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
$\left|\begin{array}{ccc}2\left(\begin{array}{cc}x-\frac{1}{2} \\ 1\end{array}\right. & x-1 & x^{2} \\ x & 1 & 0\end{array}\right|$ is equal to :
Solution:
$2 \log _{10}\left(4^{x}-2\right)=1+\log _{10}\left(4^{x}+\frac{18}{5}\right)$
$\left(4^{x}-2\right)^{2}=10\left(4^{x}+\frac{18}{5}\right)$
$\left(4^{x}\right)^{2}+4-4\left(4^{x}\right)-32=0$
$\left(4^{x}-16\right)\left(4^{x}+2\right)=0$
$4^{x}=16$
$x=2$
$\left|\begin{array}{lll}3 & 1 & 4 \\ 1 & 0 & 2 \\ 2 & 1 & 0\end{array}\right|=3(-2)-1(0-4)+4(1)$
$=-6+4+4=2$