The equation in the image appears to be:

$4 \log_3 x^2 + \log_3 x \geq 2 \sqrt{4}$

Let's break it down step by step.

1. Simplify the square root on the right-hand side: $2 \sqrt{4} = 2 \times 2 = 4.$ So the inequality becomes: $4 \log_3 x^2 + \log_3 x \geq 4.$

2. Apply logarithmic properties. First, recall that $\log_b (a^n) = n \log_b a$, so $\log_3 x^2 = 2 \log_3 x$. Substituting this into the inequality gives: $4(2 \log_3 x) + \log_3 x \geq 4.$ This simplifies to: $8 \log_3 x + \log_3 x \geq 4.$

3. Combine like terms: $9 \log_3 x \geq 4.$

4. Solve for $x$: $\log_3 x \geq \frac{4}{9}.$

5. Exponentiate both sides with base 3 to solve for $x$: $x \geq 3^{\frac{4}{9}}.$

Thus, the solution to the inequality is:

$x \geq 3^{\frac{4}{9}}.$

Would you like a more detailed explanation or further clarifications? Here are some related questions you might find helpful:

1. What are the properties of logarithms used in solving this inequality?
2. How do you handle inequalities involving logarithms?
3. How do we solve logarithmic inequalities step by step?
4. What is the significance of the base in logarithmic functions?
5. How does exponentiation help in solving logarithmic inequalities?

Tip: When solving logarithmic inequalities, always ensure that the argument of the logarithm is positive because the logarithm of a non-positive number is undefined.