JEE Main 2021 · 17 March · morning · Single correct · +4 / −1
If the fourth term in the expansion of (x + x^(log ₂x))⁷ is 4480, then the value of x where x∈N is equal to :
- (a) 3
- (b) 1
- (c) 4
- (d) 2
Answer and solution
Answer (d) 2
Solution
Maths sir · 4 steps
Write the fourth term, then take log₂ of both sides and the power tower becomes a quadratic.
- Fourth term. The fourth term is T₄ = C(7, 3)x⁴(x^(log₂ x))³ = 35 x^(4 + 3 log₂ x). Setting it equal to 4480 and dividing by 35 gives x^(4 + 3 log₂ x) = 128 = 2⁷.
- Take log base 2. Let t = log₂ x. Taking log₂ of both sides turns the exponent into a product, (4 + 3t) t = 7.
3t² + 4t - 7 = 0 ⇒ (t - 1)(3t + 7) = 0
- Pick the natural root. So t = 1 or t = -7/3. Then x = 2 or x = 2^(-7/3), and only x = 2 is a natural number. Check: 35 × 2⁴ × 2³ = 35 × 128 = 4480.
- So. The value of x is 2, option (d).
The idea to keep. When x appears in both the base and the exponent, take a log of both sides. Here log₂ fits because log₂ x is already in the power and the right side becomes a power of 2.
Answer from the NTA final answer key. Solution by PrepWiser.