Decoding a Challenging Logarithm Question from JEE Advanced 2012
JEE Advanced 2012 Solving A Tricky Logarithm Problem is a classic example of the type of question that can trip up even the most prepared candidates. The problem requires a blend of algebraic manipulation, insight into logarithmic identities, and careful handling of exponents. In this article we walk through the exact question, dissect its structure, and present a step‑by‑step solution that highlights the reasoning process a candidate should follow during the exam.
JEE Advanced 2012 Solving A Tricky Logarithm Problem: The Question
In the mathematics section of Paper 1, the question read: “Let x be a positive real number such that log₂ x + logₓ 2 = 3. Find the value(s) of x.” Candidates were asked to select the correct answer from four options, each offering a different expression for x. The problem tests familiarity with the change‑of‑base formula and the property that logₐ b = 1/log_b a.
Why This Problem Is Tricky
- Symmetry – The equation involves both log₂ x and logₓ 2, creating a reciprocal relationship that is easy to overlook.
- Implicit Exponents – The answer involves powers of 2, which may tempt students to take logarithms again or to mis‑apply exponent rules.
- Multiple Valid Roots – The quadratic formed during the solution yields two positive roots; both are mathematically valid but only one may appear in the answer choices, requiring careful verification.
Step‑by‑Step Solution
1. Introduce a new variable. Let a = log₂ x. Then x = 2ᵃ. Using the reciprocal property, we have logₓ 2 = 1/log₂ x = 1/a.
2. Rewrite the equation. Substituting the expressions for the logarithms gives: a + 1/a = 3.
3. Clear the fraction. Multiply both sides by a to obtain: a² + 1 = 3a.
4. Form the quadratic. Rearranging yields a² – 3a + 1 = 0.
5. Solve the quadratic. Using the quadratic formula, a = (3 ± √(9 – 4))/2 = (3 ± √5)/2. Because a = log₂ x > 0, both solutions are admissible.
6. Back‑substitute to find x. Since x = 2ᵃ, we obtain two possible values:
- When a = (3 + √5)/2, x = 2^((3 + √5)/2).
- When a = (3 – √5)/2, x = 2^((3 – √5)/2).
Both expressions are valid solutions. In the original exam, the answer choice that matched 2^((3 – √5)/2) was marked correct; however, the other value also satisfies the equation, illustrating the importance of verifying each answer option against the question’s constraints.
Common Pitfalls to Avoid
- Neglecting the domain. Logarithms require positive arguments. Forgetting that x > 0 can lead to extraneous solutions.
- Misapplying the reciprocal property. Confusing logₐ b with log_b a in the wrong direction leads to incorrect equations.
- Skipping verification. Selecting the first algebraic root without plugging back into the original equation can produce a seemingly correct yet invalid answer.
Exam‑Ready Tips for Logarithm Questions
- Always set a single variable for one logarithm; the other will then be its reciprocal.
- Check whether the equation can be transformed into a polynomial in that variable.
- After solving, revert to the original variable and ensure the answer fits the domain constraints.
- When multiple options appear similar, compare them using exact algebraic forms rather than approximate decimals.
Broader Takeaways for JEE Advanced
This problem exemplifies a broader class
