if `log_kx.log_5k = log_x5` , `k!=1 , k>0` , then find the value of x

Options

A. 5
B. `(24)/(5)`
C. `(26)/(5)`
D. `(37)/(5)`

Answer

To solve the equation \( \log_k x \cdot \log_5 k = \log_x 5 \), we will use the properties of logarithms.

### Step-by-Step Solution:

1. **Rewrite the logarithms using the change of base formula**:
\[ \log_k x = \frac{\log x}{\log k} \quad \text{and} \quad \log_5 k = \frac{\log k}{\log 5} \]
Substituting these into the equation gives:
\[ \frac{\log x}{\log k} \cdot \frac{\log k}{\log 5} = \log_x 5 \]

2. **Simplify the left-hand side**:
The \( \log k \) terms cancel out:
\[ \frac{\log x}{\log 5} = \log_x 5 \]

3. **Rewrite \( \log_x 5 \) using the change of base formula**:
\[ \log_x 5 = \frac{\log 5}{\log x} \]
Now our equation looks like:
\[ \frac{\log x}{\log 5} = \frac{\log 5}{\log x} \]

4. **Cross-multiply**:
\[ (\log x)^2 = (\log 5)^2 \]

5. **Take the square root of both sides**:
\[ \log x = \log 5 \quad \text{or} \quad \log x = -\log 5 \]

6. **Solve for \( x \)**:
- From \( \log x = \log 5 \):
\[ x = 5 \]
- From \( \log x = -\log 5 \):
\[ \log x = \log \frac{1}{5} \quad \Rightarrow \quad x = \frac{1}{5} \]

7. **Conclusion**:
Since \( k > 0 \) and \( k \neq 1 \), both solutions \( x = 5 \) and \( x = \frac{1}{5} \) are valid. However, the context of the problem might suggest that \( x = 5 \) is the primary solution we are looking for.

### Final Answer:
\[ x = 5 \]