The correct answer is (D):
log35 = log5 (x + 2)
log33 < log3 5 < log39
1 < log35 < 2
So, 1 < log5 (x + 2) < 2
5 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 :
If l, m, n are the pth, qth and rth terms of a G.P. respectively and l, m, n > 0, then
\[ \begin{vmatrix} \log_l p & 1 \\ \log_m q & 1 \\ \log_n r & 1 \end{vmatrix} \]