If x , y, z are in G.P. and `a^x = b^y = c^z` , then `log_b a* log_b c ` is equal to :

Options

A. 0
B. 1
C. ac
D. none of these

Answer

To solve the problem, we need to analyze the given conditions and derive the required expression step by step.

### Step-by-Step Solution:

1. **Understand the Given Conditions**: We are given that \( x, y, z \) are in Geometric Progression (G.P.) and that \( a^x = b^y = c^z \).

2. **Express the Equalities Using Logarithms**: Since \( a^x = b^y = c^z \), we can set this common value equal to some constant \( k \). Thus, we can write:
\[ a^x = k, \quad b^y = k, \quad c^z = k \]

3. **Take Logarithms**: Taking logarithms on both sides gives us:
\[ x \log a = \log k, \quad y \log b = \log k, \quad z \log c = \log k \]

4. **Express \( x, y, z \) in terms of \( k \)**:
\[ x = \frac{\log k}{\log a}, \quad y = \frac{\log k}{\log b}, \quad z = \frac{\log k}{\log c} \]

5. **Use the G.P. Condition**: Since \( x, y, z \) are in G.P., we have the condition:
\[ y^2 = xz \]

6. **Substitute the Values**: Substituting the expressions for \( x, y, z \):
\[ \left(\frac{\log k}{\log b}\right)^2 = \left(\frac{\log k}{\log a}\right) \left(\frac{\log k}{\log c}\right) \]

7. **Simplify the Equation**: This gives us:
\[ \frac{(\log k)^2}{(\log b)^2} = \frac{(\log k)^2}{\log a \cdot \log c} \]

8. **Cancel \( (\log k)^2 \)** (assuming \( k \neq 1 \)):
\[ \frac{1}{(\log b)^2} = \frac{1}{\log a \cdot \log c} \]

9. **Cross Multiply**:
\[ \log a \cdot \log c = (\log b)^2 \]

10. **Rearranging the Expression**: We need to find \( \log_b a \cdot \log_b c \):
\[ \log_b a = \frac{\log a}{\log b}, \quad \log_b c = \frac{\log c}{\log b} \]
Therefore,
\[ \log_b a \cdot \log_b c = \left(\frac{\log a}{\log b}\right) \cdot \left(\frac{\log c}{\log b}\right) = \frac{\log a \cdot \log c}{(\log b)^2} \]

11. **Substituting the Value**: From our earlier result, we know that \( \log a \cdot \log c = (\log b)^2 \):
\[ \log_b a \cdot \log_b c = \frac{(\log b)^2}{(\log b)^2} = 1 \]

### Final Answer:
\[ \log_b a \cdot \log_b c = 1 \]