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

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

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

$\text{To find: } x \text{ from } 3^{x} = 4^{x-1}$

$1.\ \text{take } \log_3 \text{ on both sides}$

$2.\ \log_3 4 = 2\log_3 2$

$3.\ \text{solve the linear equation for } x$

$y = 3^x$

$y = 4^{x-1} = \dfrac{4^x}{4}$

$\log_3(3^x) = \log_3(4^{x-1})$

$x = (x-1)\log_3 4$

$\log_3 4 = \log_3 2^2 = 2\log_3 2$

$x = 2(x-1)\log_3 2$

$x = 2x\log_3 2 - 2\log_3 2$

$x - 2x\log_3 2 = -2\log_3 2$

$x(1 - 2\log_3 2) = -2\log_3 2$

$x = \dfrac{-2\log_3 2}{1 - 2\log_3 2}$

$x = \dfrac{2\log_3 2}{2\log_3 2 - 1}$

$x \approx 4.82:\quad 3^{x} \approx 199,\quad 4^{x-1} \approx 199$

$\dfrac{2\log_3 2}{2\log_3 2 - 1} = \dfrac{2}{2-\log_2 3}$

$= \dfrac{1}{1-\log_4 3}$

$\boxed{x = \dfrac{2\log_3 2}{2\log_3 2 - 1}}$

$\text{option (1)}$