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Order the steps to solve the equation [tex]\log_3(x+2) = \log_3(2x^2 - 1)[/tex] from 1 to 6.

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[/tex] or [tex]x + 1 = 0[/tex]
5. Solve for potential solutions: [tex]x = \frac{3}{2}[/tex] or [tex]x = -1[/tex]
6. Verify the potential solutions: [tex]\frac{3}{2}[/tex] and [tex]-1[/tex]

[tex]3^{\log_3(x+2)} = 3^{\log_3(2x^2 - 1)}[/tex]



Answer :

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]