The image contains the following mathematical expression:

$(4 \log_4(3x^2)) - (5 \log_4(2x)) - (3 \log_4(3)) + (\log_4(8))$

Let's simplify this expression step by step using logarithmic properties.

Step 1: Apply the power rule of logarithms:

The power rule is $\log_b(x^n) = n \log_b(x)$, so we can rewrite the first two terms.

$4 \log_4(3x^2) = \log_4((3x^2)^4) = \log_4(81x^8)$ $5 \log_4(2x) = \log_4((2x)^5) = \log_4(32x^5)$

Step 2: Simplify the logarithmic terms:

The expression now becomes:

$\log_4(81x^8) - \log_4(32x^5) - 3 \log_4(3) + \log_4(8)$

Step 3: Combine the logarithms:

Use the quotient rule $\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)$.

$\log_4\left(\frac{81x^8}{32x^5}\right) - 3 \log_4(3) + \log_4(8)$ Simplify the fraction inside the logarithm:

$\frac{81x^8}{32x^5} = \frac{81}{32} x^3$ So, the expression is now:

$\log_4\left(\frac{81}{32} x^3\right) - 3 \log_4(3) + \log_4(8)$

Step 4: Simplify the remaining logarithms:

We know that $\log_4(8)$ and $\log_4(3)$ are constants:

$\log_4(8) = \frac{3}{2} \quad \text{and} \quad \log_4(3)$ Thus, the expression simplifies further, but to fully simplify, we'd need to approximate or calculate $\log_4(3)$. Would you like to proceed with the exact form or approximate values?

Let me know if you want more details, or if you have any questions. Here are some relative questions to expand on the topic:

1. How do logarithmic properties like the product rule apply to expressions?
2. How would the base change formula affect this logarithmic simplification?
3. What happens if the logarithmic base were different, say base 10?
4. Can you explain how to deal with logarithms in exponential equations?
5. How can natural logarithms be used in more complex calculus problems?

Tip: Always check whether logarithmic terms can be simplified using rules before plugging values or using calculators.