Prove the following
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:

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

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

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

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

$4^{x}=16$

$x=2$

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

$=-6+4+4=2$

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