57 Let $f(x)=2^{x}, g(x)=\log _{\sqrt{2}} 8 x$, then the value of $x$ that satisfies $f[g(x)]=g[f(x)]$ is

Let's translate the problem and solution step by step.

### Problem:
Let \( f(x) = 2^x \) and \( g(x) = \log_{\sqrt{2}} (8x) \). Find the value of \( x \) that satisfies \( f[g(x)] = g[f(x)] \).

### Solution:
1. **Express \( f[g(x)] \):**
   \[
   f[g(x)] = f\left( \log_{\sqrt{2}} (8x) \right)
   \]
   Since \( f(x) = 2^x \), we have:
   \[
   f\left( \log_{\sqrt{2}} (8x) \right) = 2^{\log_{\sqrt{2}} (8x)}
   \]

2. **Simplify \( 2^{\log_{\sqrt{2}} (8x)} \):**
   Recall that \( \log_{\sqrt{2}} (8x) = \frac{\log_2 (8x)}{\log_2 (\sqrt{2})} \). Since \( \log_2 (\sqrt{2}) = \frac{1}{2} \), we get:
   \[
   \log_{\sqrt{2}} (8x) = 2 \log_2 (8x)
   \]
   Therefore:
   \[
   2^{\log_{\sqrt{2}} (8x)} = 2^{2 \log_2 (8x)} = (2^{\log_2 (8x)})^2 = (8x)^2 = 64x^2
   \]

3. **Express \( g[f(x)] \):**
   \[
   g[f(x)] = g(2^x)
   \]
   Since \( g(x) = \log_{\sqrt{2}} (8x) \), we have:
   \[
   g(2^x) = \log_{\sqrt{2}} (8 \cdot 2^x) = \log_{\sqrt{2}} (8 \cdot 2^x)
   \]

4. **Simplify \( \log_{\sqrt{2}} (8 \cdot 2^x) \):**
   \[
   \log_{\sqrt{2}} (8 \cdot 2^x) = \log_{\sqrt{2}} (2^3 \cdot 2^x) = \log_{\sqrt{2}} (2^{3+x})
   \]
   Using the change of base formula:
   \[
   \log_{\sqrt{2}} (2^{3+x}) = \frac{\log_2 (2^{3+x})}{\log_2 (\sqrt{2})} = \frac{3 + x}{\frac{1}{2}} = 2(3 + x) = 6 + 2x
   \]

5. **Set \( f[g(x)] \) equal to \( g[f(x)] \):**
   \[
   64x^2 = 6 + 2x
   \]

6. **Solve the equation \( 64x^2 = 6 + 2x \):**
   \[
   64x^2 - 2x - 6 = 0
   \]
   This is a quadratic equation. We can solve it using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 64 \), \( b = -2 \), and \( c = -6 \):
   \[
   x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 64 \cdot (-6)}}{2 \cdot 64} = \frac{2 \pm \sqrt{4 + 1536}}{128} = \frac{2 \pm \sqrt{1540}}{128} = \frac{2 \pm 2\sqrt{385}}{128} = \frac{1 \pm \sqrt{385}}{64}
   \]

7. **Check the solutions:**
   \[
   x = \frac{1 + \sqrt{385}}{64} \quad \text{and} \quad x = \frac{1 - \sqrt{385}}{64}
   \]
   Since \( x \) must be positive, we discard \( x = \frac{1 - \sqrt{385}}{64} \).

Thus, the value of \( x \) that satisfies the equation is:
\[
\boxed{\frac{1 + \sqrt{385}}{64}}
\]
$57 \quad x=\frac{1+\sqrt{385}}{64} . \quad f[g(x)]=2 \log _{\sqrt{2}}^{(8 x)}=64 x^{2}, g[f(x)]=\log _{\sqrt{2}}^{(8 \cdot 2 x)}=2(x+3)$. From $64 x^{2}=2(x+3)$ we get $x=\frac{1 \pm \sqrt{385}}{64}$. Since $x>0$, we have $x=\frac{1+\sqrt{385}}{64}$.