Solve the Logarithmic Equation: Log_4(3x^2+8x-10) = Log_4(-x^2-x+12.5)

$\log_4(3x^2+8x-10)-\log_4(-x^2-x+12.5)=0$

?=x

To solve this problem, we'll apply the following steps:

• Step 1: Use logarithmic properties to rewrite $\log_4(3x^2+8x-10) - \log_4(-x^2-x+12.5) = 0$ as a single logarithm: $\log_4 \left( \frac{3x^2 + 8x - 10}{-x^2 - x + 12.5} \right) = 0$.
• Step 2: Recognize that if $\log_4 \left( \frac{3x^2 + 8x - 10}{-x^2 - x + 12.5} \right) = 0$, then $\frac{3x^2 + 8x - 10}{-x^2 - x + 12.5} = 4^0 = 1$.
• Step 3: Set up the equation: $3x^2 + 8x - 10 = -x^2 - x + 12.5$.
• Step 4: Rearrange the equation to: $3x^2 + 8x - 10 + x^2 + x - 12.5 = 0$.
• Step 5: Simplify to: $4x^2 + 9x - 22.5 = 0$.
• Step 6: Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4$, $b = 9$, $c = -22.5$.
• Step 7: Calculate: $b^2 - 4ac = 9^2 - 4 \times 4 \times (-22.5) = 81 + 360 = 441$.
• Step 8: Find $x$: $x = \frac{-9 \pm \sqrt{441}}{8} = \frac{-9 \pm 21}{8}$.
• Step 9: Compute the roots: $x_1 = \frac{-9 + 21}{8} = \frac{12}{8} = 1.5$ and $x_2 = \frac{-9 - 21}{8} = \frac{-30}{8} = -3.75$.
• Step 10: Verify these solutions satisfy the domain of original logarithmic expressions by substituting back into $3x^2 + 8x - 10 > 0$ and $-x^2 - x + 12.5 > 0$.

Therefore, the solutions to the problem are $x = -3.75, 1.5$.

The correct choice from the provided options is: $-3.75, 1.5$.