Answer :
Sure, let's solve the equation [tex]\(\log_3(x + 2) = \log_3(2x^2 - 1)\)[/tex] step by step.
1. Rewrite the equation using the property of logarithms:
Since [tex]\(\log_b(A) = \log_b(B)\)[/tex] implies [tex]\(A = B\)[/tex], we have:
[tex]\[ x + 2 = 2x^2 - 1 \][/tex]
2. Move all terms to one side of the equation to set the equation to 0:
[tex]\[ 0 = 2x^2 - x - 3 \][/tex]
3. Factor the quadratic equation:
[tex]\[ 0 = (2x - 3)(x + 1) \][/tex]
4. Set each factor to 0 to solve for [tex]\(x\)[/tex]:
[tex]\[ 2x - 3 = 0 \quad \text{or} \quad x + 1 = 0 \][/tex]
5. Solve each equation to find the potential solutions:
[tex]\[ 2x - 3 = 0 \implies x = \frac{3}{2} \][/tex]
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]
Therefore, the potential solutions are:
[tex]\[ \text{Potential solutions are } -1 \text{ and } \frac{3}{2} \][/tex]
Now, let's order the given options according to the steps we have:
1. [tex]\(x + 2 = 2x^2 - 1\)[/tex]
2. [tex]\(0 = 2x^2 - x - 3\)[/tex]
3. [tex]\(0 = (2x - 3)(x + 1)\)[/tex]
4. [tex]\(2x - 3 = 0 \quad \text{or} \quad x + 1 = 0\)[/tex]
5. Potential solutions are -1 and [tex]\(\frac{3}{2}\)[/tex]
The ordered steps are:
1. [tex]\(x + 2 = 2x^2 - 1\)[/tex]
2. [tex]\(0 = 2x^2 - x - 3\)[/tex]
3. [tex]\(0 = (2x - 3)(x + 1)\)[/tex]
4. [tex]\(2x - 3 = 0 \quad \text{or} \quad x + 1 = 0\)[/tex]
5. Potential solutions are -1 and [tex]\(\frac{3}{2}\)[/tex]
6. [tex]\(3^{\log _3(x+2)}=3^{\log _3(2x^2-1)}\)[/tex] is irrelevant since it is already assumed within the logarithmic property mentioned in step 1.
So the final ordered list is:
[tex]$\square$[/tex] [tex]$x+2=2x^2-1$[/tex] \\
[tex]$\square$[/tex] [tex]$0=2x^2-x-3$[/tex] \\
[tex]$\square$[/tex] [tex]$0=(2 x-3)(x+1)$[/tex] \\
[tex]$\square$[/tex] [tex]$2 x-3=0$[/tex] or [tex]$x+1=0$[/tex] \\
[tex]$\square$[/tex] Potential solutions are -1 and [tex]$\frac{3}{2}$[/tex].
1. Rewrite the equation using the property of logarithms:
Since [tex]\(\log_b(A) = \log_b(B)\)[/tex] implies [tex]\(A = B\)[/tex], we have:
[tex]\[ x + 2 = 2x^2 - 1 \][/tex]
2. Move all terms to one side of the equation to set the equation to 0:
[tex]\[ 0 = 2x^2 - x - 3 \][/tex]
3. Factor the quadratic equation:
[tex]\[ 0 = (2x - 3)(x + 1) \][/tex]
4. Set each factor to 0 to solve for [tex]\(x\)[/tex]:
[tex]\[ 2x - 3 = 0 \quad \text{or} \quad x + 1 = 0 \][/tex]
5. Solve each equation to find the potential solutions:
[tex]\[ 2x - 3 = 0 \implies x = \frac{3}{2} \][/tex]
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]
Therefore, the potential solutions are:
[tex]\[ \text{Potential solutions are } -1 \text{ and } \frac{3}{2} \][/tex]
Now, let's order the given options according to the steps we have:
1. [tex]\(x + 2 = 2x^2 - 1\)[/tex]
2. [tex]\(0 = 2x^2 - x - 3\)[/tex]
3. [tex]\(0 = (2x - 3)(x + 1)\)[/tex]
4. [tex]\(2x - 3 = 0 \quad \text{or} \quad x + 1 = 0\)[/tex]
5. Potential solutions are -1 and [tex]\(\frac{3}{2}\)[/tex]
The ordered steps are:
1. [tex]\(x + 2 = 2x^2 - 1\)[/tex]
2. [tex]\(0 = 2x^2 - x - 3\)[/tex]
3. [tex]\(0 = (2x - 3)(x + 1)\)[/tex]
4. [tex]\(2x - 3 = 0 \quad \text{or} \quad x + 1 = 0\)[/tex]
5. Potential solutions are -1 and [tex]\(\frac{3}{2}\)[/tex]
6. [tex]\(3^{\log _3(x+2)}=3^{\log _3(2x^2-1)}\)[/tex] is irrelevant since it is already assumed within the logarithmic property mentioned in step 1.
So the final ordered list is:
[tex]$\square$[/tex] [tex]$x+2=2x^2-1$[/tex] \\
[tex]$\square$[/tex] [tex]$0=2x^2-x-3$[/tex] \\
[tex]$\square$[/tex] [tex]$0=(2 x-3)(x+1)$[/tex] \\
[tex]$\square$[/tex] [tex]$2 x-3=0$[/tex] or [tex]$x+1=0$[/tex] \\
[tex]$\square$[/tex] Potential solutions are -1 and [tex]$\frac{3}{2}$[/tex].
