Two lines `5x+3^alphay=1` and `alphax-y=0` are perpendicular to each other for some value of `alpha` lying in the interval:

Options

A. `(-1,0)`
B. `(1,2)`
C. `(2,3)`
D. `(3,4)`

Answer

To determine the values of \(\alpha\) for which the lines \(5x + 3^{\alpha}y = 1\) and \(\alpha x - y = 0\) are perpendicular, we can follow these steps:

### Step 1: Find the slopes of the lines

1. **Convert the first line to slope-intercept form**:
\[ 5x + 3^{\alpha}y = 1 \]
Rearranging gives:
\[ 3^{\alpha}y = 1 - 5x \]
Dividing by \(3^{\alpha}\):
\[ y = -\frac{5}{3^{\alpha}}x + \frac{1}{3^{\alpha}} \]
Thus, the slope \(m_1\) of the first line is:
\[ m_1 = -\frac{5}{3^{\alpha}} \]

2. **Convert the second line to slope-intercept form**:
\[ \alpha x - y = 0 \]
Rearranging gives:
\[ y = \alpha x \]
Thus, the slope \(m_2\) of the second line is:
\[ m_2 = \alpha \]

### Step 2: Use the condition for perpendicular lines

For two lines to be perpendicular, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Substituting the slopes we found:
\[ -\frac{5}{3^{\alpha}} \cdot \alpha = -1 \]
This simplifies to:
\[ \frac{5\alpha}{3^{\alpha}} = 1 \]

### Step 3: Rearranging the equation

Rearranging gives:
\[ 5\alpha = 3^{\alpha} \]

### Step 4: Analyze the function

We need to find the values of \(\alpha\) that satisfy the equation \(5\alpha = 3^{\alpha}\). To do this, we can analyze the functions \(y = 5\alpha\) and \(y = 3^{\alpha}\).

1. **Graph the functions**:
- The line \(y = 5\alpha\) is a straight line through the origin with a slope of 5.
- The function \(y = 3^{\alpha}\) is an exponential function that increases rapidly.

2. **Finding intersections**:
- At \(\alpha = 0\), \(5\alpha = 0\) and \(3^{\alpha} = 1\).
- At \(\alpha = 1\), \(5\alpha = 5\) and \(3^{\alpha} = 3\).
- At \(\alpha = 2\), \(5\alpha = 10\) and \(3^{\alpha} = 9\).
- At \(\alpha = 3\), \(5\alpha = 15\) and \(3^{\alpha} = 27\).

### Step 5: Determine the intervals

From the values calculated:
- The line \(y = 5\alpha\) intersects \(y = 3^{\alpha}\) between \(\alpha = 2\) and \(\alpha = 3\) since:
- At \(\alpha = 2\), \(5\alpha < 3^{\alpha}\)
- At \(\alpha = 3\), \(5\alpha > 3^{\alpha}\)

Thus, the value of \(\alpha\) for which the lines are perpendicular lies in the interval \((2, 3)\).

### Final Answer
The value of \(\alpha\) lies in the interval:
\[ \boxed{(2, 3)} \]