If `2^(x)=4^(y)=8^(z)` and `(1)/(2x)+(1)/(4y)+(1)/(8z)=4` find the value of x.

Answer

To solve the problem, we start with the given equations:

1. \( 2^x = 4^y = 8^z \)
2. \( \frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = 4 \)

### Step 1: Express all terms in terms of base 2

We know that:
- \( 4 = 2^2 \)
- \( 8 = 2^3 \)

Thus, we can rewrite the first equation as:
- \( 2^x = 2^{2y} = 2^{3z} \)

Since the bases are the same, we can equate the exponents:
- \( x = 2y \)
- \( 2y = 3z \)

### Step 2: Express \( y \) and \( z \) in terms of \( x \)

From \( x = 2y \), we can express \( y \) as:
\[ y = \frac{x}{2} \]

From \( 2y = 3z \), substituting \( y \):
\[ 2\left(\frac{x}{2}\right) = 3z \]
This simplifies to:
\[ x = 3z \]
Thus, we can express \( z \) as:
\[ z = \frac{x}{3} \]

### Step 3: Substitute \( y \) and \( z \) into the second equation

Now we substitute \( y \) and \( z \) into the equation:
\[ \frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = 4 \]

Substituting \( y \) and \( z \):
\[ \frac{1}{2x} + \frac{1}{4\left(\frac{x}{2}\right)} + \frac{1}{8\left(\frac{x}{3}\right)} = 4 \]

This simplifies to:
\[ \frac{1}{2x} + \frac{1}{2x} + \frac{3}{8x} = 4 \]

### Step 4: Combine the fractions

Combining the fractions on the left side:
\[ \frac{1}{2x} + \frac{1}{2x} = \frac{2}{2x} = \frac{1}{x} \]
So we have:
\[ \frac{1}{x} + \frac{3}{8x} = 4 \]

Finding a common denominator (which is \( 8x \)):
\[ \frac{8}{8x} + \frac{3}{8x} = 4 \]
This simplifies to:
\[ \frac{11}{8x} = 4 \]

### Step 5: Solve for \( x \)

Now, we can solve for \( x \):
\[ 11 = 32x \]
\[ x = \frac{11}{32} \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{\frac{11}{32}} \]