If log_3 2, log_3(2^x - 5) and log_3(2^x - 7/ | vedclass

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(D) Given that $\log_3 2, \log_3(2^x - 5)$,and $\log_3(2^x - 7/2)$ are in $A.P.$Since they are in $A.P.$,the middle term is the arithmetic mean of the

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(D) Given that $\log_3 2, \log_3(2^x - 5)$,and $\log_3(2^x - 7/2)$ are in $A.P.$Since they are in $A.P.$,the middle term is the arithmetic mean of the other two: $2 \log_3(2^x - 5) = \log_3 2 + \log_3(2^x - 7/2)$.Using the property $\log_b a + \log_b c = \log_b(ac)$,we get: $\log_3(2^x - 5)^2 = \log_3[2(2^x - 7/2)]$.Equating the arguments: $(2^x - 5)^2 = 2(2^x) - 7$.Let $y = 2^x$. Then $(y - 5)^2 = 2y - 7$.$y^2 - 10y + 25 = 2y - 7$.$y^2 - 12y + 32 = 0$.Factoring the quadratic: $(y - 8)(y - 4) = 0$,so $y = 8$ or $y = 4$.If $y = 8$,then $2^x = 8 \Rightarrow x = 3$.If $y = 4$,then $2^x = 4 \Rightarrow x = 2$.Check the domain: For $\log_3(2^x - 5)$ to be defined,$2^x - 5 > 0$,so $2^x > 5$.If $x = 2$,$2^2 = 4$,which is not $> 5$. Thus,$x = 2$ is rejected.If $x = 3$,$2^3 = 8$,which is $> 5$. Thus,$x = 3$ is the only solution.Since $3$ is not listed in options $A, B,$ or $C$,the correct option is $D$.

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

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