Question

If the two curves \[ y=a^x \] and \[ y=b^x \] intersect at angle \(\alpha\), then \[ \tan\alpha= \] is:

• A. \(\dfrac{\log a-\log b}{1+\log a\log b}\)
• B. \(\dfrac{\log a+\log b}{1-\log a\log b}\)
• C. \(\dfrac{\pi}{4}\)
• D. \(\dfrac{\pi}{2}\)

Hint:

The angle between two curves at their point of intersection is the angle between their tangents at that point.

Answer 1

The Correct Option is A

Step 1: Find the point of intersection.
The two curves are \[ y=a^x \] and \[ y=b^x \] At the point of intersection, \[ a^x=b^x \] This is always true at \[ x=0 \] because \[ a^0=b^0=1 \] Thus, the curves intersect at \[ (0,1) \]

Step 2: Find the slope of \(y=a^x\).
For \[ y=a^x, \] differentiate with respect to \(x\): \[ \frac{dy}{dx}=a^x\log a \] At \[ x=0, \] we get \[ m_1=a^0\log a \] \[ m_1=\log a \]

Step 3: Find the slope of \(y=b^x\).
For \[ y=b^x, \] differentiate with respect to \(x\): \[ \frac{dy}{dx}=b^x\log b \] At \[ x=0, \] we get \[ m_2=b^0\log b \] \[ m_2=\log b \]

Step 4: Use the formula for angle between two curves.
The angle between two curves is the angle between their tangents at the point of intersection.
If the slopes of two tangents are \(m_1\) and \(m_2\), then \[ \tan\alpha=\frac{m_1-m_2}{1+m_1m_2} \] Substitute \[ m_1=\log a,\quad m_2=\log b \] Therefore, \[ \tan\alpha= \frac{\log a-\log b}{1+\log a\log b} \]

Step 5: Final conclusion.
Hence, \[ \boxed{\frac{\log a-\log b}{1+\log a\log b}} \]