15 The functions $f_{0}, f_{1}, f_{2}, \cdots$ are recursively defined as follows:
(1) $f_{0}(x)=x, x \in \mathbf{R}$;
(2) $f_{2 k+1}(x)=3^{f_{k}(x)}, x \in \mathbf{R}, k=0,1,2, \cdots$;
(3) $f_{2 k}(x)=2^{f_{2 k-1}(x)}, x \in \mathbf{R}, k=0,1,2, \cdots$.
Which is larger, $f_{10}(1)$ or $f_{9}(2)$? Prove your conclusion.
15 It is easy to calculate $f_{0}(1)=1, f_{1}(1)=3, f_{2}(1)=2^{3}=8, f_{3}(1)=3^{8}=6561 ; f_{0}(2)$ $=2, f_{1}(2)=3^{2}=9, f_{2}(2)=2^{9}=512$. Clearly, $f_{3}(1)>2 f_{2}(2), 2^{f_{3}(1)}>$ $2^{2 f_{2}(2)}=4^{f_{2}(2)}>3^{f_{2}(2)}$, so $f_{4}(1)>f_{3}(2)$. Suppose for some $k \geqslant 1$ we have
$$
\begin{array}{l}
f_{2 k+1}(1)>2 f_{2 k}(2),(k \geqslant 1) \\
f_{2 k}(1)>f_{2 k-1}(2),(k \geqslant 1)
\end{array}
$$

Then
$$
\begin{aligned}
f_{2 k+2}(1) & =2^{f_{2 k+1}(1)}>2^{2 f_{2 k}(2)}>3^{f_{2 k}(2)}=f_{2 k+1}(2), \\
f_{2 k+3}(1) & =3^{f_{2 k+2}(1)}>3^{f_{2 k+1}(2)}=3^{2} \cdot 3^{f_{2 k+1}(2)-2} \\
& >2^{3} \cdot 2^{f_{2 k+1}(2)^{-2}}=2 \cdot 2^{f_{2 k+1}(2)}=2 f_{2 k+2}(2) .
\end{aligned}
$$

Thus (1) holds for all natural numbers $k$. In particular, $f_{10}(1)>f_{9}(2)$.