If the angle between the curves $y = 2^x$ and $y=3^x$ is $\alpha,$ then the value of $\tan \alpha$ is equal to :

Given curves are $y = 2^x$ and $y =3^x$

The point of intersection is $3^x = 2^x \Rightarrow x = 0$

On differentiating w.r.t. $x,$ we get

$\dfrac {dy}{dx} = 2^x \log 2 = m_1$

and $\dfrac {dy}{dx} = 3^x \log 3 = m_2$

Therefore, $\tan \alpha = \dfrac {m_2 - m_1}{1 + m_1m_2}$

$= \dfrac {3^x \log 3 - 2^x \log 2 }{1+3^x \times 2^x \log 3 \times \log 2}$

At $x=0$,

$\tan \alpha = \dfrac {3^0 \log 3 - 2^0 \log 2 }{1 + 3^0 \times 2^0 \log 2 \log 3 }$

$= \dfrac { \log \dfrac {3}{2} } { 1 + \log 2 \log 3 }$