If 3^x = 4^{x-1},then x = | vedclass

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(A,B,C) Given $3^x = 4^{x-1}$.Taking $\log_3$ on both sides:$x = (x-1) \log_3 4$$x = (x-1) \cdot 2 \log_3 2$$x = 2x \log_3 2 - 2 \log_3 2$$2 \log_3 2

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$(C) \frac{1}{1 - \log_4 3}$

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(A,B,C) Given $3^x = 4^{x-1}$.Taking $\log_3$ on both sides:$x = (x-1) \log_3 4$$x = (x-1) \cdot 2 \log_3 2$$x = 2x \log_3 2 - 2 \log_3 2$$2 \log_3 2 = x(2 \log_3 2 - 1)$$x = \frac{2 \log_3 2}{2 \log_3 2 - 1}$ (Option $A$ is correct).Taking $\log_2$ on both sides:$x \log_2 3 = (x-1) \log_2 4$$x \log_2 3 = 2(x-1)$$x \log_2 3 = 2x - 2$$2 = x(2 - \log_2 3)$$x = \frac{2}{2 - \log_2 3}$ (Option $B$ is correct).Since $\log_2 3 = \log_4 3^2 = 2 \log_4 3$,we have:$x = \frac{2}{2 - 2 \log_4 3} = \frac{1}{1 - \log_4 3}$ (Option $C$ is correct).

If $3^x = 4^{x-1}$,then $x = $

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