If 1, log_9(3^{1-x} + 2), log_3(4 cdot 3^x - | vedclass

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(B) Given that $1, \log_9(3^{1-x} + 2), \log_3(4 \cdot 3^x - 1)$ are in $A.P.$Since $a, b, c$ are in $A.P.$ if $2b = a + c$,we have:$2 \log_9(3^{1-x}

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$1 - \log_3 4$

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(B) Given that $1, \log_9(3^{1-x} + 2), \log_3(4 \cdot 3^x - 1)$ are in $A.P.$Since $a, b, c$ are in $A.P.$ if $2b = a + c$,we have:$2 \log_9(3^{1-x} + 2) = \log_3(4 \cdot 3^x - 1) + 1$Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we get:$2 \log_{3^2}(3^{1-x} + 2) = \log_3(4 \cdot 3^x - 1) + \log_3 3$$\frac{2}{2} \log_3(3^{1-x} + 2) = \log_3(3(4 \cdot 3^x - 1))$$3^{1-x} + 2 = 3(4 \cdot 3^x - 1)$Let $y = 3^x$,then $3^{1-x} = \frac{3}{y}$.$\frac{3}{y} + 2 = 12y - 3$$3 + 2y = 12y^2 - 3y$$12y^2 - 5y - 3 = 0$$(4y - 3)(3y + 1) = 0$Since $y = 3^x > 0$,we have $y = \frac{3}{4}$.$3^x = \frac{3}{4} = 3^1 \cdot 3^{-\log_3 4} = 3^{1 - \log_3 4}$Therefore,$x = 1 - \log_3 4$.

If $1, \log_9(3^{1-x} + 2), \log_3(4 \cdot 3^x - 1)$ are in $A.P.$,then $x$ equals

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