The given equation is ${3^x} = {4^{x - 1}}$. To solve for $x$, we can take the logarithm of both sides. We will check each option by either manipulating a general solution for $x$ or by taking a logarithm with a base that is convenient for the option being checked.

Step 1: Solve for $x$ in a general form

Let's take the natural logarithm (ln) on both sides of the equation: $\ln({3^x}) = \ln({4^{x - 1}})$ Using the logarithm property $\log(a^b) = b \log(a)$: $x \ln(3) = (x - 1) \ln(4)$ Since $\ln(4) = \ln(2^2) = 2 \ln(2)$: $x \ln(3) = (x - 1) (2 \ln(2))$ $x \ln(3) = 2x \ln(2) - 2 \ln(2)$ Rearrange the terms to solve for $x$: $2 \ln(2) = 2x \ln(2) - x \ln(3)$ $2 \ln(2) = x (2 \ln(2) - \ln(3))$ $x = {{2 \ln(2)} \over {2 \ln(2) - \ln(3)}}$ This is a general expression for $x$. We will now check if the given options are equivalent to this expression.

Step 2: Evaluate each option

Option A: ${{2{{\log }_3}\,2} \over {2{{\log }_3}\,2 - 1}}$

We can solve the original equation by taking the logarithm to the base 3: $\log_3(3^x) = \log_3(4^{x-1})$ $x = (x-1) \log_3(4)$ $x = (x-1) \log_3(2^2) = (x-1) \cdot 2 \log_3(2)$ $x = 2x \log_3(2) - 2 \log_3(2)$ $2 \log_3(2) = 2x \log_3(2) - x$ $2 \log_3(2) = x(2 \log_3(2) - 1)$ $x = {{2{{\log }_3}\,2} \over {2{{\log }_3}\,2 - 1}}$ Thus, option A is correct.

Option B: ${2 \over {2 - {{\log }_2}\,3}}$

Let's solve the original equation by taking the logarithm to the base 2: $\log_2(3^x) = \log_2(4^{x-1})$ $x \log_2(3) = (x-1) \log_2(4)$ Since $\log_2(4) = \log_2(2^2) = 2$: $x \log_2(3) = (x-1) \cdot 2$ $x \log_2(3) = 2x - 2$ $2 = 2x - x \log_2(3)$ $2 = x(2 - \log_2(3))$ $x = {2 \over {2 - {{\log }_2}\,3}}$ Thus, option B is correct.

Option C: ${1 \over {1 - {{\log }_4}\,3}}$

Let's solve the original equation by taking the logarithm to the base 4: $\log_4(3^x) = \log_4(4^{x-1})$ $x \log_4(3) = (x-1) \log_4(4)$ Since $\log_4(4) = 1$: $x \log_4(3) = x - 1$ $1 = x - x \log_4(3)$ $1 = x(1 - \log_4(3))$ $x = {1 \over {1 - {{\log }_4}\,3}}$ Thus, option C is correct.

Option D: ${{2{{\log }_2}\,3} \over {2{{\log }_2}\,3 - 1}}$

From our derivation for option B, we found that $x = {2 \over {2 - {{\log }_2}\,3}}$. Let's check if this is equal to the expression in option D. Let $y = \log_2(3)$. We are checking if ${2 \over {2 - y}} = {{2y} \over {2y - 1}}$. Cross-multiplying gives: $2(2y - 1) = 2y(2 - y)$ $4y - 2 = 4y - 2y^2$ $-2 = -2y^2$ $y^2 = 1 \implies y = \pm 1$ This would mean $\log_2(3) = 1$ or $\log_2(3) = -1$. Neither of these is true, since $2^1 = 2 \neq 3$ and $2^{-1} = 1/2 \neq 3$. Therefore, the expression in option D is not a correct value for $x$.

Conclusion

The correct expressions for $x$ are given in options A, B, and C.