If the two curves `y = a^x` and `y =b^x` intersect at an angle `alpha`, then tan `alpha` equals

Options

A. ` (loga-logb)/(1+logalogb)`
B. `(loga+logb)/(1-logalogb)`
C. `(loga-logb)/(1-loga-logb)`
D. none of these

Answer

To solve the problem of finding the angle of intersection \( \alpha \) between the curves \( y = a^x \) and \( y = b^x \), we follow these steps:

### Step 1: Find the Intersection Point
To find the intersection point of the two curves, we set them equal to each other:

\[ a^x = b^x \]

Taking the logarithm of both sides gives:

\[ \log(a^x) = \log(b^x) \]

This simplifies to:

\[ x \log a = x \log b \]

Rearranging this, we have:

\[ x (\log a - \log b) = 0 \]

This implies either \( x = 0 \) or \( \log a - \log b = 0 \). Since we are looking for the intersection point, we take \( x = 0 \).

### Step 2: Calculate the y-coordinate at the Intersection Point
Substituting \( x = 0 \) into either curve, we find:

\[ y = a^0 = 1 \quad \text{or} \quad y = b^0 = 1 \]

Thus, the intersection point is \( P(0, 1) \).

### Step 3: Find the Slopes of the Tangents at the Intersection Point
Next, we need to find the derivatives of both functions to determine the slopes of the tangents at the intersection point.

For the first curve \( y = a^x \):

\[ \frac{dy}{dx} = a^x \log a \]

At the point \( P(0, 1) \):

\[ M_1 = a^0 \log a = \log a \]

For the second curve \( y = b^x \):

\[ \frac{dy}{dx} = b^x \log b \]

At the point \( P(0, 1) \):

\[ M_2 = b^0 \log b = \log b \]

### Step 4: Calculate the Angle of Intersection
The formula for the tangent of the angle \( \alpha \) between two curves is given by:

\[ \tan \alpha = \frac{|M_1 - M_2|}{1 + M_1 M_2} \]

Substituting \( M_1 \) and \( M_2 \):

\[ \tan \alpha = \frac{|\log a - \log b|}{1 + \log a \cdot \log b} \]

### Final Result
Thus, we conclude that:

\[ \tan \alpha = \frac{|\log a - \log b|}{1 + \log a \cdot \log b} \]