If the curves $y=a^x$ and $y=e^x$ intersect at an angle $\alpha$, then $\tan \alpha$ equals
Answer & explanation
Correct answer: option 3
The equations of the two curves are
$C_1: y=a^x$ .....(i)
and $C_2: y=e^x$ .....(ii)
At the point of intersection of these two curves, we must have
$a^x=e^x \Rightarrow\left(\frac{a}{e}\right)^x=1 \Rightarrow x=0$
Putting x = 0 in any one of the two curves, we get y = 1
Thus, the two curves intersect at P(0, 1).
Clearly, at the point P(0, 1)
$\left(\frac{d y}{d x}\right)_{C_1}=\log _e a$ and $\left(\frac{d y}{d x}\right)_{C_2}=1$
∴ $\tan \alpha=\left|\frac{\left(\frac{d y}{d x}\right)_{C_1}-\left(\frac{d y}{d x}\right)_{C_2}}{1+\left(\frac{d y}{d x}\right)_{C_1} \times\left(\frac{d y}{d x}\right)_{C_2}}\right| \Rightarrow \tan \alpha=\mid \frac{\log _e a-1}{\log _e a+1 \mid}$