1. Given equation

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

We need the sum of all roots, and it is given as $\ln p$.

2. Substitute

Let

$y = e^x$

Then

$e^{2x} = y^2, \qquad e^{-x} = \frac{1}{y}$

So the equation becomes

$y^2 - 11y - 45\left(\frac{1}{y}\right) + \frac{81}{2} = 0$

Multiply throughout by $y$:

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

Multiply by $2$ to remove fraction:

$2y^3 - 22y^2 + 81y - 90 = 0$

3. Factor the cubic

We test simple rational roots. For $y=2$:

$2(2)^3 - 22(2)^2 + 81(2) - 90 = 16 - 88 + 162 - 90 = 0$

So $(y-2)$ is a factor.

Divide:

$2y^3 - 22y^2 + 81y - 90 = (y-2)(2y^2 - 18y + 45)$

Now factor the quadratic:

$2y^2 - 18y + 45 = 0$

Discriminant:

$D = (-18)^2 - 4(2)(45) = 324 - 360 = -36 < 0$

So the quadratic has no real roots.

Since $y=e^x>0$ and must be real, the only valid real root is

$y=2$

Thus

$e^x = 2 \implies x = \ln 2$

4. Sum of all real roots

There is only one real root, so the sum of all roots is

$\ln 2$

Given that this equals $\ln p$, we get

$\ln p = \ln 2 \implies p=2$

5. Comparison with stored answer

The stored correct answer is $45$, but the derived value is $2$.

The discrepancy likely arises from incorrectly taking product of roots in $y$ or including non-real values of $y$. Since $y=e^x$ must be a positive real number, only $y=2$ is admissible.

Hence the correct integer should be $2$.