Let `agt0,cgt0,b=sqrtac,a,c` and `acne1,Ngt0`.
Prove that `log_aN/log_cN=(log_aN-log_bN)/(log_bN-log_cN)`

Answer

To prove the equation

\[ \frac{\log_a N}{\log_c N} = \frac{\log_a N - \log_b N}{\log_b N - \log_c N} \]

given that \( b = \sqrt{ac} \), \( a > 0 \), \( c > 0 \), \( ac \neq 1 \), and \( N > 0 \), we will use the change of base formula for logarithms.

### Step-by-step Solution:

1. **Change of Base**: We will apply the change of base formula for logarithms. The change of base formula states that:
\[ \log_x y = \frac{\log_k y}{\log_k x} \]
for any positive \( k \). We will use base \( N \) for our calculations.

Thus, we can express the logarithms as:
\[ \log_a N = \frac{\log N}{\log a}, \quad \log_c N = \frac{\log N}{\log c}, \quad \log_b N = \frac{\log N}{\log b} \]

2. **Substituting into the Left-Hand Side (LHS)**:
\[ \frac{\log_a N}{\log_c N} = \frac{\frac{\log N}{\log a}}{\frac{\log N}{\log c}} = \frac{\log c}{\log a} \]

3. **Substituting into the Right-Hand Side (RHS)**:
\[ \text{RHS} = \frac{\log_a N - \log_b N}{\log_b N - \log_c N} \]
Substituting the expressions from step 1:
\[ = \frac{\frac{\log N}{\log a} - \frac{\log N}{\log b}}{\frac{\log N}{\log b} - \frac{\log N}{\log c}} \]

4. **Simplifying the Numerator**:
\[ \text{Numerator} = \frac{\log N}{\log a} - \frac{\log N}{\log b} = \log N \left( \frac{1}{\log a} - \frac{1}{\log b} \right) = \log N \cdot \frac{\log b - \log a}{\log a \cdot \log b} \]

5. **Simplifying the Denominator**:
\[ \text{Denominator} = \frac{\log N}{\log b} - \frac{\log N}{\log c} = \log N \left( \frac{1}{\log b} - \frac{1}{\log c} \right) = \log N \cdot \frac{\log c - \log b}{\log b \cdot \log c} \]

6. **Putting it Together**:
Now substituting back into the RHS:
\[ \text{RHS} = \frac{\log N \cdot \frac{\log b - \log a}{\log a \cdot \log b}}{\log N \cdot \frac{\log c - \log b}{\log b \cdot \log c}} = \frac{\frac{\log b - \log a}{\log a}}{\frac{\log c - \log b}{\log c}} = \frac{\log b - \log a}{\log c - \log b} \]

7. **Using the Property of \( b \)**:
Since \( b = \sqrt{ac} \), we know:
\[ \log b = \frac{1}{2}(\log a + \log c) \]
This means:
\[ \log b - \log a = \frac{1}{2}(\log a + \log c) - \log a = \frac{1}{2}(\log c - \log a) \]
and
\[ \log c - \log b = \log c - \frac{1}{2}(\log a + \log c) = \frac{1}{2}(\log c - \log a) \]

8. **Final Simplification**:
Thus, we can rewrite the RHS:
\[ \text{RHS} = \frac{\frac{1}{2}(\log c - \log a)}{\frac{1}{2}(\log c - \log a)} = \frac{\log c}{\log a} \]

9. **Conclusion**:
Since both LHS and RHS equal \( \frac{\log c}{\log a} \), we have proved that:
\[ \frac{\log_a N}{\log_c N} = \frac{\log_a N - \log_b N}{\log_b N - \log_c N} \]