If the sum of all the roots of the equation e ^ 2x - 11 e ^ x - 45...

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

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

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

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 $P = 45$