Question

If the two curves $y = a^{x}$ and $y = b^{x}$ intersect at $\alpha$, then $\tan\alpha$

Equal

• A. $\frac{\log a - \log b}{1 + \log a\log b}$
• B. $\frac{\log a + \log b}{1 - \log a\log b}$
• C. $\frac{\log a - \log b}{1 - \log a\log b}$
• D. None of these

Answer 1

Answer: (A)

$\frac{\log a - \log b}{1 + \log a\log b}$

Answer 2

Clearly the point of intersection of curves is (0, 1)

Now, slope of tangent of first curve, $m_{1} = \frac{dy}{dx} = a^{x}\log a$

⇒ $\left( \frac{dy}{dx} \right)_{(0,1)} = m_{1} = \log a$

Slope of tangent of second curve, $m_{2} = \frac{dy}{dx} = b^{x}\log b$

⇒ $m_{2} = \left( \frac{dy}{dx} \right)_{(0,1)} = \log b$

$\therefore$ $\tan\alpha = \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} = \frac{\log a - \log b}{1 + \log a\log b}$.