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

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{e}^{5(\ln x)^2+3}=\mathrm{x}^8 \ & \Rightarrow \ell \mathrm{ne}^{5(\ln x)^2+3}=\ell \mathrm{nn}^8 \ & \Rightarrow 5(\ln \mathrm{x})^2+3=8 \ell \mathrm{nx} \ & (\ell \mathrm{nx}=\mathrm{t}) \ & \Rightarrow 5 \mathrm{t}^2-8 \mathrm{t}+3=0 \ & \quad \mathrm{t}_1+\mathrm{t}_2=\frac{8}{5} \ & \quad \ln \mathrm{x}_1 \mathrm{x}_2=\frac{8}{5} \ & \mathrm{x}_1 \mathrm{x}_2=\mathrm{e}^{8 / 5}\end{aligned}$

The product of all solutions of the equation $\mathrm{e}^{5\left(\log _{\mathrm{e}} x\right)^2+3}=x^8, x \gt…

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