Which of the following equation (s) has (have) only natural solution(s)
(A) `6.9^(1//k)-13.6^(1//k)+6.4^(1//k)=0`
(B) `4x.sqrt(8^(x-1))=4`

Answer

To determine which of the given equations has only natural solutions, we will analyze each equation step by step.

### Given Equations:
(A) \( 6 \cdot 9^{1/k} - 13 \cdot 6^{1/k} + 6 \cdot 4^{1/k} = 0 \)
(B) \( 4x \cdot \sqrt{8^{x-1}} = 4 \)

### Step-by-Step Solution:

#### For Equation (A):
1. **Rewrite the equation**:
We start with the equation:
\[ 6 \cdot 9^{1/k} - 13 \cdot 6^{1/k} + 6 \cdot 4^{1/k} = 0 \]

2. **Substitute \(9\) and \(4\)**:
Rewrite \(9\) as \(3^2\) and \(4\) as \(2^2\):
\[ 6 \cdot (3^2)^{1/k} - 13 \cdot 6^{1/k} + 6 \cdot (2^2)^{1/k} = 0 \]

3. **Simplify the terms**:
This can be rewritten as:
\[ 6 \cdot 3^{2/k} - 13 \cdot 6^{1/k} + 6 \cdot 2^{2/k} = 0 \]

4. **Factor out common terms**:
Factor out \(6\):
\[ 6 \left( 3^{2/k} - \frac{13}{6} \cdot 6^{1/k} + 2^{2/k} \right) = 0 \]

5. **Set the inner expression to zero**:
We need to solve:
\[ 3^{2/k} - \frac{13}{6} \cdot 6^{1/k} + 2^{2/k} = 0 \]

6. **Test natural number values for \(k\)**:
- For \(k = 1\):
\[ 3^2 - \frac{13}{6} \cdot 6 + 2^2 = 9 - 13 + 4 = 0 \quad \text{(valid)} \]
- For \(k = -1\):
\[ 3^{-2} - \frac{13}{6} \cdot 6^{-1} + 2^{-2} \quad \text{(not a natural number)} \]

7. **Conclusion for Equation (A)**:
The only valid natural solution is \(k = 1\).

#### For Equation (B):
1. **Rewrite the equation**:
Start with:
\[ 4x \cdot \sqrt{8^{x-1}} = 4 \]

2. **Simplify the equation**:
Divide both sides by \(4\):
\[ x \cdot \sqrt{8^{x-1}} = 1 \]

3. **Rewrite \(8\)**:
Rewrite \(8\) as \(2^3\):
\[ x \cdot \sqrt{(2^3)^{x-1}} = 1 \]
This simplifies to:
\[ x \cdot 2^{(3(x-1))/2} = 1 \]

4. **Isolate \(x\)**:
Rearranging gives:
\[ x = \frac{1}{2^{(3(x-1))/2}} \]

5. **Test natural number values for \(x\)**:
- For \(x = 1\):
\[ 1 \cdot \sqrt{8^{1-1}} = 1 \quad \text{(valid)} \]
- Check other natural numbers (e.g., \(x = 2\)):
\[ 2 \cdot \sqrt{8^{2-1}} = 2 \cdot \sqrt{8} \quad \text{(not valid)} \]

6. **Conclusion for Equation (B)**:
The only valid natural solution is \(x = 1\).

### Final Conclusion:
Both equations (A) and (B) have only natural solutions.