Even and Odd Powers.

Let $f(x) = x^{2n + 1}$ and $g(x) = \log_{b}(x) = \dfrac{\ln(x)}{\ln(b)}$ for $x > 0$

$\implies f'(a) = (2n + 1)a^{2n} = g'(a) = \dfrac{1}{a\ln(b)} \implies$ $a^{2n + 1} = \dfrac{1}{(2n + 1)\ln(b)} \implies a = (\dfrac{1}{(2n + 1)\ln(b)})^{\frac{1}{2n + 1}}$

and

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

$\large b = e^{\frac{1}{(2n + 1)e}} \implies a = e^{\frac{1}{2n + 1}} \implies a^{2n + 1} = e$

$\implies \large A:(e^{\frac{1}{2n + 1}}, e)$ and using the symmetry about the origin we have $\large A':(-e^{\frac{1}{2n + 1}}, -e)$

$\implies \large m_{AA'} = e^{\frac{2n}{2n + 1}} \implies y = e^{\frac{2n}{2n + 1}}x$

$\large \implies A_{n} = 2\displaystyle\int_{0}^{e^{\frac{1}{2n + 1}}} (e^{\frac{2n}{2n + 1}}x - x^{2n + 1}) \:\ dx = (\dfrac{n}{n + 1})e^{\frac{2n + 2}{2n + 1}}$

$\implies \large L = \lim_{n \rightarrow \infty} A_{n} = e\lim_{n \rightarrow \infty} (1 - \dfrac{1}{n + 1})e^{\frac{1}{2n + 1}} = e$ .

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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 $a^{2n} = e$ .

$\implies \large B(e^{\frac{1}{2n}}, e)$ and using the symmetry about the $y$ axis we have

$\large B'(-e^{\frac{1}{2n}}, e)$ and line $\large BB'$ is $\large y = e \implies$ $\large A^{*}_{n} = \displaystyle 2\int_{0}^{e^{\frac{1}{2n}}} (e - x^{2n}) \:\ dx = 2(\dfrac{2n}{2n + 1})e^{\frac{2n + 1}{2n}}$

$\implies \large L^{*} = \lim_{n \rightarrow \infty} A^{*}_{n} = 2e\lim_{n \rightarrow \infty} (1 - \dfrac{1}{2n + 1})e^{\frac{1}{2n}} = 2e$

$\therefore \dfrac{L^{*}}{L} = \boxed{2}$ .