If x, y ,z are in G.P. and `a^x = b^y = c^z` , then :

Options

A. `log_b a = log_c b`
B. `log_b a = log_a c`
C. `log_c b = log_a c `
D. none of these

Answer

To solve the problem, we need to show that if \( x, y, z \) are in geometric progression (G.P.) and \( a^x = b^y = c^z \), then it follows that \( y^2 = xz \).

### Step-by-Step Solution:

1. **Understanding the Conditions**:
Since \( x, y, z \) are in G.P., we know that:
\[ y^2 = xz \]

2. **Using the Given Equality**:
We have the equality:
\[ a^x = b^y = c^z \]
Let's denote this common value as \( k \). Thus, we can write:
\[ a^x = k, \quad b^y = k, \quad c^z = k \]

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

4. **Expressing x, y, z in terms of k**:
From the logarithmic equations, we can express \( x, y, z \) as:
\[ x = \frac{\log k}{\log a}, \quad y = \frac{\log k}{\log b}, \quad z = \frac{\log k}{\log c} \]

5. **Substituting into the G.P. Condition**:
Now substituting these values into the G.P. condition \( y^2 = xz \):
\[ \left(\frac{\log k}{\log b}\right)^2 = \left(\frac{\log k}{\log a}\right) \left(\frac{\log k}{\log c}\right) \]

6. **Simplifying the Equation**:
This simplifies to:
\[ \frac{(\log k)^2}{(\log b)^2} = \frac{(\log k)^2}{\log a \cdot \log c} \]
Assuming \( \log k \neq 0 \), we can cancel \( (\log k)^2 \) from both sides:
\[ \frac{1}{(\log b)^2} = \frac{1}{\log a \cdot \log c} \]

7. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ \log a \cdot \log c = (\log b)^2 \]

8. **Rearranging the Result**:
This can be rearranged to show that:
\[ \frac{\log a}{\log b} = \frac{\log b}{\log c} \]
Which implies:
\[ \log_a b = \log_b c \]

### Conclusion:
Thus, we have shown that if \( x, y, z \) are in G.P. and \( a^x = b^y = c^z \), then indeed \( y^2 = xz \).