If the curves $y=a^x$ and $y=e^x$ intersect at an angle $\alpha$, then $\tan \alpha$ equals 

$\left|\frac{\log _e a-1}{\log _e a+1}\right|$

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 $\left(\frac{d y}{d x}\right)_{C_2}=1$

∴   $\tan \alpha=\left|\frac{\left(\frac{d y}{d x}\right)_{C_1}-\left(\frac{d y}{d x}\right)_{C_2}}{1+\left(\frac{d y}{d x}\right)_{C_1} \times\left(\frac{d y}{d x}\right)_{C_2}}\right| \Rightarrow \tan \alpha=\mid \frac{\log _e a-1}{\log _e a+1 \mid}$