Cramer's Theorem question

As you suggest this is a straightforward application of Cramer's theorem. Recall that Cramer's theorem tells us that $$\lim_{n \to \infty} \frac{1}{n} P(\overline{X_n} > x) = -\Lambda^*(x)$$ where the $X_i$ are iid random variables with logarithmic moment generating function $\Lambda(t)$, and $\Lambda^*(x) = \text{Sup}_{t \in \mathbb{R}}(tx - \Lambda(t))$ is the Legendre transform. Sometimes it is very difficult to compute this, but when $\Lambda(t)$ is a well known smooth function of $t$ the Legendre transform is easy to compute. In this case the generating function $E[e^{tX}] = \frac{1}{2}(e^t + e^{-t})$, whence the logarithmic generating function $\Lambda(t) = Log(\frac{1}{2}(e^t + e^{-t}))$, we use calculus to take the Legendre transform when $x = \frac{1}{2}$: $$ \partial_t\left(\frac{t}{2} -  Log\left(\frac{1}{2}(e^t + e^{-t})\right)\right) = 0$$ if and only if $$ \frac{1}{2} = \frac{e^t - e^{-t}}{e^t + e^{-t}}$$ solving for $t$ we have $$\frac{3}{2} e^{-t} = \frac{1}{2}e^t$$ or $$e^{2t} = 3$$ so finally we see that $t = \frac{ln(3)}{2}$. Now plugging in we see that $$-\Lambda^*(\frac{1}{2}) = -\left(\frac{ln(3)}{4} - ln\left(\frac{\sqrt{3} + \frac{1}{\sqrt{3}}}{2}\right) \right)$$ or, after some manipulation using the properties of the logarithm $$ = ln(2) - \frac{3}{4} ln(3).$$