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

Now, slope of tangent of first curve

$m_1=\frac{dy}{dx}=a^x\log a$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(0,1)}=m_1=\log a$

Slope of tangent of second curve,

$m_2=\frac{dy}{dx}=b^x\log b$

$\Rightarrow m_2={\left(\frac{dy}{dx}\right)}_{(0,1)}=\log b$

$\therefore \tan \alpha =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\frac{\log a-\log b}{1+\log a\log b}$