Key Concepts and Formulas

• Substitution for Exponential Equations: Simplifying exponential equations by substituting a variable for the exponential term (e.g., $y = e^x$).
• Properties of Exponents and Logarithms: $e^A = B \iff A = \log_e B$ (for $B>0$) and $e^a \cdot e^b = e^{a+b}$.
• Real Roots Constraint for $e^x$: For $x$ to be real, $e^x > 0$.

Step-by-Step Solution

1. Transforming the Exponential Equation into a Polynomial

• Why? The goal is to eliminate the negative exponent and fraction, transforming the equation into a standard polynomial form by multiplying by $2e^x$.
• Given equation: $e^{2x} - 11e^x - 45e^{-x} + \frac{81}{2} = 0$
• Multiply by $2e^x$: $2e^x(e^{2x}) - 2e^x(11e^x) - 2e^x(45e^{-x}) + 2e^x(\frac{81}{2}) = 0$ $2e^{3x} - 22e^{2x} - 90 + 81e^x = 0$ $2e^{3x} - 22e^{2x} + 81e^x - 90 = 0$
• Why? To simplify the equation further, we substitute $y = e^x$. Note that since we are looking for real roots $x$, we must have $y > 0$.
• Let $y = e^x$. Substituting $y$ into the equation: $2(e^x)^3 - 22(e^x)^2 + 81(e^x) - 90 = 0$ $2y^3 - 22y^2 + 81y - 90 = 0$

2. Finding the Real Roots of the Polynomial in $y$

• Why? We need to find the positive real roots of the cubic polynomial in $y$ because only these correspond to real values of $x$.
• Let $P(y) = 2y^3 - 22y^2 + 81y - 90$. We look for rational roots by testing factors of 90 divided by factors of 2.
• Test $y = 2$: $P(2) = 2(2)^3 - 22(2)^2 + 81(2) - 90 = 16 - 88 + 162 - 90 = 0$
• Since $P(2) = 0$, $y = 2$ is a root.
• Why? Knowing one root allows us to factor the cubic polynomial.
• Divide $2y^3 - 22y^2 + 81y - 90$ by $(y - 2)$: $(y - 2)(2y^2 - 18y + 45) = 0$
• Now, we need to find the roots of the quadratic $2y^2 - 18y + 45 = 0$.
• Calculate the discriminant $\Delta = b^2 - 4ac$: $\Delta = (-18)^2 - 4(2)(45) = 324 - 360 = -36$
• Why? A negative discriminant indicates complex conjugate roots.
• Since $\Delta < 0$, the quadratic has complex roots. These do not yield real values for $x$.
• Therefore, $y = 2$ is the only positive real root for the polynomial in $y$.

3. Finding the Real Root(s) of the Original Equation in $x$

• Why? Convert the real root of $y$ back to an $x$ root using $y = e^x$.
• Using $y = e^x$ and $y = 2$: $e^x = 2$
• Why? Solve for $x$ by taking the natural logarithm.
• $x = \log_e 2$
• This is the only real root of the original equation.

4. Calculating the Sum of Roots and Finding $p$

• Why? The problem asks for the sum of all roots, implying real roots in this context.
• The sum of all real roots is $x = \log_e 2$.
• The problem states the sum of all roots is $\log_e p$.
• Comparing: $\log_e p = \log_e 2$
• Therefore: $p = 2$

Common Mistakes & Tips

• Distinguish Real vs. Complex Roots: Always remember the domain constraints when using substitutions like $y = e^x$. A real $x$ requires $y > 0$.
• Careful Algebra: Pay close attention to signs and exponents during algebraic manipulations to avoid errors.

Summary

By substituting $y = e^x$, we transformed the given exponential equation into a cubic polynomial. Recognizing that $y$ must be a positive real number for $x$ to be real, we found only one positive real root for $y$, which led to a single real root for $x$. The sum of the real roots allowed us to determine the value of $p$.

The final answer is $\boxed{2}$.