If log_3 2, log_3(2^x - 5) and log_3(2^x - fr | vedclass

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(D) Given that $\log_3 2, \log_3(2^x - 5)$ and $\log_3(2^x - \frac{7}{2})$ are in $A.P.$By the property of $A.P.$,$2b = a + c$,so:$2 \log_3(2^x - 5) =

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(D) Given that $\log_3 2, \log_3(2^x - 5)$ and $\log_3(2^x - \frac{7}{2})$ are in $A.P.$By the property of $A.P.$,$2b = a + c$,so:$2 \log_3(2^x - 5) = \log_3 2 + \log_3(2^x - \frac{7}{2})$Using the property $\log a + \log b = \log(ab)$:$\log_3(2^x - 5)^2 = \log_3(2(2^x - \frac{7}{2}))$$(2^x - 5)^2 = 2^{x+1} - 7$Let $2^x = y$. Then $(y - 5)^2 = 2y - 7$$y^2 - 10y + 25 = 2y - 7$$y^2 - 12y + 32 = 0$$(y - 8)(y - 4) = 0$So,$y = 8$ or $y = 4$.If $2^x = 8$,then $x = 3$.If $2^x = 4$,then $x = 2$.Check the domain of the logarithmic terms:For $x = 2$,$\log_3(2^2 - 5) = \log_3(-1)$,which is undefined.For $x = 3$,$\log_3(2^3 - 5) = \log_3(3) = 1$ and $\log_3(2^3 - 3.5) = \log_3(4.5)$,which are defined.Thus,$x = 3$ is the only solution.

If $\log_3 2, \log_3(2^x - 5)$ and $\log_3(2^x - \frac{7}{2})$ are in $A.P.$,then $x$ is equal to

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