The product of all solutions of the equation e^5( _ex)^2+3=x^8,x0, is :

Question

The product of all solutions of the equation ${e}^{5{({\log }_{e}x)}^{2}+3}={x}^{8},x>0$, is :

Accepted Answer

We begin with the equation:

$ e^{5(\log_e x)^2 + 3} = x^8, \; x > 0. $

Equating the exponents, we have:

$ 5(\log_e x)^2 + 3 = \log_e x^8 = 8 \log_e x. $

Let $ t = \log_e x $. Substituting, the equation becomes:

$ 5t^2 + 3 = 8t. $

Rewriting this as a quadratic equation:

$ 5t^2 - 8t + 3 = 0. $

Factoring the quadratic:

$ 5t^2 - 5t - 3t + 3 = 0, $

$ (5t - 3)(t - 1) = 0. $

Thus, the solutions for $ t $ are:

$ t = 1 $ implies $ \log_e x = 1 $, giving $ x = e $.

$ t = \frac{3}{5} $ implies $ \log_e x = \frac{3}{5} $, giving $ x = e^{\frac{3}{5}} $.

The product of all solutions is:

$ e^1 	imes e^{\frac{3}{5}} = e^{1 + \frac{3}{5}} = e^{\frac{8}{5}}. $

Solutions · 1