The domain of f(x) = (log ₍ₓ₊₁₎(x - 2))/(e^(2log ₑx) - (2x + 3)),x ∈ ℝ is

Solution

Maths sir · 4 steps

List every condition the logs and the denominator impose, then keep only the x that pass all of them.

1. Collect conditions. Each piece of f(x) brings its own restriction. A log needs a positive argument and a positive base not equal to 1, and the denominator must not be zero.

   Piece	Condition	Gives
   Base of log₍ₓ₊₁₎	x + 1 > 0 and x + 1 ≠ 1	x > -1, x ≠ 0
   Argument of log₍ₓ₊₁₎	x - 2 > 0	x > 2
   logₑ x in the denominator	x > 0	x > 0
   Denominator not zero	x² - 2x - 3 ≠ 0	x ≠ 3, x ≠ -1

2. Simplify the denominator. For x > 0, e^(2 logₑ x) = x². So the denominator is x² - (2x + 3) = (x - 3)(x + 1), which is zero at x = 3 and x = -1.
3. Intersect. The strictest condition is x > 2. It already guarantees x > -1, x ≠ 0 and x > 0. The only point left to remove is x = 3.
4. So. The domain is (2, ∞) - {3}, which is option (c).

The idea to keep. For a domain question, write down every restriction separately: log base, log argument, hidden logs like e^(2 logₑ x), and zero denominators. Then take the intersection.

Answer from the NTA final answer key. Solution by PrepWiser.