Basic Mathematics & Logarithm
Basic of Mathematics
nta_pyq_2025_jan
Grade 11
Question:
The product of all solutions of the equation $e^{5(\log_e x)^2 + 3} = x^{8},\ x>0$, is :
$e^{8/5}$
$e^{6/5}$
$e^{2}$
$e$
Step-by-Step Solution
Key Concept: Take $\log_e$ of both sides to convert the exponential equation into a quadratic in $t=\log_e x$. Then the product of the original roots is $e^{t_1+t_2}$ (sum of roots of the quadratic) — Vieta's directly gives the product of $x$-solutions without finding them individually.
Take natural log on both sides:
$$\log_e\!\left(e^{5(\log_e x)^2 + 3}\right) = \log_e\!\left(x^{8}\right)\Longrightarrow 5(\log_e x)^2 + 3 = 8\,\log_e x.$$
Let $t = \log_e x$. The equation becomes
$$5t^{2} - 8t + 3 = 0.$$
By Vieta's, $t_1 + t_2 = \dfrac{8}{5}$. Hence
$$\log_e(x_1 x_2) = t_1 + t_2 = \frac{8}{5}\ \Longrightarrow\ x_1 x_2 = e^{8/5}.$$
Correct Answer: 1