Answer:
Step-by-step explanation:
[tex](x^2-3x+1)^{(2x^2+x-6)}=1\ we \ take \ the \ logarithm\\\\(2x^2+x-6)*ln(x^2-3x+1)=0\\\\(2x^2+x-6)=0\ or\ x^2-3x+1=1\\\\1)\\2x^2+x-6=0\\2x^2+4x-3x-6=0\\2x(x+2)-3(x+2)=0\\(x+2)(2x-3)\\x=-2\ or\ x=\dfrac{3}{2} \\\\2)\\x^2-3x=0\\x(x-3)=0\\x=0\ or\ x=3\\\\sol=\{-2,0,\dfrac{3}{2},3\}\\[/tex]
