Your expression seems to involve logarithms, but it is a bit unclear. It looks like you're working with:

• Base-2 logarithms, possibly written as $\log_2$
• There seems to be some reference to "64" and fractions such as $\frac{1}{6}$, but the format is ambiguous.
  Let me interpret it as follows, based on what I think you're aiming for:

Interpretation:

$\log_2 + \log_2 64 - \log_2 \left(\frac{1}{6}\right)$

If this is correct, let’s solve it step by step.

---

Step 1: Simplify $\log_2 64$

Since $64 = 2^6$, we know: $\log_2 64 = 6$

---

Step 2: Simplify $\log_2 \left(\frac{1}{6}\right)$

Using logarithm properties: $\log_2 \left(\frac{1}{6}\right) = \log_2 1 - \log_2 6 = 0 - \log_2 6 = -\log_2 6$

We leave $\log_2 6$ as is, or approximate it: $\log_2 6 \approx 2.585$ Thus: $-\log_2 6 \approx -2.585$

---

Step 3: Combine Terms

Now the expression is: $\log_2 + 6 - (-2.585)$

Since there is no value provided for the first $\log_2$, it may be a typo. Assuming it's intended to be zero: $0 + 6 + 2.585 = 8.585$

---

Final Answer:

$\boxed{8.585}$

If my interpretation differs from your intent, please clarify! I'll be happy to adjust the solution.

---

Would you like further details or have any questions?
Here are five related questions that expand on this topic:

1. How do we solve logarithms with different bases, like $\log_3(27)$?
2. What are the key logarithmic properties for solving complex problems?
3. How do logarithms relate to exponential functions?
4. Can you express $\log_2(6)$ exactly or only approximately?
5. How do negative arguments inside logarithms behave?

Tip: Remember that $\log_b(xy) = \log_b(x) + \log_b(y)$ helps with simplifying products inside logarithms.