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

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(B) Since the given numbers are in $A.P.$, the middle term is the arithmetic mean of the other two terms.$2 \log_9(3^{1-x} + 2) = \log_3(4 \cdot 3^x -

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$1 - \log_3 4$

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(B) Since the given numbers are in $A.P.$, the middle term is the arithmetic mean of the other two terms.$2 \log_9(3^{1-x} + 2) = \log_3(4 \cdot 3^x - 1) + 1$Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$, we get:$2 \cdot \frac{1}{2} \log_3(3^{1-x} + 2) = \log_3(4 \cdot 3^x - 1) + \log_3 3$$\log_3(3^{1-x} + 2) = \log_3(3(4 \cdot 3^x - 1))$$3^{1-x} + 2 = 12 \cdot 3^x - 3$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 $y = \frac{3}{4}$.$3^x = \frac{3}{4} \implies x = \log_3 \left(\frac{3}{4}\right) = \log_3 3 - \log_3 4 = 1 - \log_3 4$.

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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