If `2^(x)=4^(y)=8^(z)` and `((1)/(2x)+(1)/(4y)+(1)/(6z))=(24)/(7)`, then find the value of z.

Answer

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

1. **Given Equations**:
\[ 2^x = 4^y = 8^z \]
We can express \(4\) and \(8\) in terms of base \(2\):
\[ 4 = 2^2 \quad \text{and} \quad 8 = 2^3 \]
Therefore, we can rewrite the equations as:
\[ 2^x = (2^2)^y = 2^{2y} \quad \text{and} \quad 2^x = (2^3)^z = 2^{3z} \]

2. **Equating Exponents**:
Since the bases are the same, we can equate the exponents:
\[ x = 2y \quad \text{(1)} \]
\[ x = 3z \quad \text{(2)} \]

3. **Expressing \(y\) and \(x\) in terms of \(z\)**:
From equation (2), we can express \(x\) in terms of \(z\):
\[ x = 3z \]
Substituting \(x\) from equation (2) into equation (1):
\[ 3z = 2y \implies y = \frac{3z}{2} \quad \text{(3)} \]

4. **Substituting into the given condition**:
Now we substitute \(x\) and \(y\) into the condition:
\[ \frac{1}{2x} + \frac{1}{4y} + \frac{1}{6z} = \frac{24}{7} \]
Substituting \(x = 3z\) and \(y = \frac{3z}{2}\):
\[ \frac{1}{2(3z)} + \frac{1}{4\left(\frac{3z}{2}\right)} + \frac{1}{6z} = \frac{24}{7} \]

5. **Simplifying each term**:
- The first term:
\[ \frac{1}{2(3z)} = \frac{1}{6z} \]
- The second term:
\[ \frac{1}{4\left(\frac{3z}{2}\right)} = \frac{1}{6z} \]
- The third term remains:
\[ \frac{1}{6z} \]

Now we can combine these:
\[ \frac{1}{6z} + \frac{1}{6z} + \frac{1}{6z} = \frac{3}{6z} = \frac{1}{2z} \]

6. **Setting up the equation**:
Now we have:
\[ \frac{1}{2z} = \frac{24}{7} \]

7. **Cross-multiplying to solve for \(z\)**:
Cross-multiplying gives:
\[ 7 = 48z \implies z = \frac{7}{48} \]

Thus, the value of \(z\) is:
\[ \boxed{\frac{7}{48}} \]