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4-2.Quadratic Equations and Inequations
medium
The product of all solutions of the equation $e ^{5\left(\log _{ e } x \right)^2+3}= x ^8, x >0$, is :
A$e^{8 / 5}$
B$e^{6 / 5}$
C$e^2$
De
(JEE MAIN-2025)
Solution
$ e ^{5(\operatorname{lnx})^2+3}= x ^8$
$\Rightarrow \ell \ln e ^{5(\operatorname{lnx})^2+3}=\ell n x ^8$
$\Rightarrow 5(\ln x )^2+3=8 \ln x$
$(\ell \ln x = t )$
$\Rightarrow 5 t ^2-8 t +3=0$
$ t _1+ t _2=\frac{8}{5}$
$ \ln x _1 x _2=\frac{8}{5}$
$ x _1 x _2= e ^{8 / 5}$
$\Rightarrow \ell \ln e ^{5(\operatorname{lnx})^2+3}=\ell n x ^8$
$\Rightarrow 5(\ln x )^2+3=8 \ln x$
$(\ell \ln x = t )$
$\Rightarrow 5 t ^2-8 t +3=0$
$ t _1+ t _2=\frac{8}{5}$
$ \ln x _1 x _2=\frac{8}{5}$
$ x _1 x _2= e ^{8 / 5}$
Standard 11
Mathematics
