Question

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 $\begin{vmatrix}2\left(x-\frac{1}{2}\right) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0\end{vmatrix}$ is equal to :

Answer 1

$2 \log _{10}\left(4^{x}-2\right)=1+\log _{10}\left(4^{x}+\frac{18}{5}\right)$
$\left(4^{x}-2\right)^{2}=10\left(4^{x}+\frac{18}{5}\right) $
$\left(4^{x}\right)^{2}+4-4(4^ x)-32=0 $
$ \left(4^{x}-16\right)\left(4^{x}+2\right)=0$
$4^{x}=16 $
$x=2$
$\begin{vmatrix}3&1&4\\ 1&0&2\\ 2&1&0\end{vmatrix}=3\left(-2\right)-1\left(0-4\right)+4\left(1\right)$
$=-6+4+4=2$