Solve Logarithmic Equation: 1/2 log₃(x⁴) = log₃(3x² + 5x + 1)

To solve the equation $\frac{1}{2}\log_3(x^4) = \log_3(3x^2 + 5x + 1)$, we will first use the power property of logarithms.

• Step 1: Apply the power property to the left side: $\frac{1}{2}\log_3(x^4) = \log_3(x^4)^{\frac{1}{2}} = \log_3(x^2)$.

• Step 2: Now, equating the arguments on both sides, we have: $x^2 = 3x^2 + 5x + 1$.

• Step 3: Rearrange the equation to form a standard quadratic: $0 = 2x^2 + 5x + 1$ or $2x^2 + 5x + 1 = 0$.

• Step 4: Solve the quadratic using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = 5$, and $c = 1$.

• Step 5: Substitute the coefficients into the quadratic formula:

• $\begin{aligned} x &= \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \\ &= \frac{-5 \pm \sqrt{25 - 8}}{4} \\ &= \frac{-5 \pm \sqrt{17}}{4} \end{aligned}$

Since we need the solutions to keep the arguments of the logarithms positive, we ensure that 3x^2 + 5x + 1 > 0 for values of $x$ from our solution set.

Thus, the solutions satisfying these conditions are given by $x = -\frac{5}{4} \pm \frac{\sqrt{17}}{4}$. Therefore, the correct answer is choice 1: $-\frac{5}{4} \pm \frac{\sqrt{17}}{4}$.