Multiple choice

The sum of all the real roots of the equation (e2x – 4)(6e2x – 5ex + 1) = 0 is

  1. loge3

  2. –loge3

  3. loge6

  4. –loge6
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

(e^2x - 4)(6e^2x - 5e^x + 1) = 0. Factors: (e^x - 2)(e^x + 2)(2e^x - 1)(3e^x - 1) = 0. Real roots for e^x: e^x = 2, e^x = 1/2, e^x = 1/3. x = ln(2), x = ln(1/2) = -ln(2), x = ln(1/3) = -ln(3). Sum = ln(2) - ln(2) - ln(3) = -ln(3).

Keep practicing — related questions