"If 1, log10(4x $-$ 2) and log10$\left( {{4^x} + {{18} \over 5}} \right)$ are in arithmetic progression for a real number x, then the value of the determinant $$\left| {\matrix{ {2\left( {x - {1 \over 2}} \right)} & {x - 1} & {{x^2}} \cr 1 & 0 & x \cr x & 1 & 0 \cr } } \right|$$ is equal to :"

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Accepted Answer

"1, $lo{g_{10}}({4^x} - 2),\,lo{g_{10}}\left( {{4^x} + {{18} \over 5}} \right)$ in AP.

$\therefore$ 2$\times$$lo{g_{10}}({4^x} - 2) = 1 + \,lo{g_{10}}\left( {{4^x} + {{18} \over 5}} \right)$

$$lo{g_{10}}{({4^x} - 2)^2} = \,lo{g_{10}}\left( {10.\left( {{4^x} + {{18} \over 5}} \right)} \right)$$

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

${({4^x})^2} + 4 - {4.4^x} = {10.4^x} + 36$

${({4^x})^2} - {14.4^x} - 32 = 0$

${({4^x})^2} + {2.4^x} - {16.4^x} - 32 = 0$

${4^x}({4^x} + 2) - 16.({4^x} + 2) = 0$

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

4^x = -2 (Not Possible)

Or 4^x = 16

$\Rightarrow$ x = 2

Therefore $$\left| {\matrix{ {2(x - 1/2)} & {x - 1} & {{x^2}} \cr 1 & 0 & x \cr x & 1 & 0 \cr } } \right|$$

$$ = \left| {\matrix{ 3 & 1 & 4 \cr 1 & 0 & 2 \cr 2 & 1 & 0 \cr } } \right|$$

$= 3( - 2) - 1(0 - 4) + 4(1 - 0)$

$= - 6 + 4 + 4 = 2$"

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If 1, log10(4x $-$ 2) and log10$\left( {{4^x} + {{18} \over 5}} \right)$ are in arithmetic progression for a real number x, then the value of the determinant $$\left| {\matrix{ {2\left( {x - {1 \over 2}} \right)} & {x - 1} & {{x^2}} \cr 1 & 0 & x \cr x & 1 & 0 \cr } } \right|$$ is equal to :

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