The product of all solutions of the equation | vedclass

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Question

$ e ^{5(\operatorname{lnx})^2+3}= x ^8$
$\Rightarrow \ell \ln e ^{5(\operatorname{lnx})^2+3}=\ell n x ^8$
$\Rightarrow 5(\ln x )^2+3=8 \ln x$
$(\el

Vedclass · Accepted Answer

$e^{8 / 5}$

Vedclass · Answer

$ 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}$

The product of all solutions of the equation $e ^{5\left(\log _{ e } x \right)^2+3}= x ^8, x >0$, is :

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