If the curves `y= a^(x) and y=e^(x)` intersect at and angle `alpha, " then " tan alpha` equals

Options

A. `|(log_(e)a)/(1+log_(e)a)|`
B. `|(1+log_(e)a)/(1+log_(e)a)|`
C. `|(log_(e)a-1)/(log_(e)a+1)|`
D. none of these

Explanation

To solve the problem of finding \( \tan \alpha \) where the curves \( y = a^x \) and \( y = e^x \) intersect at an angle \( \alpha \), we can follow these steps:

### Step 1: Find the slopes of the curves
We need to differentiate both curves to find their slopes at the point of intersection.

1. For the curve \( y = a^x \):
\[ m_1 = \frac{dy}{dx} = a^x \ln a \]

2. For the curve \( y = e^x \):
\[ m_2 = \frac{dy}{dx} = e^x \]

### Step 2: Use the formula for the tangent of the angle between two curves
The formula for the tangent of the angle \( \alpha \) between two curves is given by:
\[ \tan \alpha = \frac{m_1 - m_2}{1 + m_1 m_2} \]

### Step 3: Substitute the slopes into the formula
Substituting the values of \( m_1 \) and \( m_2 \) into the formula:
\[ \tan \alpha = \frac{a^x \ln a - e^x}{1 + (a^x \ln a)(e^x)} \]

### Step 4: Find the point of intersection
To find the point of intersection, we set the two equations equal to each other:
\[ a^x = e^x \]
Taking the natural logarithm of both sides:
\[ x \ln a = x \]
This implies:
\[ \ln a = 1 \quad \text{or} \quad x = 0 \]
Thus, the curves intersect when \( x = 0 \).

### Step 5: Calculate the y-coordinate at the point of intersection
Substituting \( x = 0 \) into either equation:
\[ y = a^0 = 1 \quad \text{and} \quad y = e^0 = 1 \]
So the point of intersection is \( (0, 1) \).

### Step 6: Substitute \( x = 0 \) into the tangent formula
Now, substituting \( x = 0 \) into the expression for \( \tan \alpha \):
\[ \tan \alpha = \frac{a^0 \ln a - e^0}{1 + (a^0 \ln a)(e^0)} \]
This simplifies to:
\[ \tan \alpha = \frac{\ln a - 1}{1 + \ln a} \]

### Final Result
Thus, we have:
\[ \tan \alpha = \frac{\ln a - 1}{\ln a + 1} \]