If $1,{\log }_{10}\left(4^x-2\right)$ 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 $\left|\begin{array}{ccc}2\left(x-\frac{1}{2}\right)& x-1& x^2\\ 1& 0& x\\ x& 1& 0\end{array}\right|$ is equal to:

If three numbers $a, b, c$ are in arithmetic progression, then $2b=a+c.$

Given $1, {\log }_{10}\left(4^x-2\right), {\log }_{10}\left(4^x+\frac{18}{5}\right)$ are in arithmetic progression, then we have

$2{\log }_{10}\left(4^x-2\right)=1+{\log }_{10}\left(4^x+\frac{18}{5}\right)$

Using ${\log }_{10}{m}^n=n{\log }_{10}m,$ we get

${\log }_{10}{\left(4^x-2\right)}^2={\log }_{10}\left(10\right)+{\log }_{10}\left(4^x+\frac{18}{5}\right)$

Using ${\log }_{10}\left(m\right)+{\log }_{10}\left(n\right)={\log }_{10}\left(mn\right),$ we get

${\log }_{10}{\left(4^{\mathrm{x}}-2\right)}^2={\log }_{10}\left(10\left(4^{\mathrm{x}}+\frac{18}{5}\right)\right)$

$\Rightarrow {\left(4^{\mathrm{x}}-2\right)}^2=10\left(4^{\mathrm{x}}+\frac{18}{5}\right)$

$\Rightarrow {\left(4^{\mathrm{x}}\right)}^2+4-4·4^{\mathrm{x}}=10·4^{\mathrm{x}}+36$

$\Rightarrow {\left(4^x\right)}^2-14\left(4^x\right)-32=0$

$\Rightarrow \left(4^x-16\right)\left(4^{\mathrm{x}}+2\right)=0$

$\Rightarrow 4^{\mathrm{x}}=16$ as $4^x+2\ne 0$

$\Rightarrow x=2$

Now, $\left|\begin{array}{ccc}2\left(x-\frac{1}{2}\right)& x-1& x^2\\ 1& 0& x\\ x& 1& 0\end{array}\right|$

$=\left|\begin{array}{ccc}3& 1& 4\\ 1& 0& 2\\ 2& 1& 0\end{array}\right|$

$=3\left(0-2\right)-1\left(0-4\right)+4\left(1-0\right)$

$=-6+4+4=2.$