To solve the equation [tex]\(\log_3(x + 2) = \log_3(2x^2 - 1)\)[/tex], we need to determine the steps in the correct order. Here is the detailed, step-by-step solution:
1. Step 1: Exponentiate both sides using base 3.
Given:
[tex]\[
\log_3(x + 2) = \log_3(2x^2 - 1)
\][/tex]
Raise both sides to the power of 3:
[tex]\[
3^{\log_3(x + 2)} = 3^{\log_3(2x^2 - 1)}
\][/tex]
2. Step 2: Simplify the equation by removing the logarithms.
Since [tex]\(3^{\log_3 A} = A\)[/tex] for any [tex]\(A > 0\)[/tex], we simplify to:
[tex]\[
x + 2 = 2x^2 - 1
\][/tex]
3. Step 3: Rearrange the equation into a standard quadratic form.
Move all terms to one side of the equation to set it to zero:
[tex]\[
x + 2 - (2x^2 - 1) = 0
\][/tex]
Simplify:
[tex]\[
0 = 2x^2 - x - 3
\][/tex]
4. Step 4: Factor the quadratic equation.
Factor the quadratic equation [tex]\(2x^2 - x - 3 = 0\)[/tex] into two binomials:
[tex]\[
0 = (2x - 3)(x + 1)
\][/tex]
5. Step 5: Set each factor equal to zero and solve for [tex]\(x\)[/tex].
Solve each factor individually:
[tex]\[
2x - 3 = 0 \quad \text{or} \quad x + 1 = 0
\][/tex]
Solving these individually gives potential solutions:
[tex]\[
x = \frac{3}{2} \quad \text{or} \quad x = -1
\][/tex]
6. Step 6: State the potential solutions.
The potential solutions of the quadratic equation are:
[tex]\[
x = -1 \quad \text{and} \quad x = \frac{3}{2}
\][/tex]
Hence, the correct order of steps to solve the equation [tex]\(\log _3(x+2)=\log _3\left(2 x^2-1\right)\)[/tex] is:
1. [tex]\(3^{\log_3(x+2)}=3^{\log_3(2 x^2-1)}\)[/tex]
2. [tex]\(x+2=2 x^2-1\)[/tex]
3. [tex]\(0=2 x^2-x-3\)[/tex]
4. [tex]\(0=(2 x-3)(x+1)\)[/tex]
5. [tex]\(2 x-3=0\)[/tex] or [tex]\(x+1=0\)[/tex]
6. Potential solutions are -1 and [tex]\(\frac{3}{2}\)[/tex]
