3x+1}}}{\sqrt{4^{2x+1}}} = 2$

$\sqrt{\frac{4^{3x+1}}{4^{2x+1}}} = 2$

$\sqrt{4^{(3x+1)-(2x+1)}} = 2$

$\sqrt{4^{3x+1-2x-1}} = 2$

$\sqrt{4^x} = 2$

$(4^x)^{\frac{1}{2}} = 2$

$4^{\frac{x}{2}} = 2$

$(2^2)^{\frac{x}{2}} = 2$

$2^{2 \cdot \frac{x}{2}} = 2$

$2^x = 2^1$

Since the bases are equal, the exponents must be equal.

$x = 1$

3. Final Answer

$x = 1$