For a natural number x > 1, let p = logₓ (x + 1) and q = log₍ₓ₊₁₎ (x + 2). Which one of the following is correct about p and q?

Introduction / Context:
This problem is a conceptual logarithm inequality question. Instead of asking for a numeric value, it compares two expressions involving logarithms with slightly different bases and arguments. Such questions test your understanding of how logarithmic functions behave as the base and argument change, and they are common in higher level aptitude and entrance exams.

Given Data / Assumptions:

• x is a natural number with x > 1.
• p = logₓ (x + 1).
• q = log₍ₓ₊₁₎ (x + 2).
• We must decide whether p < q, p = q, p > q, or if the relation cannot be determined.

Concept / Approach:
For bases greater than 1, logarithmic functions are increasing in their argument and decreasing in their base. In p, the base is x and the argument is x + 1. In q, the base has increased to x + 1 and the argument to x + 2. We can compare p and q by writing both in terms of natural logarithms and carefully analysing how the ratios behave. Another powerful tool is to test several values of x to observe a pattern and then reason more generally.

Step-by-Step Solution:
Write p and q using natural logs: p = ln(x + 1) / ln x and q = ln(x + 2) / ln(x + 1). Evaluate numerically for x = 2. Then p = ln 3 / ln 2 ≈ 1.585 and q = ln 4 / ln 3 ≈ 1.262. So p > q for x = 2. For x = 3, p = ln 4 / ln 3 ≈ 1.262 while q = ln 5 / ln 4 ≈ 1.113, again p > q. For x = 4, p = ln 5 / ln 4 ≈ 1.161 while q = ln 6 / ln 5 ≈ 1.113, still p > q. These consistent results suggest that p is greater than q for all natural x > 1. To see why, notice that ln(x + 1) / ln x decreases slowly with x, but ln(x + 2) / ln(x + 1) is evaluated at a larger base and is pulled down more, making q smaller.

Verification / Alternative check:
We can reason qualitatively: for large x, (x + 1) is only slightly larger than x, so p is close to 1 from above. Meanwhile, x + 2 is only slightly larger than x + 1, so q is also close to 1 but with a slightly larger base in the denominator, which pushes q a bit lower than p. The numerical checks at several integer values confirm this consistent inequality p > q.

Why Other Options Are Wrong:
p < q: Contradicted by direct substitution for x = 2, 3, 4, all giving p > q.
p = q: If they were equal for all x, numeric testing would show equality, but it does not.
can't be determined: The pattern is clear and supported both numerically and conceptually, so the relationship is determinable as p > q.

Common Pitfalls:
A frequent mistake is to treat the increase in argument (from x + 1 to x + 2) as dominant and ignore the effect of increasing the base. Another error is assuming that logs with nearby arguments and bases automatically become equal. It is essential to remember that for base greater than 1, increasing the base while also increasing the argument does not guarantee any specific order without proper analysis.

Final Answer:
For all natural numbers x > 1, we have p > q.