Question

order the steps to solve the equation $log_{3}(x + 2) = log_{3}(2x^{2} - 1)$ from 1 to 6.$0 = (2x - 3)(x + 1)$$0 = 2x^{2} - x -3$potential solutions are $-1$ and $\frac{3}{2}$.$2x - 3 = 0$ or $x + 1 = 0$$x + 2 = 2x^{2} - 1$$3^{log_{3}(x + 2)} = 3^{log_{3}(2x^{2} - 1)}$

Explanation:

Step1: Eliminate log base 3

$3^{\log_3(x + 2)} = 3^{\log_3(2x^2 - 1)}$

Step2: Simplify using log property

$x + 2 = 2x^2 - 1$

Step3: Rearrange to standard quadratic

$0 = 2x^2 - x - 3$

Step4: Factor the quadratic

$0 = (2x - 3)(x + 1)$

Step5: Set factors equal to 0

$2x - 3 = 0 \text{ or } x + 1 = 0$

Step6: Solve for potential solutions

Potential solutions are $-1$ and $\frac{3}{2}$.

Answer:

1. $3^{\log_3(x + 2)} = 3^{\log_3(2x^2 - 1)}$
2. $x + 2 = 2x^2 - 1$
3. $0 = 2x^2 - x - 3$
4. $0 = (2x - 3)(x + 1)$
5. $2x - 3 = 0 \text{ or } x + 1 = 0$
6. Potential solutions are $-1$ and $\frac{3}{2}$.