If the sum of all the roots of the equation e | vedclass

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Question

$\left.e^{2 x}-11 e^{x}-45 e^{-x}+\frac{81}{2}=0\right]$

$\left.\left(e^{x}\right)^{3}-11\left(e^{x}\right)^{2}-45+\frac{81 e^{x}}{2}=0\right]$

Vedclass · Accepted Answer

$45$

Vedclass · Answer

$\left.e^{2 x}-11 e^{x}-45 e^{-x}+\frac{81}{2}=0\right]$

$\left.\left(e^{x}\right)^{3}-11\left(e^{x}\right)^{2}-45+\frac{81 e^{x}}{2}=0\right]$

$e^{x}=t$

$2 t^{3}-22 t^{2}+81 t-90=0$

$t_{1} t_{2} t_{3}=45$

$e^{x_{1}} \cdot e^{x_{2}} \cdot e^{x_{3}}=45$

$e^{x_{1}+x_{2}+x_{3}}=45$

$\log _{e} e^{x_{1}+x_{2}+x_{3}}=\log _{e} 45$

$x_{1}+x_{2}+x_{3}=\log _{e} 45$

$\log _{e} P=\log _{e} 45$

$P =45$

If the sum of all the roots of the equation $e^{2 x}-11 e^{x}-45 e^{-x}+\frac{81}{2}=0$ is $\log _{ e } P$, then $p$ is equal to

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