Tolong dijwab yah....pleaseee

Dengan menggunakan ragam sifat eksponen:

[tex]$\begin{align}\sqrt{8^{x^2-4x+3}}&=\frac1{32^{x-1}} \\ \sqrt{(2^3)^{x^4-4x+3}}&=\frac1{(2^5)^{x-1}} \\ \sqrt{2^{3(x^2-4x+3)}}&=(2^{5(x-1)})^{-1} \\ 2^{\frac32(x^2-4x+3)}&=2^{-5(x-1)} \\ \frac32(x^2-4x+3)&=-5(x-1) \\ 3(x^2-4x+3)&=-10(x-1) \\ 3x^2-12x+9&=-10x+10 \\ 3x^2-2x-1&=0 \\ (3x+1)(x-1)&=0\end{align}[/tex]

Diperoleh solusi x = {-1/3, 1}