We have \(\log_x(x^2 + 12) = 4\)
\(⇒ x ^2 +12=x ^4 \)
\(⇒x^ 4 −x^ 2 −12=0 \)
\(x^2(x^2 - 4) + 3(x^2 - 4) = 0\)
\((x ^2 −4)(x^ 2 +3)=0\)
given that x is a positive real number, then x = 2.
Given \(3\log_y{x} = 1\)
\(\log_y{x} = \frac{1}{3}\)
\(⇒\) \(x = y^\frac{1}{3}\)
\(⇒\) \(y=x^ 3 \)
\(⇒\) \(y = 8.\)
\(⇒\) \(x + y = 2 + 8 = 10.\)
The product of all solutions of the equation \(e^{5(\log_e x)^2 + 3 = x^8, x > 0}\) , is :