You've provided an expression involving logarithms to be simplified. Let's start by simplifying the inner expression and then proceed to solve it.

The expression given is:

$\frac{1}{2} (\log{225} + 8\log{6} - 3\log{121})$

Step-by-step Simplification:

1. Simplify each logarithm:

   • $\log{225}$ can be rewritten using the property of logarithms $\log{a^b} = b \log{a}$, so: $\log{225} = \log{15^2} = 2\log{15}$
   • Similarly for $\log{6}$ and $\log{121}$: $8\log{6} = \log{6^8}$ $3\log{121} = \log{121^3} = \log{11^6}$

2. Combine the logarithms:

   • We can combine the logs using the properties $\log{a} + \log{b} = \log{(ab)}$ and $\log{a} - \log{b} = \log{\left(\frac{a}{b}\right)}$: $2\log{15} + \log{6^8} - \log{11^6} = \log{(15^2 \cdot 6^8)} - \log{11^6} = \log{\left(\frac{15^2 \cdot 6^8}{11^6}\right)}$

3. Evaluate the fraction: We need to evaluate $\frac{15^2 \cdot 6^8}{11^6}$ to find its logarithm.

Let's calculate this value using the formulas we derived.The simplified inner expression of the logarithms is:

$-6 \log{11} + 2 \log{15} + 8 \log{6}$

Since the constants are retained as part of the logarithmic terms, we can proceed by simplifying the outer fraction:

$\frac{1}{2} (-6 \log{11} + 2 \log{15} + 8 \log{6})$

Now, let's simplify this further to find the final numerical value.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like to revisit this problem later, or is there another question or topic you'd like to explore? Here are some related topics you might find interesting:

1. Properties of logarithms and how they are used in simplifying expressions.
2. Applications of logarithmic functions in real-world scenarios.
3. Different bases of logarithms (natural log, common log) and their properties.
4. How to solve exponential equations using logarithms.
5. The concept of logarithmic scales and how they are used in scientific measurements.

Tip: When dealing with logarithmic simplifications, remember to factorize numbers wherever possible to make simplifications easier.