If 1, log _9(3^{1-x}+2), log _3(4 cdot 3^x-1) | 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 $2b = a + c$ for an $A.P.$, we have:$2 \log _9(3^{1-x}+2) = 1 + \log

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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 $2b = a + c$ for an $A.P.$, we have:$2 \log _9(3^{1-x}+2) = 1 + \log _3(4 \cdot 3^x-1)$Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$, we get $\log_9(y) = \frac{1}{2} \log_3(y)$:$2 \cdot \frac{1}{2} \log _3(3^{1-x}+2) = \log _3 3 + \log _3(4 \cdot 3^x-1)$$\log _3(3^{1-x}+2) = \log _3(3(4 \cdot 3^x-1))$$3^{1-x}+2 = 12 \cdot 3^x - 3$Let $3^x = t$. Then $\frac{3}{t} + 2 = 12t - 3$$3 + 2t = 12t^2 - 3t$$12t^2 - 5t - 3 = 0$$(4t - 3)(3t + 1) = 0$Since $t = 3^x > 0$, we have $t = \frac{3}{4}$.$3^x = \frac{3}{4} \Rightarrow x = \log_3 \left(\frac{3}{4}\right) = \log_3 3 - \log_3 4 = 1 - \log_3 4$.

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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