If {log _3}2,;{log _3}({2^x} - 5) and {log _3 | vedclass

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(d) ${\log _3}2,\;{\log _3}({2^x} - 5)$ and ${\log _3}\left( {{2^x} - \frac{7}{2}} \right)$ are in $A.P.$

$ \Rightarrow $$2{\log _3}({2^x} - 5) = {

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None of these

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(d) ${\log _3}2,\;{\log _3}({2^x} - 5)$ and ${\log _3}\left( {{2^x} - \frac{7}{2}} \right)$ are in $A.P.$

$ \Rightarrow $$2{\log _3}({2^x} - 5) = {\log _3}\left[ {(2)\,\left( {{2^x} - \frac{7}{2}} \right)} \right]$

$ \Rightarrow $ ${({2^x} - 5)^2} = {2^{x + 1}} - 7$

$ \Rightarrow $${2^{2x}} - 12\;.\;{2^x} - 32 = 0$

$ \Rightarrow $ $x = 2,\;3$

But $x = 2$ does not hold, hence $x = 3$.

If ${\log _3}2,\;{\log _3}({2^x} - 5)$ and ${\log _3}\left( {{2^x} - \frac{7}{2}} \right)$ are in $A.P.$, then $x$ is equal to

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