If $$1, \log_{10}(4^x - 2)$$ and $$\log_{10}\left(4^x + \frac{18}{5}\right)$$ are in arithmetic progression for a real number $$x$$ then the value of the determinant $$\begin{vmatrix} 2(x-\frac{1}{2}) & x-1 & x^2 \\ 1 & 0 & x \\ x & 1 & 0 \end{vmatrix}$$ is equal to ________.

Since $$1$$, $$\log_{10}(4^x - 2)$$, and $$\log_{10}\left(4^x + \frac{18}{5}\right)$$ are in arithmetic progression, the middle term equals the average of the other two: $$2\log_{10}(4^x - 2) = 1 + \log_{10}\left(4^x + \frac{18}{5}\right)$$.

This gives $$\log_{10}(4^x - 2)^2 = \log_{10}\left(10 \cdot \left(4^x + \frac{18}{5}\right)\right)$$, so $$(4^x - 2)^2 = 10\left(4^x + \frac{18}{5}\right)$$.

Let $$t = 4^x$$. Then $$(t-2)^2 = 10t + 36$$, giving $$t^2 - 4t + 4 = 10t + 36$$, which simplifies to $$t^2 - 14t - 32 = 0$$.

Factoring: $$(t-16)(t+2) = 0$$, so $$t = 16$$ or $$t = -2$$. Since $$4^x > 0$$, we need $$t = 16$$, giving $$4^x = 16 = 4^2$$, so $$x = 2$$.

We verify: $$4^x - 2 = 14 > 0$$ and $$4^x + \frac{18}{5} = \frac{98}{5} > 0$$, so the logarithms are defined.

Now we evaluate the determinant with $$x = 2$$: $$\begin{vmatrix} 2(2-\frac{1}{2}) & 2-1 & 2^2 \\ 1 & 0 & 2 \\ 2 & 1 & 0 \end{vmatrix} = \begin{vmatrix} 3 & 1 & 4 \\ 1 & 0 & 2 \\ 2 & 1 & 0 \end{vmatrix}$$.

Expanding along the first row: $$3(0 \cdot 0 - 2 \cdot 1) - 1(1 \cdot 0 - 2 \cdot 2) + 4(1 \cdot 1 - 0 \cdot 2) = 3(-2) - 1(-4) + 4(1) = -6 + 4 + 4 = 2$$.

The value of the determinant is $$2$$.