If $1, {\log }_{10}\left(4^{\mathrm{x}}-2\right)$ and ${\log }_{10}\left(4^{\mathrm{x}}+\frac{18}{5}\right)$ are in arithmetic progression for a real number x, then the value of the determinant $\left|\begin{array}{ccc}2\left(\mathrm{x}-\frac{1}{2}\right)& \mathrm{x}-1& {\mathrm{x}}^2\\ 1& 0& \mathrm{x}\\ \mathrm{x}& 1& 0\end{array}\right|$ is equal to:

$$\begin{gathered} 2{\log }_{10}^{(4^{\mathrm{x}}-2)}={\log }_{10}^{10}+{\log }_{10}^{\left(4^{\mathrm{x}}+\frac{18}{5}\right)} \\ {\left(4^{\mathrm{x}}-2\right)}^2=10\left(4^{\mathrm{x}}+\frac{18}{5}\right) \\ {4}^{\mathrm{x}}=\mathrm{y} \Rightarrow {\mathrm{y}}^2+4-4\mathrm{y}=10\mathrm{y}+36 \\ {\mathrm{y}}^2-14\mathrm{y}-32=0 \\ (\mathrm{y}-16) (\mathrm{y}+2)=0 \\ \mathrm{y}=16, -2 \\ {4}^{\mathrm{x}}=16=4^2 \Rightarrow \mathrm{x}=2 \end{gathered}$$

$$\begin{gathered} \left|\begin{array}{ccc}3& 1& 4\\ 1& 0& 2\\ 2& 1& 0\end{array}\right| \\ =3(0-2)-1(0-4)+4(1-0) \\ \begin{array}{c}=-6+4+4=2\end{array} \end{gathered}$$