Consider the equation [tex]\log_3(x + 2)=\log_3(2x^2-1).[/tex]
1 step (option 6): rise 3 to the power of left and right sides
[tex]3^{\log_3(x + 2)}=3^{\log_3(2x^2-1)},\\ \\x+2=2x^2-1.[/tex]
Thus, you get equation from option 5.
2 step (option 2): rewrite the equation in one side
[tex]2x^2-x-2-1=0,\\ \\2x^2-x-3=0.[/tex]
3 step (option 1): factor the previous equation
[tex]D=(-1)^2-4\cdot 2\cdot (-3)=25,\ \sqrt{D}=5,\\ \\x_1=\dfrac{1-5}{4}=-1,\ x_2=\dfrac{1+5}{4}=\dfrac{3}{2},\\ \\2x^2-x-3=2\left(x-\dfrac{3}{2}\right)(x+1)=(2x-3)(x+1).[/tex]
Then the equatios becomes
[tex](2x-3)(x+1)=0.[/tex]
4 step (option 4): If product is equal to zero, then either [tex]2x-3=0[/tex] or [tex]x+1=0.[/tex]
5 step (option 3): Then Potential solutions are −1 and 3/2. You can check them substituting into the initial equation:
for [tex]x=-1:[/tex] [tex]\log_3(-1+ 2)=\log_31=0,\\ \\ \log_3(2\cdot (-1)^2-1)=\log_3 1=0.[/tex]
for [tex]x=\dfrac{3}{2}:[/tex] [tex]\log_3\left(\dfrac{3}{2}+ 2\right)=\log_3\dfrac{7}{2},\\ \\ \log_3(2\cdot \left(\dfrac{3}{2}\right)^2-1)=\log_3 \dfrac{7}{2}.[/tex]
Both are roots.
Answer: correct order of steps is 6, 5, 2, 1, 4, 3.
on Edg. the order of the answers would be 4,3,6,5,2,1
