Evaluate 1/(log_2 π) + 1/(log_6 π).

Introduction / Context:
Use the reciprocal log identity: 1/log_a b = log_b a (a, b > 0, a, b ≠ 1). This lets us express the sum in a common base and combine using log addition rules.

Given Data / Assumptions:

• Terms: 1/(log_2 π) and 1/(log_6 π).
• π ≈ 3.1416; bases 2 and 6 are valid (≠ 1).

Concept / Approach:

• 1/log_a b = log_b a ⇒ 1/log_2 π = log_π 2; 1/log_6 π = log_π 6.
• Sum ⇒ log_π 2 + log_π 6 = log_π (12).

Step-by-Step Solution:

Verification / Alternative check:
Compute approximately: ln 12 / ln π ≈ 2.4849 / 1.1447 ≈ 2.17 > 1.

Why Other Options Are Wrong:

• “Less than 1” contradicts 12 > π.
• “Between 5 and 6” grossly overestimates.
• “None of these” is incorrect since “greater than 1” is true.

Final Answer:
greater then 1