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Let \(y=2^{-x}\). Then \(y^2+y^2+y=1\Rightarrow2y^2+y-1=0\) and this has solutions \(y={-1\pm3\over4}=1/2,-1\). Clearly \(-1\) is extraneous so \(y=1/2\) which gives us \(x=1\).
Let \(s\) be the side length of the smaller square. The larger square has side length \(s\sqrt{2}\) so its area is \(2s^2\), twice the area of the smaller square which is \(s^2\). So the answer is 2.
For this problem, we can use the triangle area formula with angle \(\theta\) between the sides \(a\) and \(b\) \[{1\over2}ab\sin(\theta)\] In this case, \(a=3\) and \(b=4\), which can be used for both the area 3 triangle and the \(x\) triangle. Since the 2 squares (we assume they are squares) have right angles, the angle for the area 3 triangle and the area for the \(x\) triangle sum to \(180^\circ\). This means that \(\sin(\theta)\) will be the same in both cases since \[\sin(\theta)=\sin(180^\circ-\theta)\] so we must also have \(x=3\).