Given $y=a^x, y=b^x$
$\mathrm{a}^{\mathrm{x}}=\mathrm{b}^{\mathrm{x}} \Rightarrow \mathrm{x}=\mathrm{o}$
$\therefore$ curves $\mathrm{y}=\mathrm{a}^{\mathrm{x}}$ and $\mathrm{y}=\mathrm{b}^{\mathrm{x}}$ intersects at $\mathrm{x}=\mathrm{O}, \mathrm{y}=1$
Let $\theta_1$ be the angle made by the tangent of $\mathrm{y}=\mathrm{a}^{\mathrm{x}}$ at $\mathrm{x}=0$ with $\mathrm{x}$ axis
$\begin{aligned} & \therefore \text { slope }=m_1=\tan \theta_1=\left.\frac{d y}{d x}\right|_{x=0}=\left.a^x \log a\right|_{x=0} \\ & =\log a\end{aligned}$
Let $\theta_2$ be the angle made by the tangent of $y=b^x$ at $x=0$ with $x$ axis
$\begin{aligned} & \therefore \text { slope }=m_2=\tan \theta_2=\left.\frac{d y}{d x}\right|_{x=0}=\left.b^x \log b\right|_{x=0} \\ & =\log b\end{aligned}$
Given angle between curves $2 \alpha$
$\begin{aligned} & \theta_1-\theta_2=\alpha \\ & \therefore \tan \alpha=\frac{\tan \theta_1-\tan \theta_2}{1+\tan \theta_1 \tan \theta_2} \\ & =\frac{m_1-m_2}{1+m_1 m_2} \\ & \therefore \tan a=\frac{\log a-\log b}{1+\log a \log b}\end{aligned}$