Question

If $1,{\log }_{10}\left(4^x-2\right)$ and ${\log }_{10}\left[4^x+\left(\frac{18}{5}\right)\right]$ 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

STEP 1: Use the property of arithmetic progression.

As $1,{\log }_{10}\left(4^x-2\right)$ and ${\log }_{10}\left[4^x+\left(\frac{18}{5}\right)\right]$ are in arithmetic progression for a real number $x$. Hence

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

STEP 2: Using properties of log function.

$$\begin{gathered} \Rightarrow 2{\log }_{10}\left(4^x-2\right)=1+{\log }_{10}\left[4^x+\left(\frac{18}{5}\right)\right] \\ \Rightarrow {\log }_{10}{\left(4^x-2\right)}^2={\log }_{10}10+{\log }_{10}\left[4^x+\left(\frac{18}{5}\right)\right] \\ \Rightarrow {\log }_{10}{\left(4^x-2\right)}^2={\log }_{10}\left[10\left(4^x+\left(\frac{18}{5}\right)\right)\right] \\ \Rightarrow {\left(4^x-2\right)}^2=10\left(4^x+\left(\frac{18}{5}\right)\right) \end{gathered}$$

STEP 3: Simplify $\begin{array}{rcl}{\left(4^x-2\right)}^{\mathbf{2}}& \mathbf{=}& \mathbf{10}\left(4^x+\left(\frac{18}{5}\right)\right)\end{array}$.

$$\begin{gathered} \Rightarrow {\left(4^x-2\right)}^2=10\left(4^x+\left(\frac{18}{5}\right)\right) \\ \Rightarrow {\left(4^x\right)}^2+4-4.4^x=10.4^x+36 \\ \Rightarrow {\left(4^x\right)}^2-14.4^x-32=0 \\ \Rightarrow {\left(4^x\right)}^2+2.4^x-16.4^x-32=0 \\ \Rightarrow 4^x\left(4^x+2\right)-16\left(4^x+2\right)=0 \\ \Rightarrow \left(4^x+2\right)\left(4^x-16\right)=0 \end{gathered}$$

Hence

$$\begin{gathered} 4^x+2=0or{4}^x-16=0 \\ {4}^x=-2or{4}^x=16 \end{gathered}$$

STEP 4: Rejecting value of ${\mathbf{4}}^{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{2}$.

Since $4^x$ is a strictly increasing function with positive values only. Hence $4^x$ cant take negative values.

Thus $4^x=-2$ is rejected.

Hence $4^x=16\Rightarrow 4^x=4^2\Rightarrow x=2$.

STEP 5: Put the value $x=2$ in the determinant $\left|\begin{array}{ccc}2\left[x-\frac{1}{2}\right]& x-1& x^2\\ 1& 0& x\\ x& 1& 0\end{array}\right|$.

$\begin{array}{rcl}\left|\begin{array}{ccc}2\left[x-\frac{1}{2}\right]& x-1& x^2\\ 1& 0& x\\ x& 1& 0\end{array}\right|& =& \left|\begin{array}{ccc}2\left[2-\frac{1}{2}\right]& 2-1& 2^2\\ 1& 0& 2\\ 2& 1& 0\end{array}\right|\\ & =& \left|\begin{array}{ccc}4-1& 1& 4\\ 1& 0& 2\\ 2& 1& 0\end{array}\right|\\ & =& \left|\begin{array}{ccc}3& 1& 4\\ 1& 0& 2\\ 2& 1& 0\end{array}\right|\end{array}$

STEP 6: Solve the determinant $\begin{array}{rcl}& & \left|\begin{array}{ccc}3& 1& 4\\ 1& 0& 2\\ 2& 1& 0\end{array}\right|\end{array}$.

$\begin{array}{rcl}\left|\begin{array}{ccc}3& 1& 4\\ 1& 0& 2\\ 2& 1& 0\end{array}\right|& =& 3\left(0-2\right)-1\left(0-4\right)+4\left(1-0\right)\\ & =& 3\left(-2\right)-1\left(-4\right)+4\left(1\right)\\ & =& -6+4+4\\ & =& 2\end{array}$

Hence, 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 $\mathbf{2}\mathbf{.}$