The Lyapunov exponent of the fully chaotic logistic map equals ln 2

Step 1 (the conjugacy). Substitute $x=\sin^2(\pi y)$. Using the double-angle identity, $4x(1-x)=4\sin^2(\pi y)\cos^2(\pi y)=\sin^2(2\pi y)$. If we also set $y' = T(y)$ for the doubling map $T(y)=2y \bmod 1$, then $\sin^2(\pi y') = \sin^2(2\pi y)$ matches exactly the right-hand side above, so $f(h(y)) = h(T(y))$ with $h(y)=\sin^2(\pi y)$: the logistic map at r = 4 is smoothly conjugate to the doubling map.

Step 2 (Lyapunov exponent of the doubling map). The doubling map is piecewise linear with $T'(y)=2$ everywhere it is differentiable, so along every orbit $\tfrac1n\sum_{i=0}^{n-1}\ln|T'(y_i)| = \ln 2$ exactly, for every $n$ — the limit is trivially $\ln 2$.

Step 3 (transporting the exponent through the conjugacy). Differentiating $f(h(y))=h(T(y))$ by the chain rule gives $f'(h(y))h'(y)=h'(T(y))T'(y)$, i.e. $\ln|f'(x)| = \ln|h'(T(y))| + \ln|T'(y)| - \ln|h'(y)|$ at $x=h(y)$. Summing this identity along an orbit $y_0,y_1,\dots,y_{n-1}$ and dividing by $n$, the middle terms telescope: $\tfrac1n\sum_{i=0}^{n-1}\ln|f'(x_i)| = \ln2+\tfrac1n\big[\ln|h'(y_n)|-\ln|h'(y_0)|\big]$.

Step 4 (the boundary term vanishes). Since $h'(y)=\pi\sin(2\pi y)$ is bounded (its magnitude never exceeds $\pi$), the bracketed term $\ln|h'(y_n)|-\ln|h'(y_0)|$ stays bounded as $n\to\infty$ for any orbit that avoids the countably many zeros of $h'$ — a set of Lebesgue measure zero. Dividing a bounded quantity by $n$ sends it to $0$, so the Lyapunov exponent of the logistic map at r = 4 equals $\ln 2$ for almost every initial condition.