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

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(C) Given that the terms $\log _{5} 2, \log _{5}(2^{x}-3)$,and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$For three terms $a, b, c$ to be in $A.P.

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$3$

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(C) Given that the terms $\log _{5} 2, \log _{5}(2^{x}-3)$,and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$For three terms $a, b, c$ to be in $A.P.$,the condition is $2b = a + c$.Substituting the given values: $2 \log _{5}(2^{x}-3) = \log _{5} 2 + \log _{5}(\frac{17}{2}+2^{x-1})$.Using the property $\log m + \log n = \log(mn)$,we get: $\log _{5}(2^{x}-3)^{2} = \log _{5}(2 \cdot (\frac{17}{2}+2^{x-1}))$.Removing the logarithm: $(2^{x}-3)^{2} = 17 + 2 \cdot 2^{x-1}$.Since $2 \cdot 2^{x-1} = 2^{x}$,the equation becomes: $(2^{x}-3)^{2} = 17 + 2^{x}$.Let $2^{x} = y$. Then $(y-3)^{2} = 17 + y$.$y^{2} - 6y + 9 = 17 + y \Rightarrow y^{2} - 7y - 8 = 0$.Factoring the quadratic: $(y-8)(y+1) = 0$.So,$y = 8$ or $y = -1$.Since $y = 2^{x}$,$2^{x} = 8$ gives $x = 3$. $2^{x} = -1$ is not possible for real $x$.Thus,$x = 3$.

If $\log _{5} 2, \log _{5}(2^{x}-3)$ and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$,then the value of $x$ is:

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