Given, $$3^x=4^{x-1}$$

Taking $$\log$$ on both side with base 3

$$\begin{aligned} & \Rightarrow \log _3 3^x=\log _3\left(4^{x-1}\right) \\ & \Rightarrow x \log _3 3=(x-1) \log _3 4 \\ & \Rightarrow \quad x .1=(x-1) \log _3 4 \\ & \Rightarrow \quad x=\frac{\log _3 4}{\log _3 4-1} \\ & \Rightarrow \quad x=\frac{1}{1-\frac{1}{\log _3 4}} \\ \end{aligned}$$

$$\text { According to base changing rule } \frac{1}{\log _3 4}=\log _4 3$$

$$\begin{array}{ll} \therefore & x=\frac{1}{1-\log _4 3} \quad \text{... (i)}\\ \Rightarrow & x=\frac{1}{1-\log _{2^2} 3} \\ \Rightarrow & x=\frac{1}{1-\frac{1}{2} \log _2 3} \quad\left[\log _{a^m} x=\frac{1}{m} \log _a x\right] \\ \Rightarrow & x=\frac{2}{2-\log _2 3} \quad \text{... (ii)} \end{array}$$

$$\begin{aligned} & \Rightarrow \quad x=\frac{2}{2-\frac{1}{\log _3 2}} \\ & \Rightarrow \quad x=\frac{2 \log _3 2}{2 \log _3 2-1} \quad \text{... (iii)} \end{aligned}$$

From equation (i), (ii) and (iii), it is clear that option (A), (B) and (C) are correct.

Hints:

(i) Base changing Rule $$\log _b^a=\frac{1}{\log _a^b}$$

(ii) $$\log _a x^n=n \log _a x$$

(iii) $$\log _{a^m} x=\frac{1}{m} \log _a x$$