Solve Base-13 Logarithmic Inequality: log₁₃(2x² + 3) ≤ log₁₃7 - log₁₃x²

To solve this problem, we'll follow these steps:

• Step 1: Simplify the logarithmic expressions using $\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$.
• Step 2: Compare the simplified expressions' arguments.
• Step 3: Solve the algebraic inequality involving the arguments.

Now, let's work through each step:

Step 1: Use the property $\log_{13} A - \log_{13} B = \log_{13}\left(\frac{A}{B}\right)$ to simplify:

$\log_{13}(2x^2+3) - \log_{13}2 = \log_{13}\left(\frac{2x^2+3}{2}\right)$

$\log_{13}7 - \log_{13}x^2 = \log_{13}\left(\frac{7}{x^2}\right)$

Step 2: This gives the inequality:

$\log_{13}\left(\frac{2x^2+3}{2}\right) \le \log_{13}\left(\frac{7}{x^2}\right)$

Since the logarithm function is monotonically increasing, we can drop the logs and solve:

$\frac{2x^2+3}{2} \le \frac{7}{x^2}$

Multiplying through by $2x^2$, to eliminate fractions, ensures none of the values of $x$ is zero, which would cause division by zero:

$(2x^2 + 3)x^2 \le 14$

Expanding gives a quadratic inequality:

$2x^4 + 3x^2 - 14 \le 0$

Step 3: Substitute $u = x^2$ to transform into quadratic form:

$2u^2 + 3u - 14 \le 0$

Find the critical points by solving the equation $2u^2 + 3u - 14 = 0$:

$u = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-14)}}{2 \cdot 2}$

$u = \frac{-3 \pm \sqrt{9 + 112}}{4}$

$u = \frac{-3 \pm \sqrt{121}}{4}$

$u = \frac{-3 \pm 11}{4}$

This gives the roots $u = 2$ and $u = -\frac{7}{2}$. Only non-negative values for $u$ make sense since $u = x^2$, so consider:

$x^2 = u \le 2$

Thus, $-\sqrt{2} \le x \le \sqrt{2}$.

Therefore, the solution to the problem is $-\sqrt{2} \le x \le \sqrt{2}$.