Multiple choice

If ${x_1}{\rm{ and }}\ {{{x}}_2}$ are the root of the equation ${e^\cfrac32}.{x^{2\ell nx}} = {x^4}$ , then the product of the roots of the equation

  1. ${e^2}$
  2. $e$
  3. $\dfrac{e^3}{2}$
  4. ${e^{-2}}$
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A Correct answer
Explanation

Interpreting the equation as e^(3/2) * x^(2 ln x) = x^4 and taking natural logarithms gives 2(ln x)^2 - 4 ln x + 3/2 = 0. The sum of the two values of ln x is 2, so the product of the roots is e^2.

AI explanation

Divide both sides of the equation (e^(3/2))*x^(2 ln x) = x^4 by x^4 to get (e^(3/2))*x^(2 ln x - 4) = 1. Taking the natural logarithm of both sides results in 3/2 + (2 ln x - 4)(ln x) = 0. Letting y = ln x, this becomes the quadratic equation 2y^2 - 4y + 3/2 = 0. For this equation in y, the sum of its roots is 4/2 = 2. Since the original roots are x1 = e^y1 and x2 = e^y2, their product is x1 x2 = e^(y1 + y2) = e^2.

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