Basic Mathematics & Logarithm
Exponential and Logarithmic Equations
nta_pyq_2025_apr
Grade 11
Question:
The product of all the 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: Substitute $t = \ln x$ to convert the equation into a quadratic in $t$, then use Vieta's formulas to find $t_1+t_2$; the product of the original solutions is $e^{t_1+t_2}$.
Let $t = \ln x$. The equation becomes $e^{5t^2+3} = e^{8t}$, so
$$5t^2 - 8t + 3 = 0.$$
By Vieta's formulas, $t_1 + t_2 = \dfrac{8}{5}$.
Product of solutions $= x_1 x_2 = e^{\ln x_1}\cdot e^{\ln x_2} = e^{t_1+t_2} = e^{8/5}$.
Correct Answer: 1