The product of all solutions of the equation e 5 (log e x ) 2 +3 = x 8 , x 0 , is :

Mathematics · Logrithms

JEE Main 2025 — 22 January, Morning Shift — Question 9

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

$e^{5(\ln x)^{2}+3}=x^{8}$

$\Rightarrow \ell n \mathrm{e}^{5(\mathrm{fnx})^{2}+3}=\ell n x^{8}$

$\Rightarrow 5(\ln x)^{2}+3=8 \ln x$

$(\ln \mathrm{x}=\mathrm{t})$

$\Rightarrow 5 \mathrm{t}^{2}-8 \mathrm{t}+3=0$

$\mathrm{t}_{1}+\mathrm{t}_{2}=\frac{8}{5}$

$\operatorname{lnx_{1}} \mathrm{X}_{2}=\frac{8}{5}$

$x_{1} x_{2}=e^{8 / 5}$

Answer key and solution verified before publishing.

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