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  • If the sum of all the roots of the equation <img alt="\mathrm{e}^{2 x}-11 \mathrm{e}^{x}-45 \mathrm{e}^{-x}+\frac{81}{2}=0$ is $\log _{\mathrm{e}} \mathrm{p}" src="https://entrancecorner.oncodecogs

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

Option: 1

45


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The equation can be rewritten as

\mathrm{2 e^{3 x}-22 e^{2 x}+81 e^{x}-90=0}

Let the roots are \mathrm{\alpha ,\beta ,\gamma }

\mathrm{Let \: e^{x}=t, e^{\alpha}=t_{1}, e^{\beta}=t_{2}, e^{\gamma}=t_{3}}\\

\mathrm{2 t^{3}-22 t^{2}+81 t-90=0}

\mathrm{\text { Product of roots }=t, t_{2} t_{3}=e^{\alpha} \cdot e^{\beta} \cdot e^{\gamma}=\frac{90}{2}=45}

\mathrm{\Rightarrow e^{\alpha+\beta+\gamma}=45} \\

\mathrm{\Rightarrow \alpha+\beta+\gamma=\log _{e} 45}

\mathrm{So\: p=45}

Hence the answer is \mathrm{45}

Posted by

Ritika Kankaria

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