The correct answer is (D):log35 = log5 (x + 2) log33 < log3 5 < log391 < log35 < 2 So, 1 < log5 (x + 2) < 25 1 < x + 2 < 52 3 < x < 23
The product of all solutions of the equation \(e^{5(\log_e x)^2 + 3 = x^8, x > 0}\) , is :