If 1,log_{10}(4^{x}-2) and log_{10}(4^{x}+fra | vedclass

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(D) Given that $1$,$\log_{10}(4^{x}-2)$,and $\log_{10}(4^{x}+\frac{18}{5})$ are in arithmetic progression,we have:$2 \log_{10}(4^{x}-2) = 1 + \log_{10

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$2$

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(D) Given that $1$,$\log_{10}(4^{x}-2)$,and $\log_{10}(4^{x}+\frac{18}{5})$ are in arithmetic progression,we have:$2 \log_{10}(4^{x}-2) = 1 + \log_{10}(4^{x}+\frac{18}{5})$$\log_{10}(4^{x}-2)^{2} = \log_{10}(10) + \log_{10}(4^{x}+\frac{18}{5})$$(4^{x}-2)^{2} = 10(4^{x}+\frac{18}{5})$$(4^{x})^{2} - 4(4^{x}) + 4 = 10(4^{x}) + 36$$(4^{x})^{2} - 14(4^{x}) - 32 = 0$Let $y = 4^{x}$,then $y^{2} - 14y - 32 = 0$$(y-16)(y+2) = 0$Since $4^{x} > 0$,we have $4^{x} = 16$,which implies $x = 2$.Now,substitute $x=2$ into the determinant:$\left|\begin{array}{ccc} 2(2-\frac{1}{2}) & 2-1 & 2^{2} \ 1 & 0 & 2 \ 2 & 1 & 0 \end{array}ight| = \left|\begin{array}{ccc} 3 & 1 & 4 \ 1 & 0 & 2 \ 2 & 1 & 0 \end{array}ight|$$= 3(0-2) - 1(0-4) + 4(1-0)$$= -6 + 4 + 4 = 2$.

If $1$,$\log_{10}(4^{x}-2)$ and $\log_{10}(4^{x}+\frac{18}{5})$ are in arithmetic progression for a real number $x$,then the value of the determinant $\left|\begin{array}{ccc} 2(x-\frac{1}{2}) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0 \end{array}\right|$ is equal to ...... .

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