If log_{3} 2, log_{3} (2^{x} - 5) and log_{3} | vedclass

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(D) Given that $\log_{3} 2, \log_{3} (2^{x} - 5), \log_{3} (2^{x} - \frac{7}{2})$ are in $AP$. Let $a = 2^{x}$. The terms are $\log_{3} 2, \log_{3} (a

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(D) Given that $\log_{3} 2, \log_{3} (2^{x} - 5), \log_{3} (2^{x} - \frac{7}{2})$ are in $AP$. Let $a = 2^{x}$. The terms are $\log_{3} 2, \log_{3} (a - 5), \log_{3} (a - 3.5)$. Since they are in $AP$,$2 \log_{3} (a - 5) = \log_{3} 2 + \log_{3} (a - 3.5)$. Using the property $\log m + \log n = \log (mn)$ and $n \log m = \log (m^{n})$,we get: $\log_{3} (a - 5)^{2} = \log_{3} [2(a - 3.5)]$. $(a - 5)^{2} = 2a - 7$. $a^{2} - 10a + 25 = 2a - 7$. $a^{2} - 12a + 32 = 0$. $(a - 8)(a - 4) = 0$. So,$a = 8$ or $a = 4$. If $2^{x} = 8$,then $x = 3$. If $2^{x} = 4$,then $x = 2$. Since neither $3$ nor $2$ matches the given options,the correct answer is $D$.

If $\log_{3} 2, \log_{3} (2^{x} - 5)$ and $\log_{3} (2^{x} - \frac{7}{2})$ are in an Arithmetic Progression $(AP)$,then $x = \dots$

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