If `(log)_k xdot(log)_5k=(log)_x5,k!=1,k >0,` then `x` is equal to `k` (b) 1/5 (c) 5 (d) none of these

Options

A. k
B. `1//5`
C. 5
D. none of these

Explanation

To solve the equation \( \log_k x \cdot \log_5 k = \log_x 5 \), we will follow these steps:

### Step 1: Rewrite the logarithmic expressions
Using the change of base formula, we can rewrite the logarithms:
\[ \log_k x = \frac{\log x}{\log k} \quad \text{and} \quad \log_5 k = \frac{\log k}{\log 5} \]
Substituting these into the equation gives:
\[ \frac{\log x}{\log k} \cdot \frac{\log k}{\log 5} = \log_x 5 \]

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

### Step 3: Rewrite \( \log_x 5 \)
Using the change of base formula again, we can express \( \log_x 5 \) as:
\[ \log_x 5 = \frac{\log 5}{\log x} \]
Thus, we can rewrite our equation as:
\[ \frac{\log x}{\log 5} = \frac{\log 5}{\log x} \]

### Step 4: Cross-multiply
Cross-multiplying gives:
\[ (\log x)^2 = (\log 5)^2 \]

### Step 5: Take the square root
Taking the square root of both sides results in:
\[ \log x = \log 5 \quad \text{or} \quad \log x = -\log 5 \]

### Step 6: Solve for \( x \)
From \( \log x = \log 5 \):
\[ x = 5 \]
From \( \log x = -\log 5 \):
\[ \log x = \log \frac{1}{5} \implies x = \frac{1}{5} \]

### Conclusion
Thus, the possible values for \( x \) are \( 5 \) and \( \frac{1}{5} \).

### Final Answer
The correct options are:
- (b) \( \frac{1}{5} \)
- (c) \( 5 \)