To solve the problem step by step, we start with the given equations and relationships.
### Step 1: Set the common value
We have the equations:
\[ 2^x = 4^y = 8^z \]
We can express \(4\) and \(8\) in terms of base \(2\):
\[ 4 = 2^2 \quad \text{and} \quad 8 = 2^3 \]
Thus, we can rewrite the equations as:
\[ 2^x = 2^{2y} = 2^{3z} \]
Let’s set all of them equal to a common variable \(k\):
\[ 2^x = k, \quad 4^y = k, \quad 8^z = k \]
### Step 2: Express \(x\), \(y\), and \(z\) in terms of \(k\)
From \(2^x = k\), we have:
\[ x = \log_2(k) \]
From \(4^y = k\), we have:
\[ (2^2)^y = k \implies 2^{2y} = k \implies 2y = \log_2(k) \implies y = \frac{1}{2} \log_2(k) \]
From \(8^z = k\), we have:
\[ (2^3)^z = k \implies 2^{3z} = k \implies 3z = \log_2(k) \implies z = \frac{1}{3} \log_2(k) \]
### Step 3: Substitute \(x\), \(y\), and \(z\) into the product equation
We are given that:
\[ xyz = 288 \]
Substituting the expressions for \(x\), \(y\), and \(z\):
\[
\log_2(k) \cdot \frac{1}{2} \log_2(k) \cdot \frac{1}{3} \log_2(k) = 288
\]
This simplifies to:
\[
\frac{1}{6} (\log_2(k))^3 = 288
\]
Multiplying both sides by \(6\):
\[
(\log_2(k))^3 = 1728
\]
Taking the cube root:
\[
\log_2(k) = 12 \quad \Rightarrow \quad k = 2^{12} = 4096
\]
### Step 4: Find \(x\), \(y\), and \(z\)
Now substituting back to find \(x\), \(y\), and \(z\):
\[
x = \log_2(4096) = 12
\]
\[
y = \frac{1}{2} \log_2(4096) = \frac{12}{2} = 6
\]
\[
z = \frac{1}{3} \log_2(4096) = \frac{12}{3} = 4
\]
### Step 5: Calculate \( \frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} \)
Now we substitute \(x\), \(y\), and \(z\) into the expression:
\[
\frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = \frac{1}{2 \cdot 12} + \frac{1}{4 \cdot 6} + \frac{1}{8 \cdot 4}
\]
Calculating each term:
\[
\frac{1}{24} + \frac{1}{24} + \frac{1}{32}
\]
Finding a common denominator, which is \(96\):
\[
\frac{1}{24} = \frac{4}{96}, \quad \frac{1}{24} = \frac{4}{96}, \quad \frac{1}{32} = \frac{3}{96}
\]
Adding these fractions:
\[
\frac{4 + 4 + 3}{96} = \frac{11}{96}
\]
### Final Answer
Thus, the value of \( \frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} \) is:
\[
\frac{11}{96}
\]
