It's All Logs

Using the symmetry about the $y$ axis we have:

$f(x) = x^{2n}$ and $g(x) = \log_{b}(x) = \dfrac{\ln(x)}{\ln(b)} \implies f'(a) = 2na^{2n - 1} = g'(a) = \dfrac{1}{a\ln(b)} \implies a^{2n} = \dfrac{1}{2n\ln(b)} \implies a = (\dfrac{1}{2n\ln(b)})^{\frac{1}{2n}}$

and

$a^{2n} = \dfrac{\ln(a)}{\ln(b)} \implies \dfrac{1}{2n\ln(b)} = \dfrac{\ln(\dfrac{1}{2n\ln(b)})}{2n\ln(b)}$

$\implies 1 = \ln(\dfrac{1}{2n\ln(b)}) \implies$ $2n\ln(b) = \dfrac{1}{e} \implies \ln(b) = \dfrac{1}{2ne} \implies b = e^{\frac{1}{2ne}} \implies a = e^{\frac{1}{2n}}$ and $a^{2n} = e$ .

Using $A(e^{\frac{1}{2n}}, e) \implies \boxed{y - e = 2ne^{\frac{2n - 1}{2n}}(x - e^{\frac{1}{2n}})}$

and using symmetry about the $y$ axis we have $A'(-e^{\frac{1}{2n}}, e) \implies \boxed{y - e = -2ne^{\frac{2n - 1}{2n}}(x + e^{\frac{1}{2n}})}$

Solving the two equations above we obtain $B(0,(1 - 2n)e)$

Using points $A, A', B$ and $C(0,e)$ we have $AA' = 2e^{\frac{1}{2n}}$ and $BC = 2ne \implies A_{\triangle{AA'B}} = 2ne^{\frac{2n + 1}{2n}}$

and

$A_{p} = \displaystyle\int_{-e^{\frac{1}{2n}}}^{e^{\frac{1}{2n}}} x^{2n} = \dfrac{2e^{\frac{2n + 1}{2n}}}{2n + 1}$

$\implies \dfrac{A_{\triangle{AA'B}}}{A_{p}} = n(2n + 1) = 2n^2 + n = n + 18 \implies 2(n - 3)(n + 3) = 0$ and $n \neq -3 \implies n = \boxed{3}$ .