Solving Exponential Equation Involving Powers of Three and Four

To solve the given exponential equation: $3^x = 4^{x-1}$

We can rewrite the right-hand side using the properties of exponents: $3^x = \frac{4^x}{4}$

Rearranging the terms to isolate the variable $x$: $\frac{4^x}{3^x} = 4$ $\left(\frac{4}{3}\right)^x = 4$

Taking the logarithm with base $4$ on both sides: $\log_4\left(\left(\frac{4}{3}\right)^x\right) = \log_4(4)$ $x \log_4\left(\frac{4}{3}\right) = 1$ $x (\log_4 4 - \log_4 3) = 1$ $x (1 - \log_4 3) = 1$

Solving for $x$: $x = \frac{1}{1 - \log_4 3}$ This matches Option (C).

Next, converting the logarithm in the denominator to base $2$ using the change-of-base formula $\log_4 3 = \frac{\log_2 3}{\log_2 4} = \frac{\log_2 3}{2}$: $x = \frac{1}{1 - \frac{\log_2 3}{2}} = \frac{2}{2 - \log_2 3}$ This matches Option (B).

Now, converting $\log_2 3$ to base $3$ using the reciprocal property $\log_2 3 = \frac{1}{\log_3 2}$: $x = \frac{2}{2 - \frac{1}{\log_3 2}} = \frac{2}{\frac{2\log_3 2 - 1}{\log_3 2}} = \frac{2\log_3 2}{2\log_3 2 - 1}$ This matches Option (A).

Checking Option (D): $\frac{2\log_2 3}{2\log_2 3 - 1} \neq \frac{2}{2 - \log_2 3}$ Hence, Option (D) is incorrect.

Therefore, the correct options are A, B, and C.