QJEEM21Ma7e62: JEE Main 2021 Mathematics - Class 12 JEEM21Ma7e62 Solution: If 1, log 10 (4 x 2) and log 10 are in arithmetic progression…

Question

Verified Solution • 2025–26 Session moderate

If 1, log₁₀(4^x

-

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

| \begin{matrix} 2 (x - \frac{1}{2}) & x - 1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0 \end{matrix} |

is equal to :

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

| \begin{matrix} 2 (x - 1 / 2) & x - 1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0 \end{matrix} |

= | \begin{matrix} 3 & 1 & 4 \\ 1 & 0 & 2 \\ 2 & 1 & 0 \end{matrix} |

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

= - 6 + 4 + 4 = 2

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