Solve Logarithmic Inequality: log₀.₂₅7 + log₀.₂₅(1/3) < log₀.₂₅(x²)

$\log_{0.25}7+\log_{0.25}\frac{1}{3}<\log_{0.25}x^2$

$x=\text{?}$

Let's solve the inequality step-by-step:

Step 1: Apply the sum of logarithms property.
We have:

This simplifies to:

Step 2: Use the property of logarithms indicating that if bases are the same and the inequality involves $\log_b(a) < \log_b(b)$, where $b < 1$, it implies:

Since $0.25 < 1$, the inequality $\log_{0.25}\left(\frac{7}{3}\right) < \log_{0.25}(x^2)$ implies:

Step 3: Simplify the inequality:
$x^2 < \frac{7}{3}$

Since $x^2 < \frac{7}{3}$, this implies:

Thus, the domain of $x$ based on the restriction of positive numbers for logarithm and quadratic expression is:

Therefore, the correct solution is $-\sqrt{\frac{7}{3}} < x < 0, 0 < x < \sqrt{\frac{7}{3}}$.

Thus, the choice that corresponds to this solution is Choice 1.