If log _{3} 2, log _{3}(2^{x}-5), log _{3}(2^ | vedclass

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(C) Given that $\log _{3} 2, \log _{3}(2^{x}-5), \log _{3}(2^{x}-\frac{7}{2})$ are in an arithmetic progression $(AP)$.For three terms $a, b, c$ to be

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$3$

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(C) Given that $\log _{3} 2, \log _{3}(2^{x}-5), \log _{3}(2^{x}-\frac{7}{2})$ are in an arithmetic progression $(AP)$.For three terms $a, b, c$ to be in $AP$,$2b = a + c$.Therefore,$2 \log _{3}(2^{x}-5) = \log _{3} 2 + \log _{3}(2^{x}-\frac{7}{2})$.Using the property $\log a + \log b = \log(ab)$,we get $\log _{3}(2^{x}-5)^{2} = \log _{3}[2(2^{x}-\frac{7}{2})]$.$(2^{x}-5)^{2} = 2(2^{x}-\frac{7}{2})$.Let $2^{x} = t$. Then $(t-5)^{2} = 2t - 7$.$t^{2} - 10t + 25 = 2t - 7$.$t^{2} - 12t + 32 = 0$.$(t-4)(t-8) = 0$.So,$t = 4$ or $t = 8$.If $2^{x} = 4$,then $x = 2$. However,for $\log _{3}(2^{x}-5)$ to be defined,$2^{x}-5 > 0$,so $4-5 = -1$,which is not allowed.If $2^{x} = 8$,then $x = 3$. Here $8-5 = 3 > 0$ and $8-3.5 = 4.5 > 0$,which is valid.Thus,$x = 3$.

If $\log _{3} 2, \log _{3}(2^{x}-5), \log _{3}(2^{x}-\frac{7}{2})$ are in an arithmetic progression,then the value of $x$ is equal to $.....$

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