$4 \cdot 70$ The sequence of functions $\left\{f_{n}(x)\right\}$ is recursively defined by:
$$
\left\{\begin{array}{l}
f_{1}(x)=\sqrt{x^{2}+48} \\
f_{n+1}(x)=\sqrt{x^{2}+6 f_{n}(x)}
\end{array} \text { for each } n \geqslant 1 .\right.
$$

Find all real solutions to the equation $f_{n}(x)=2 x$.
［Solution］From the given information, we have
$$
f_{n}(x)>0 \quad(n=1,2, \cdots),
$$

Therefore, the equation $f_{n}(x)=2 x$ has only positive solutions.
First, we use mathematical induction to prove that $x=4$ is a solution to the equation
$$
f_{n}(x)=2 x
$$

In fact, when $n=1$, substituting $x=4$ into the left side of (1) yields $f_{1}(4)=$ $\sqrt{4^{2}+48}=8$; substituting into the right side of (1) yields $2 \times 4=8$. Therefore, $x=4$ is a solution to $f_{1}(x)=2 x$.
Assume that $x=4$ is a solution to $f_{k}(x)=2 x$. That is, $f_{k}(4)=8$.
Then
$$
f_{k+1}(4)=\sqrt{4^{2}+6 f_{k}(4)}=\sqrt{4^{2}+48}=8=2 \times 4 .
$$

Therefore, $x=4$ is also a solution to $f_{k+1}(x)=2 x$.
By the principle of mathematical induction, for any natural number $n$, $x=4$ is a solution to equation (1).
Next, we use mathematical induction to prove that the function $\frac{f_{n}(x)}{x}$ is a monotonically decreasing function on $(0,+\infty)$.
In fact, when $n=1$,
$$
\frac{f_{1}(x)}{x}=\sqrt{1+\frac{48}{x^{2}}}
$$

It is clearly a monotonically decreasing function on $(0,+\infty)$.
Assume that $\frac{f_{k}(x)}{x}$ is a monotonically decreasing function on $(0,+\infty)$. Then
$$
\frac{f_{k+1}(x)}{x}=\sqrt{1+\frac{6}{x} \cdot \frac{f_{k}(x)}{x}}
$$

is also a monotonically decreasing function on $(0,+\infty)$.
By the principle of mathematical induction, for any natural number $n$, the function $\frac{f_{n}(x)}{x}$ is a monotonically decreasing function on $(0,+\infty)$. Therefore, $\frac{f_{n}(x)}{x}=2$ has a unique solution $x=4$ on $(0,+\infty)$.