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…

Given $e^{2x}-11e^x-\frac{45}{e^x}+\frac{81}{2}=0$

$e^{3x}-11e^{2x}+\frac{81}{2}{e}^x-45=0$

Let $e^x=t$, so $t^3-11t^2+\frac{81}{2}t-45=0$

Product of roots $t_1{t}_2{t}_3=45$

i.e. $e^{x_1}\times e^{x_2}\times e^{x_3}=45$ 

$e^{x_1+x_2+x_3}=45$

$x_1+x_2+x_3={\log }_{2}45$  

Given sum of root $=\ln \left|P\right|$

So $P=45$