If the sum of all the roots of the equation $e^{2 x}-11 e^{x}-45 e^{-x}$ $+81 / 2=0$ is $\log _{e} p

Question

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

Professor PC Thomas Classes · Answer

(45) : $e^{2 x}-11 e^{x}-45 e^{-x}+\frac{81}{2}=0$\ $\Rightarrow \quad e^{3 x}-11 e^{2 x}+\frac{81}{2} e^{x}-45=0$ $$\begin{equation*}\Rightarrow \quad\left(e^{x}ight)^{3}-11\left(e^{x}ight)^{2}+\frac{81}{2} e^{x}-45=0 	ag{i} \end{equation*}$$ $e^{\alpha_{1}}, e^{\alpha_{2}}, e^{\alpha_{3}}$ will be roots of (i)\ Using product of roots, $$ \begin{aligned} & \Rightarrow \quad e^{\alpha_{1}} \cdot e^{\alpha_{2}} \cdot e^{\alpha_{3}}=45 \ & \Rightarrow \quad \alpha_{1}+\alpha_{2}+\alpha_{3}=\ln \left({ }_{1}ight. \ & \Rightarrow \quad p=45 \end{aligned} $$