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

Answer 1

Answer: 2

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