Correct answer: ±3
Explanation:
Given: \(\log_{3} x^2 = 2\)
Using logarithmic identity: \(\log_b a = c \Rightarrow a = b^c\)
So, \(x^2 = 3^2 = 9\)
Taking square root on both sides: \(x = \pm 3\)
So, x = ±3. Since ±3 is not listed directly, among the options, only 3 and -3 are valid.
The product of all solutions of the equation \(e^{5(\log_e x)^2 + 3 = x^8, x > 0}\) , is :