Animated Solution for Mathematics - Quadratic Equations: If the sum of all the roots of the equation $e^{2x}-11e^x-45e^{-x}+\frac{81}{2}=0$ is ${\log }_{e}P$, then $P$ is equal to ______.

We are presented with the following transcendental equation:

To simplify this expression, we eliminate the negative exponent and the fraction by multiplying the entire equation by $2e^x$. This strategic transformation yields:

By introducing the substitution $t=e^x$, we convert the transcendental equation into a standard cubic polynomial in terms of $t$:

Let the roots of this cubic equation be $t_1,t_2,$ and $t_3$. These roots correspond to the exponential values $e^{x_1},e^{x_2},$ and $e^{x_3}$, where $x_1,x_2,x_3$ are the roots of the original equation.

According to Vieta's formulas, for a cubic equation of the form $a{t}^3+b{t}^2+ct+d=0$, the product of the roots is given by $-\frac{d}{a}$. In our specific case, $a=2$ and $d=-90$.

Calculating the product of the roots:

Since $t_i=e^{x_i}$, we can express the product as:

To isolate the sum of the roots $x_1+x_2+x_3$, we take the natural logarithm of both sides:

Comparing this result to the form ${\log }_{e}P$, we identify the value of $P$. The final result is: $P=45$