If `2^(x)=3^(y)=6^(-z)`, then the value of `1/x+1/y+1/z-`

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To solve the equation \(2^x = 3^y = 6^{-z}\), we can set all three expressions equal to a common variable \(k\).

### Step-by-Step Solution:

1. **Set the equations equal to \(k\)**:
\[ 2^x = k, \quad 3^y = k, \quad 6^{-z} = k \]

2. **Express \(x\), \(y\), and \(z\) in terms of \(k\)**:
- From \(2^x = k\):
\[ x = \frac{\log k}{\log 2} \]
- From \(3^y = k\):
\[ y = \frac{\log k}{\log 3} \]
- From \(6^{-z} = k\):
\[ -z = \frac{\log k}{\log 6} \implies z = -\frac{\log k}{\log 6} \]

3. **Substitute \(x\), \(y\), and \(z\) into the expression \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \)**:
\[ \frac{1}{x} = \frac{\log 2}{\log k}, \quad \frac{1}{y} = \frac{\log 3}{\log k}, \quad \frac{1}{z} = -\frac{\log 6}{\log k} \]

4. **Combine the fractions**:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{\log 2}{\log k} + \frac{\log 3}{\log k} - \frac{\log 6}{\log k} \]
\[ = \frac{\log 2 + \log 3 - \log 6}{\log k} \]

5. **Use the property of logarithms**:
\[ \log 6 = \log(2 \cdot 3) = \log 2 + \log 3 \]
Therefore:
\[ \log 2 + \log 3 - \log 6 = \log 2 + \log 3 - (\log 2 + \log 3) = 0 \]

6. **Final result**:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{0}{\log k} = 0 \]

Thus, the value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \) is \(0\).