Given the equation \( 2^x = 3^y = 6^z = k \), where \( k \) is some constant, we need to find the value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \).

1. **Express in terms of \( k \):**

   - \( 2^x = k \) implies \( x = \frac{\log k}{\log 2} \)
   - \( 3^y = k \) implies \( y = \frac{\log k}{\log 3} \)
   - \( 6^z = k \) implies \( z = \frac{\log k}{\log 6} \)

2. **Find \( \frac{1}{x} \), \( \frac{1}{y} \), and \( \frac{1}{z} \):**

   - \( \frac{1}{x} = \frac{\log 2}{\log k} \)
   - \( \frac{1}{y} = \frac{\log 3}{\log k} \)
   - \( \frac{1}{z} = \frac{\log 6}{\log k} \)

3. **Sum up \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \):**

   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{\log 2}{\log k} + \frac{\log 3}{\log k} + \frac{\log 6}{\log k}
   \]

   This can be factored as:

   \[
   \frac{\log 2 + \log 3 + \log 6}{\log k}
   \]

   Using the logarithm property \( \log ab = \log a + \log b \) and noting that \( \log 6 = \log (2 \times 3) = \log 2 + \log 3 \):

   \[
   \log 2 + \log 3 + \log 6 = \log 2 + \log 3 + (\log 2 + \log 3) = 2(\log 2 + \log 3)
   \]

4. **Conclusion:**

   Since the numerator simplifies to \( 2(\log 2 + \log 3) \), and the denominator is \( \log k \):

   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{2(\log 2 + \log 3)}{\log k}
   \]

   Given that \( \log k = \log 6^z = z\log 6 = z(\log 2 + \log 3) \), the expression simplifies further:

   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{2(\log 2 + \log 3)}{z(\log 2 + \log 3)} = \frac{2}{z}
   \]

Finally, substitute \( z

Explore how to solve the equation 2^x = 3^y = 6^z and determine the sum 1/x + 1/y + 1/z. Understand the application of logarithms and exponentiation in this advanced mathematical problem.

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