To solve the expression \( ⁵\log (\sqrt{11} + 6) + ⁵\log (\sqrt{11} - 6) \), we can apply the logarithm addition rule, which states:

\[
\log_b(x) + \log_b(y) = \log_b(x \cdot y)
\]

Here, \( b = 5 \), \( x = \sqrt{11} + 6 \), and \( y = \sqrt{11} - 6 \).

So,

\[
⁵\log (\sqrt{11} + 6) + ⁵\log (\sqrt{11} - 6) = ⁵\log \left[ (\sqrt{11} + 6) \cdot (\sqrt{11} - 6) \right]
\]

Next, use the difference of squares formula:

\[
(\sqrt{11} + 6) \cdot (\sqrt{11} - 6) = (\sqrt{11})^2 - 6^2 = 11 - 36 = -25
\]

Thus, the expression simplifies to:

\[
⁵\log (-25)
\]

However, the logarithm of a negative number is undefined in the set of real numbers. Therefore, the expression \( ⁵\log (-25) \) is undefined in the real number system.

### Would you like further details or have any questions?

Here are five related questions you might find interesting:
1. How does the logarithm function behave with complex numbers?
2. What is the difference between natural logarithms and logarithms with other bases?
3. How can we compute logarithms for non-integer bases, like 5?
4. What are the common properties of logarithms that simplify complex expressions?
5. How can logarithmic functions be used in real-world applications?

**Tip:** When dealing with logarithms, always check the domain of the function, especially when dealing with negative inputs or unusual bases.

Explore the expression ⁵log (√11+6) + ⁵log (√11-6) and understand why it is undefined. Learn about logarithm properties, the domain of logarithmic functions, and how to apply the difference of squares formula in logarithmic calculations. Suitable for Grades 10-12 students.

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