If the curves $y=a^x$ and $y=b^x$ intersect at an angle α, then tanα is equal to

Question

If the curves $y=a^x$ and $y=b^x$ intersect at an angle $\alpha$, then $\tan \alpha$ is equal to

(A) $\frac{a-b}{1+a b}$

(B) $\frac{\log a-\log b}{1+\log a \log b}$

(C) $\frac{a+b}{1-a b}$

(D) $\frac{\log a+\log b}{1-\log a \log b}$

Answer 1

SOLUTION — Point of intersection of curves is $(0,1)$.

Now, slope of tangent of the curve $y=a^x$ is

$m_1=\frac{d y}{d x}=a^x \log a$

$\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(0,1)}=m_1=\log a$

Slope of tangent of the curve $y=b^x$ is

$\begin{aligned}m_2 & =\frac{d y}{d x}=b^x \log b \\\Rightarrow \quad m_2 & =\left(\frac{d y}{d x}\right)_{(0,1)}=\log b\end{aligned}$

Angle between two intersecting curve is given by

$\tan \alpha=\frac{m_1-m_2}{1+m_1 m_2}=\frac{\log a-\log b}{1+\log a \log b}$

So, The correct option is (B).