1. Let the three terms in A.P. be $a_1=1,\, a_2=\log_9(3^{1-x}+2),\, a_3=\log_3(4\cdot 3^x-1).$

2. For three numbers in A.P., the middle term is the average of the other two: $2a_2=a_1+a_3.$ So, $2\log_9(3^{1-x}+2)=1+\log_3(4\cdot 3^x-1).$

3. Convert the base $9$ logarithm to base $3$: $\log_9 A=\frac{\log_3 A}{\log_3 9}=\frac{1}{2}\log_3 A.$ Hence, $2\log_9(3^{1-x}+2)=\log_3(3^{1-x}+2).$ Therefore the A.P. condition becomes $\log_3(3^{1-x}+2)=1+\log_3(4\cdot 3^x-1).$

4. Write $1$ as $\log_3 3$: $\log_3(3^{1-x}+2)=\log_3 3+\log_3(4\cdot 3^x-1)$ $=\log_3\big(3(4\cdot 3^x-1)\big).$

5. Since logarithms with the same base are equal, their arguments are equal: $3^{1-x}+2=3(4\cdot 3^x-1).$ Simplify: $3^{1-x}+2=12\cdot 3^x-3$ $3^{1-x}+5=12\cdot 3^x.$

6. Let $t=3^x>0.$ Then $3^{1-x}=\frac{3}{3^x}=\frac{3}{t}.$ So the equation becomes $\frac{3}{t}+5=12t.$ Multiply by $t$: $3+5t=12t^2$ $12t^2-5t-3=0.$

7. Solve the quadratic: $12t^2-5t-3=(4t-3)(3t+1)=0.$ So, $t=\frac{3}{4}\quad \text{or} \quad t=-\frac{1}{3}.$ But $t=3^x>0$, so $t=\frac{3}{4}.$

8. Thus, $3^x=\frac{3}{4}$ $x=\log_3\left(\frac{3}{4}\right)=\log_3 3-\log_3 4=1-\log_3 4.$

9. Check options:

   • A: $\log_3 4$ ❌
   • B: $1-\log_3 4$ ✅
   • C: $1-\log_4 3$ ❌
   • D: $\log_4 3$ ❌

Therefore, the correct answer is B.